From 876a2ab3455cddd29304ab685c5828c31f0ca0cc Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Mon, 5 May 2025 14:22:47 -0500 Subject: [PATCH 01/37] issue-49 commit 1 eigenvalue problem 1 part 1 --- linear-algebra/eigenvalue-problem.md | 8 ++++++++ 1 file changed, 8 insertions(+) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 6ce7c9b14..6109b08ca 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -1 +1,9 @@ # Eigenvalue problem + +For all nxn square matrix \underline{\underline{A}}, there is a scalar \lamda +and vector \underline{x} such that : + + +\begin{align} +\underline{\underline{A}} \underline{x} =\lamda\underline{x} +\end{align} From 2d220ac22728bcc825ab4a095a479d1ddeee19d2 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Mon, 5 May 2025 14:34:53 -0500 Subject: [PATCH 02/37] issue49 commit to fix commit 1 --- linear-algebra/eigenvalue-problem.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 6109b08ca..326a1bc65 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -1,6 +1,6 @@ # Eigenvalue problem -For all nxn square matrix \underline{\underline{A}}, there is a scalar \lamda +For an \(n \times n\) square matrix \underline{\underline{A}} , there is a scalar \lamda and vector \underline{x} such that : From f3ec3640316a51c92ece7e381634e20446a6ef15 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 15:47:10 -0500 Subject: [PATCH 03/37] issue 49 times edit n lambda edit check --- linear-algebra/eigenvalue-problem.md | 10 ++++++++-- 1 file changed, 8 insertions(+), 2 deletions(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 9cb621cdb..a93440b3e 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -1,13 +1,19 @@ # Eigenvalue problem -For an \(n \times n\) square matrix \underline{\underline{A}} , there is a scalar \lamda +For an n\timesn square matrix $\underline{\underline{A}$} , there is a scalar \lambda and vector \underline{x} such that : \begin{align} -\underline{\underline{A}} \underline{x} =\lamda\underline{x} +\underline{\underline{A}} \underline{x} =\lambda\underline{x} \end{align} + + + + + + ## Skill builder problems Find the eigenvalues and eigenvectors From 722cfb1b178f7aebafd8547aad8fe39f646c6b85 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 15:59:24 -0500 Subject: [PATCH 04/37] issue 49 small delete to see if resolve conflict in pull --- linear-algebra/eigenvalue-problem.md | 13 ------------- 1 file changed, 13 deletions(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index a93440b3e..7828db19f 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -1,18 +1,5 @@ # Eigenvalue problem -For an n\timesn square matrix $\underline{\underline{A}$} , there is a scalar \lambda -and vector \underline{x} such that : - - -\begin{align} -\underline{\underline{A}} \underline{x} =\lambda\underline{x} -\end{align} - - - - - - ## Skill builder problems From 755a37a62dc97b1ea22581f5bb072d4620a54823 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 16:21:09 -0500 Subject: [PATCH 05/37] issue 49 commit for 1st part see if will show up --- linear-algebra/eigenvalue-problem.md | 11 +++++++++++ 1 file changed, 11 insertions(+) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 7828db19f..bce6e2e5d 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -1,5 +1,16 @@ # Eigenvalue problem +For an n\timesn square matrix $\underline{\underline{A}$} , there is a scalar \lambda +and vector \underline{x} such that : + + +\begin{align} +\underline{\underline{A}} \underline{x} =\lamda\underline{x} +\underline{\underline{A}} \underline{x} =\lambda\underline{x} +\end{align} + + + ## Skill builder problems From a42d1ee75d943d42baff44eb3d26315752c2eb13 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 16:32:21 -0500 Subject: [PATCH 06/37] issue49 commit for equations part 1 --- linear-algebra/eigenvalue-problem.md | 5 ++++- 1 file changed, 4 insertions(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 2bc739176..288bd3284 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -1,12 +1,15 @@ # Eigenvalue problem +<<<<<<< HEAD +For an $n\times n$ square matrix $\underline{\underline{A}$ , there is a scalar $\lambda$ +and vector $\underline{x}$ such that : For an n\timesn square matrix $\underline{\underline{A}$} , there is a scalar \lambda and vector \underline{x} such that : + \begin{align} -\underline{\underline{A}} \underline{x} =\lamda\underline{x} \underline{\underline{A}} \underline{x} =\lambda\underline{x} \end{align} From da12a136d1e3938e65334ed09665079718d40fef Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 16:37:52 -0500 Subject: [PATCH 07/37] issue 49 trying to fix merge conflict again for >>head --- linear-algebra/eigenvalue-problem.md | 7 ++++--- 1 file changed, 4 insertions(+), 3 deletions(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 288bd3284..8cc7eddc1 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -1,8 +1,5 @@ # Eigenvalue problem -<<<<<<< HEAD -For an $n\times n$ square matrix $\underline{\underline{A}$ , there is a scalar $\lambda$ -and vector $\underline{x}$ such that : For an n\timesn square matrix $\underline{\underline{A}$} , there is a scalar \lambda and vector \underline{x} such that : @@ -16,6 +13,10 @@ and vector \underline{x} such that : + + + + ## Multiple eigenvalues An *n* x *n* matrix has *n* eigenvalues, but they may not be distinct! For From c435beacc90b70a47449ba0c3e62fb076d16d9c2 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 16:43:25 -0500 Subject: [PATCH 08/37] issue49 solving if text is shown correct now --- linear-algebra/eigenvalue-problem.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 8cc7eddc1..772d44779 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -1,8 +1,8 @@ # Eigenvalue problem -For an n\timesn square matrix $\underline{\underline{A}$} , there is a scalar \lambda -and vector \underline{x} such that : +For an *n* x *n* square matrix $ \vv{A} $ , there is a scalar $\lambda$ +and vector $\underline{x}$ such that : From 4169bde1221a9f1546d4e809ceeead01a24ccd8d Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 16:50:36 -0500 Subject: [PATCH 09/37] issue 49 part 2 first draft --- linear-algebra/eigenvalue-problem.md | 8 +++++++- 1 file changed, 7 insertions(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 772d44779..11f194188 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -7,11 +7,17 @@ and vector $\underline{x}$ such that : \begin{align} -\underline{\underline{A}} \underline{x} =\lambda\underline{x} +\vv{A} \underline{x} =\lambda\underline{x} \end{align} +We want \underline{nontrivial} solutions (\underline{x} \ne \underline{0}). To find them: +\begin{align} +\vv{A} \underline{x} - \lambda \underline{x} = \underline{0} + +\vv{A} \underline{x} - \lambda \vv{I} \underline{x} = \underline{0} +(\vv{A} - \lambda \vv{I} )\underline{x} = \underline{0} From 01889d91f43eebfbc8e15d1c9162337c4cb3d31b Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 16:56:16 -0500 Subject: [PATCH 10/37] issue 49 part 2 fix is 0 the problem --- linear-algebra/eigenvalue-problem.md | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 11f194188..76eddb8b7 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -10,14 +10,14 @@ and vector $\underline{x}$ such that : \vv{A} \underline{x} =\lambda\underline{x} \end{align} -We want \underline{nontrivial} solutions (\underline{x} \ne \underline{0}). To find them: +We want $\underline{nontrivial}$ solutions $(\underline{x} \ne \underline{0})$. To find them: \begin{align} -\vv{A} \underline{x} - \lambda \underline{x} = \underline{0} +\vv{A} \underline{x} - \lambda \underline{x} = 0 -\vv{A} \underline{x} - \lambda \vv{I} \underline{x} = \underline{0} +\vv{A} \underline{x} - \lambda \vv{I} \underline{x} = 0 -(\vv{A} - \lambda \vv{I} )\underline{x} = \underline{0} +(\vv{A} - \lambda \vv{I} )\underline{x} = 0 From 0fd1afed14468db386a1f55ac05be0cc018c1c19 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 16:57:57 -0500 Subject: [PATCH 11/37] issue49 missing end align quick fix --- linear-algebra/eigenvalue-problem.md | 3 +-- 1 file changed, 1 insertion(+), 2 deletions(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 76eddb8b7..519f2f4b6 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -18,8 +18,7 @@ We want $\underline{nontrivial}$ solutions $(\underline{x} \ne \underline{0})$. \vv{A} \underline{x} - \lambda \vv{I} \underline{x} = 0 (\vv{A} - \lambda \vv{I} )\underline{x} = 0 - - +\end{align} From f5fd368c0d120aa0e70aef12753e9e965d3a7a8a Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 17:03:42 -0500 Subject: [PATCH 12/37] issue 49 part 3 draft --- linear-algebra/eigenvalue-problem.md | 11 ++++++++--- 1 file changed, 8 insertions(+), 3 deletions(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 519f2f4b6..09f9c34ba 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -13,13 +13,18 @@ and vector $\underline{x}$ such that : We want $\underline{nontrivial}$ solutions $(\underline{x} \ne \underline{0})$. To find them: \begin{align} -\vv{A} \underline{x} - \lambda \underline{x} = 0 +\vv{A} \underline{x} - \lambda \underline{x} = \underline{0} \\ -\vv{A} \underline{x} - \lambda \vv{I} \underline{x} = 0 +\vv{A} \underline{x} - \lambda \vv{I} \underline{x} = \underline{0} \\ -(\vv{A} - \lambda \vv{I} )\underline{x} = 0 +(\vv{A} - \lambda \vv{I} )\underline{x} = \underline{0} \end{align} +If $\vv{A} - \lambda \vv{I}$ is invertible, we have only $\lambda = \underline{0}$ as a solution. Hence, we want this to be singular! This occurs when the determinant is zero: + +\begin{align} +$|\vv{A}- \lambda \vv{I}= 0$ +\end{align} ## Multiple eigenvalues From 0ca40827a630bc916b5bfab8ba8c4522cb1ce54f Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 17:13:52 -0500 Subject: [PATCH 13/37] issue 49 part 4 draft end O page 1 --- linear-algebra/eigenvalue-problem.md | 12 +++++++++++- 1 file changed, 11 insertions(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 09f9c34ba..c82df121e 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -23,9 +23,19 @@ We want $\underline{nontrivial}$ solutions $(\underline{x} \ne \underline{0})$. If $\vv{A} - \lambda \vv{I}$ is invertible, we have only $\lambda = \underline{0}$ as a solution. Hence, we want this to be singular! This occurs when the determinant is zero: \begin{align} -$|\vv{A}- \lambda \vv{I}= 0$ +|\vv{A}- \lambda \vv{I}= 0 \end{align} +This determinant creates a $\underline{ characteristic polynomial}$ for $\lambda$ that can be solved. Then, the $\underline{x}$ that corresponds to a given $\lambda$ can be determined. + +Ex: +\begin{align} +\vv{A} = \begin{bmatrix} -5 & 2\\ 2 & 2 \end{bmatrix} + +|\vv{A} -\lambda \vv{I} = \begin{bmatrix} -5-\lambda & 2 \\ 2 & -2-\lambda \end{bmatrix} = (\lambda + 5)(\lambda +2) -4 + +=\lambda^2 +7\lambda + 6 = (\lambda +1)(\lambda +6)=0 \to \lambda = -1, -6 +\end{align} ## Multiple eigenvalues From 9ef69f5b274c7b0a7231173efe41d09182ee5480 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 17:17:22 -0500 Subject: [PATCH 14/37] issue 49 part 4 quick fix --- linear-algebra/eigenvalue-problem.md | 7 ++++--- 1 file changed, 4 insertions(+), 3 deletions(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index c82df121e..addde8f78 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -26,17 +26,18 @@ If $\vv{A} - \lambda \vv{I}$ is invertible, we have only $\lambda = \underline{0 |\vv{A}- \lambda \vv{I}= 0 \end{align} -This determinant creates a $\underline{ characteristic polynomial}$ for $\lambda$ that can be solved. Then, the $\underline{x}$ that corresponds to a given $\lambda$ can be determined. +This determinant creates a $\underline{ characteristic polynomial}$ for $\lambda$ that can be solved. Then, the $\underline{x}$ that corresponds to a given $\lambda$ can be determined. Ex: \begin{align} \vv{A} = \begin{bmatrix} -5 & 2\\ 2 & 2 \end{bmatrix} - +\\ |\vv{A} -\lambda \vv{I} = \begin{bmatrix} -5-\lambda & 2 \\ 2 & -2-\lambda \end{bmatrix} = (\lambda + 5)(\lambda +2) -4 - +\\ =\lambda^2 +7\lambda + 6 = (\lambda +1)(\lambda +6)=0 \to \lambda = -1, -6 \end{align} + ## Multiple eigenvalues An *n* x *n* matrix has *n* eigenvalues, but they may not be distinct! For From a6a7fc9568da47429857210d5ef8a6aa37724e1d Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 17:22:29 -0500 Subject: [PATCH 15/37] issue 49 part 4 fix 2 --- linear-algebra/eigenvalue-problem.md | 7 +++---- 1 file changed, 3 insertions(+), 4 deletions(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index addde8f78..ca4ac18ba 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -26,13 +26,12 @@ If $\vv{A} - \lambda \vv{I}$ is invertible, we have only $\lambda = \underline{0 |\vv{A}- \lambda \vv{I}= 0 \end{align} -This determinant creates a $\underline{ characteristic polynomial}$ for $\lambda$ that can be solved. Then, the $\underline{x}$ that corresponds to a given $\lambda$ can be determined. +This determinant creates a $\underline{ characteristic polynomial}$ for $\lambda$ that can be solved. Then, the $\underline{x}$ that corresponds to a given $\lambda$ can be determined. -Ex: -\begin{align} +Ex:\begin{align} \vv{A} = \begin{bmatrix} -5 & 2\\ 2 & 2 \end{bmatrix} \\ -|\vv{A} -\lambda \vv{I} = \begin{bmatrix} -5-\lambda & 2 \\ 2 & -2-\lambda \end{bmatrix} = (\lambda + 5)(\lambda +2) -4 +|\vv{A} -\lambda \vv{I} = \begin{bmatrix} -5-\lambda & 2 \\ 2 & -2-\lambda \end{bmatrix} = (-5 -\lambda)(-2-\lambda)-2\cdot2 =(\lambda + 5)(\lambda +2) -4 \\ =\lambda^2 +7\lambda + 6 = (\lambda +1)(\lambda +6)=0 \to \lambda = -1, -6 \end{align} From 2ca4384066cf53d3f4ee08926071a9039a43ddb7 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 17:27:35 -0500 Subject: [PATCH 16/37] issue 49 part 4 fix 3 --- linear-algebra/eigenvalue-problem.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index ca4ac18ba..40b26cbee 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -28,7 +28,8 @@ If $\vv{A} - \lambda \vv{I}$ is invertible, we have only $\lambda = \underline{0 This determinant creates a $\underline{ characteristic polynomial}$ for $\lambda$ that can be solved. Then, the $\underline{x}$ that corresponds to a given $\lambda$ can be determined. -Ex:\begin{align} +Ex: +\begin{align} \vv{A} = \begin{bmatrix} -5 & 2\\ 2 & 2 \end{bmatrix} \\ |\vv{A} -\lambda \vv{I} = \begin{bmatrix} -5-\lambda & 2 \\ 2 & -2-\lambda \end{bmatrix} = (-5 -\lambda)(-2-\lambda)-2\cdot2 =(\lambda + 5)(\lambda +2) -4 From 0ebe968958ca52e335fece0d05994a8bcefb0e18 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 17:36:54 -0500 Subject: [PATCH 17/37] issue 49 part 5 first draft --- linear-algebra/eigenvalue-problem.md | 9 ++++++++- 1 file changed, 8 insertions(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 40b26cbee..2d81c3fb6 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -34,9 +34,16 @@ Ex: \\ |\vv{A} -\lambda \vv{I} = \begin{bmatrix} -5-\lambda & 2 \\ 2 & -2-\lambda \end{bmatrix} = (-5 -\lambda)(-2-\lambda)-2\cdot2 =(\lambda + 5)(\lambda +2) -4 \\ -=\lambda^2 +7\lambda + 6 = (\lambda +1)(\lambda +6)=0 \to \lambda = -1, -6 +=\lambda^2 +7\lambda + 6 = (\lambda +1)(\lambda +6)=0 \to \underline{\lambda = -1, -6} \end{align} +$\lambda=-1$: +\begin{align} +\begin{bmatrix} -5 -(-1) & 2 \\ 2 & -2-(-1) \end{bmatrix} \underline{x}= \begin{bmatrix} -4 & 2 \\ 2 & 1 \end{bmatrix} \underline{x} =0 +\\ +\begin{bmatrix} -4 & 2 & 0 \\ 2 & 1 & 0 \end{bmatrix} \to \begin{bmatrix} -4 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to \begin{bmatrix} 1 & -1/2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to x_1 -x_2 /2 =0 +x_2 free +\end{align} ## Multiple eigenvalues From efee3b1d063a39a85e7ef6b1b3b2677bf38ea6d4 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 17:40:43 -0500 Subject: [PATCH 18/37] issue 49 part 5 fix 1 see if it aligns better --- linear-algebra/eigenvalue-problem.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 2d81c3fb6..2aeccc537 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -41,7 +41,7 @@ $\lambda=-1$: \begin{align} \begin{bmatrix} -5 -(-1) & 2 \\ 2 & -2-(-1) \end{bmatrix} \underline{x}= \begin{bmatrix} -4 & 2 \\ 2 & 1 \end{bmatrix} \underline{x} =0 \\ -\begin{bmatrix} -4 & 2 & 0 \\ 2 & 1 & 0 \end{bmatrix} \to \begin{bmatrix} -4 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to \begin{bmatrix} 1 & -1/2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to x_1 -x_2 /2 =0 +\begin{bmatrix} -4 & 2 & 0 \\ 2 & 1 & 0 \end{bmatrix} \to \begin{bmatrix} -4 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to \begin{bmatrix} 1 & -1/2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to x_1 -x_2 /2 =0 \\ x_2 free \end{align} From c79ced090be0dd543036d3ec2fbc4d144cef725b Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 17:45:35 -0500 Subject: [PATCH 19/37] issue 49 part 5 draft --- linear-algebra/eigenvalue-problem.md | 9 +++++++++ 1 file changed, 9 insertions(+) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 2aeccc537..d53045c62 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -45,6 +45,15 @@ $\lambda=-1$: x_2 free \end{align} +This solution is not unique, which makes sense from the equation. We may then choose x_1 or x_2 so that the eigenvector has a nice norm or values. e.g., +\begin{align} +x_1 =1, x_2 = 2 \to \underline{x}= \begin{bmatrix} 1 \\ 2 +\end{align} + +Repeat for $\lambda = -6$: +\begin{align} +\begin{bmatrix} -5 -(-6) & 2 \\ 2 & -2-(-6) \end{bmatrix} \underline{x} = \begin{bmatrix} 1 & 2 \\ 2 & 4 \underline{x} =0 + ## Multiple eigenvalues An *n* x *n* matrix has *n* eigenvalues, but they may not be distinct! For From 73fa920c1a64330b5d707f45cedb3f90bad2d2a2 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 17:47:09 -0500 Subject: [PATCH 20/37] issue 49 part5 fix 1 --- linear-algebra/eigenvalue-problem.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index d53045c62..16e47e4b4 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -53,6 +53,8 @@ x_1 =1, x_2 = 2 \to \underline{x}= \begin{bmatrix} 1 \\ 2 Repeat for $\lambda = -6$: \begin{align} \begin{bmatrix} -5 -(-6) & 2 \\ 2 & -2-(-6) \end{bmatrix} \underline{x} = \begin{bmatrix} 1 & 2 \\ 2 & 4 \underline{x} =0 +\end{align} + ## Multiple eigenvalues From b75d514e7330878396f101769f75f7ec7cc42a57 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 17:47:51 -0500 Subject: [PATCH 21/37] issue 49 part 5 fix 3 --- linear-algebra/eigenvalue-problem.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 16e47e4b4..5ca70fd31 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -47,7 +47,7 @@ x_2 free This solution is not unique, which makes sense from the equation. We may then choose x_1 or x_2 so that the eigenvector has a nice norm or values. e.g., \begin{align} -x_1 =1, x_2 = 2 \to \underline{x}= \begin{bmatrix} 1 \\ 2 +x_1 =1, x_2 = 2 \to \underline{x}= \begin{bmatrix} 1 \\ 2 end{bmatrix} \end{align} Repeat for $\lambda = -6$: From 07f817246f1cd02da695ee369f540d705919d764 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 17:48:38 -0500 Subject: [PATCH 22/37] issue 49 matrix fix part 5 --- linear-algebra/eigenvalue-problem.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 5ca70fd31..dea5501db 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -52,7 +52,7 @@ x_1 =1, x_2 = 2 \to \underline{x}= \begin{bmatrix} 1 \\ 2 end{bmatrix} Repeat for $\lambda = -6$: \begin{align} -\begin{bmatrix} -5 -(-6) & 2 \\ 2 & -2-(-6) \end{bmatrix} \underline{x} = \begin{bmatrix} 1 & 2 \\ 2 & 4 \underline{x} =0 +\begin{bmatrix} -5 -(-6) & 2 \\ 2 & -2-(-6) \end{bmatrix} \underline{x} = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}\underline{x} =0 \end{align} From 45b9e7e65d5d9355e00e2fca9a7916dcd765f258 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 17:52:41 -0500 Subject: [PATCH 23/37] issue 49 part 5 matrix fix 2 --- linear-algebra/eigenvalue-problem.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index dea5501db..ba7d3f628 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -47,7 +47,7 @@ x_2 free This solution is not unique, which makes sense from the equation. We may then choose x_1 or x_2 so that the eigenvector has a nice norm or values. e.g., \begin{align} -x_1 =1, x_2 = 2 \to \underline{x}= \begin{bmatrix} 1 \\ 2 end{bmatrix} +x_1 =1, x_2 = 2 \to \underline{x}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} \end{align} Repeat for $\lambda = -6$: From 538e2abc31f00722912820ea1d2419049c235835 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 17:59:46 -0500 Subject: [PATCH 24/37] issue 49 last part draft --- linear-algebra/eigenvalue-problem.md | 11 +++++++++++ 1 file changed, 11 insertions(+) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index ba7d3f628..0e7950f81 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -55,6 +55,17 @@ Repeat for $\lambda = -6$: \begin{bmatrix} -5 -(-6) & 2 \\ 2 & -2-(-6) \end{bmatrix} \underline{x} = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}\underline{x} =0 \end{align} +Row 2 is 2x Row 1, so we can just use Row 1!(This will be a pattern for 2x2 matricies). + +\begin{align} +x_1 +2x_2 =0 \to \underline{x}=\begin{bmatrix} 2 \\ -1 \end{bmatrix} +\end{align} + +Hence, the eigenvalues and eigenvectors are: +\begin{align} +\lambda_1 , \underline{x_1}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} and \lambda_2 = -6, \underline{x_2}= \begin{bmatrix} 2 \\ -1 \end{bmatrix} +\end{align} + ## Multiple eigenvalues From b0d6d2eb373693842ae21450267e2bfb6b0fed4e Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 18:05:50 -0500 Subject: [PATCH 25/37] issue 49 last part fix 1 --- linear-algebra/eigenvalue-problem.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 0e7950f81..34bc86797 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -63,7 +63,7 @@ x_1 +2x_2 =0 \to \underline{x}=\begin{bmatrix} 2 \\ -1 \end{bmatrix} Hence, the eigenvalues and eigenvectors are: \begin{align} -\lambda_1 , \underline{x_1}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} and \lambda_2 = -6, \underline{x_2}= \begin{bmatrix} 2 \\ -1 \end{bmatrix} +\lambda_1 = -1 , \underline{x_1}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} \, \text{and} \, \lambda_2 = -6, \underline{x_2}= \begin{bmatrix} 2 \\ -1 \end{bmatrix} \end{align} From 28422d2336639db6350f62d51fbd53428f109eff Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 18:07:50 -0500 Subject: [PATCH 26/37] issue 49 last part spacing fix --- linear-algebra/eigenvalue-problem.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 34bc86797..85a0ed653 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -63,7 +63,7 @@ x_1 +2x_2 =0 \to \underline{x}=\begin{bmatrix} 2 \\ -1 \end{bmatrix} Hence, the eigenvalues and eigenvectors are: \begin{align} -\lambda_1 = -1 , \underline{x_1}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} \, \text{and} \, \lambda_2 = -6, \underline{x_2}= \begin{bmatrix} 2 \\ -1 \end{bmatrix} +\lambda_1 = -1 , \underline{x_1}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} \quad \text{and} \quad \lambda_2 = -6, \underline{x_2}= \begin{bmatrix} 2 \\ -1 \end{bmatrix} \end{align} From 23fdbe75d6a7c0c159a7faeea9c8b5cfeca7e739 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 18:14:06 -0500 Subject: [PATCH 27/37] issue 49 last quickfix 1 --- linear-algebra/eigenvalue-problem.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 85a0ed653..9ba1e5a87 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -23,7 +23,7 @@ We want $\underline{nontrivial}$ solutions $(\underline{x} \ne \underline{0})$. If $\vv{A} - \lambda \vv{I}$ is invertible, we have only $\lambda = \underline{0}$ as a solution. Hence, we want this to be singular! This occurs when the determinant is zero: \begin{align} -|\vv{A}- \lambda \vv{I}= 0 +|\vv{A}- \lambda \vv{I}|= 0 \end{align} This determinant creates a $\underline{ characteristic polynomial}$ for $\lambda$ that can be solved. Then, the $\underline{x}$ that corresponds to a given $\lambda$ can be determined. From 28f28f79cf98ac36aa53c47fd39f1d1c727c5dc0 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 18:15:07 -0500 Subject: [PATCH 28/37] issue49 quickfix 2 spacing experiment --- linear-algebra/eigenvalue-problem.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 9ba1e5a87..c5758de37 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -26,7 +26,7 @@ If $\vv{A} - \lambda \vv{I}$ is invertible, we have only $\lambda = \underline{0 |\vv{A}- \lambda \vv{I}|= 0 \end{align} -This determinant creates a $\underline{ characteristic polynomial}$ for $\lambda$ that can be solved. Then, the $\underline{x}$ that corresponds to a given $\lambda$ can be determined. +This determinant creates a $\underline{ characteristic \, polynomial}$ for $\lambda$ that can be solved. Then, the $\underline{x}$ that corresponds to a given $\lambda$ can be determined. Ex: \begin{align} From 04b6ef4c7e54730b7f4e3c28b7d15835c191e227 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 18:21:38 -0500 Subject: [PATCH 29/37] issue 49 quick fix 3 --- linear-algebra/eigenvalue-problem.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index c5758de37..6e0ff317c 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -32,7 +32,7 @@ Ex: \begin{align} \vv{A} = \begin{bmatrix} -5 & 2\\ 2 & 2 \end{bmatrix} \\ -|\vv{A} -\lambda \vv{I} = \begin{bmatrix} -5-\lambda & 2 \\ 2 & -2-\lambda \end{bmatrix} = (-5 -\lambda)(-2-\lambda)-2\cdot2 =(\lambda + 5)(\lambda +2) -4 +|\vv{A} -\lambda \vv{I}| = \begin{bmatrix} -5-\lambda & 2 \\ 2 & -2-\lambda \end{bmatrix} = (-5 -\lambda)(-2-\lambda)-2\cdot2 =(\lambda + 5)(\lambda +2) -4 \\ =\lambda^2 +7\lambda + 6 = (\lambda +1)(\lambda +6)=0 \to \underline{\lambda = -1, -6} \end{align} @@ -41,7 +41,7 @@ $\lambda=-1$: \begin{align} \begin{bmatrix} -5 -(-1) & 2 \\ 2 & -2-(-1) \end{bmatrix} \underline{x}= \begin{bmatrix} -4 & 2 \\ 2 & 1 \end{bmatrix} \underline{x} =0 \\ -\begin{bmatrix} -4 & 2 & 0 \\ 2 & 1 & 0 \end{bmatrix} \to \begin{bmatrix} -4 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to \begin{bmatrix} 1 & -1/2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to x_1 -x_2 /2 =0 \\ +\begin{bmatrix} -4 & 2 & 0 \\ 2 & -1 & 0 \end{bmatrix} \to \begin{bmatrix} -4 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to \begin{bmatrix} 1 & -1/2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to x_1 -\frac{x_2}{2} =0 \\ x_2 free \end{align} From dc0db952f15f4fd5c4d3542bcbb9d521b63acf41 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 18:24:20 -0500 Subject: [PATCH 30/37] issue49 quickfix 5 --- linear-algebra/eigenvalue-problem.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 6e0ff317c..708280779 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -30,9 +30,9 @@ This determinant creates a $\underline{ characteristic \, polynomial}$ for $\lam Ex: \begin{align} -\vv{A} = \begin{bmatrix} -5 & 2\\ 2 & 2 \end{bmatrix} +\vv{A} = \begin{bmatrix} -5 & 2\\ 2 & -2 \end{bmatrix} \\ -|\vv{A} -\lambda \vv{I}| = \begin{bmatrix} -5-\lambda & 2 \\ 2 & -2-\lambda \end{bmatrix} = (-5 -\lambda)(-2-\lambda)-2\cdot2 =(\lambda + 5)(\lambda +2) -4 +| \vv{A} -\lambda \vv{I} | = \begin{bmatrix} -5-\lambda & 2 \\ 2 & -2-\lambda \end{bmatrix} = (-5 -\lambda)(-2-\lambda)-2\cdot2 =(\lambda + 5)(\lambda +2) -4 \\ =\lambda^2 +7\lambda + 6 = (\lambda +1)(\lambda +6)=0 \to \underline{\lambda = -1, -6} \end{align} From c23074d7a26ca9b7b66c36c2db1334471ca92ae7 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 18:30:06 -0500 Subject: [PATCH 31/37] issue49 qickfix 6 x subscript --- linear-algebra/eigenvalue-problem.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 708280779..77a9e8616 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -45,7 +45,7 @@ $\lambda=-1$: x_2 free \end{align} -This solution is not unique, which makes sense from the equation. We may then choose x_1 or x_2 so that the eigenvector has a nice norm or values. e.g., +This solution is not unique, which makes sense from the equation. We may then choose $x_1$ or $x_2$ so that the eigenvector has a nice norm or values. e.g., \begin{align} x_1 =1, x_2 = 2 \to \underline{x}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} \end{align} From 55bc2e133ef1a63e260ec74b56d05fbaea345bb9 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Thu, 8 May 2025 18:34:34 -0500 Subject: [PATCH 32/37] issue49 quickfix 7 small typo --- linear-algebra/eigenvalue-problem.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 77a9e8616..835f6f3fe 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -39,7 +39,7 @@ Ex: $\lambda=-1$: \begin{align} -\begin{bmatrix} -5 -(-1) & 2 \\ 2 & -2-(-1) \end{bmatrix} \underline{x}= \begin{bmatrix} -4 & 2 \\ 2 & 1 \end{bmatrix} \underline{x} =0 +\begin{bmatrix} -5 -(-1) & 2 \\ 2 & -2-(-1) \end{bmatrix} \underline{x}= \begin{bmatrix} -4 & 2 \\ 2 & -1 \end{bmatrix} \underline{x} =0 \\ \begin{bmatrix} -4 & 2 & 0 \\ 2 & -1 & 0 \end{bmatrix} \to \begin{bmatrix} -4 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to \begin{bmatrix} 1 & -1/2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to x_1 -\frac{x_2}{2} =0 \\ x_2 free From fd69abf8652da28d41355733b91a451646f4fa0d Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Fri, 9 May 2025 11:17:43 -0500 Subject: [PATCH 33/37] issue49 first post review fix test --- linear-algebra/eigenvalue-problem.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 835f6f3fe..1010306c0 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -2,7 +2,7 @@ For an *n* x *n* square matrix $ \vv{A} $ , there is a scalar $\lambda$ -and vector $\underline{x}$ such that : +and vector $\vv{x}$ such that : From 4a91e44f55151b177644b1476482fc7136c4fb48 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Fri, 9 May 2025 11:23:21 -0500 Subject: [PATCH 34/37] issue49 replace all underlined x with \vv for bold instead --- linear-algebra/eigenvalue-problem.md | 20 ++++++++++---------- 1 file changed, 10 insertions(+), 10 deletions(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 1010306c0..5ed2f1c01 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -7,17 +7,17 @@ and vector $\vv{x}$ such that : \begin{align} -\vv{A} \underline{x} =\lambda\underline{x} +\vv{A} \vv{x} =\lambda\vv{x} \end{align} -We want $\underline{nontrivial}$ solutions $(\underline{x} \ne \underline{0})$. To find them: +We want $\underline{nontrivial}$ solutions $(\vv{x} \ne \underline{0})$. To find them: \begin{align} -\vv{A} \underline{x} - \lambda \underline{x} = \underline{0} \\ +\vv{A} \vv{x} - \lambda \vv{x} = \underline{0} \\ -\vv{A} \underline{x} - \lambda \vv{I} \underline{x} = \underline{0} \\ +\vv{A} \vv{x} - \lambda \vv{I} \vv{x} = \underline{0} \\ -(\vv{A} - \lambda \vv{I} )\underline{x} = \underline{0} +(\vv{A} - \lambda \vv{I} )\vv{x} = \underline{0} \end{align} If $\vv{A} - \lambda \vv{I}$ is invertible, we have only $\lambda = \underline{0}$ as a solution. Hence, we want this to be singular! This occurs when the determinant is zero: @@ -26,7 +26,7 @@ If $\vv{A} - \lambda \vv{I}$ is invertible, we have only $\lambda = \underline{0 |\vv{A}- \lambda \vv{I}|= 0 \end{align} -This determinant creates a $\underline{ characteristic \, polynomial}$ for $\lambda$ that can be solved. Then, the $\underline{x}$ that corresponds to a given $\lambda$ can be determined. +This determinant creates a $\underline{ characteristic \, polynomial}$ for $\lambda$ that can be solved. Then, the $\vv{x}$ that corresponds to a given $\lambda$ can be determined. Ex: \begin{align} @@ -39,7 +39,7 @@ Ex: $\lambda=-1$: \begin{align} -\begin{bmatrix} -5 -(-1) & 2 \\ 2 & -2-(-1) \end{bmatrix} \underline{x}= \begin{bmatrix} -4 & 2 \\ 2 & -1 \end{bmatrix} \underline{x} =0 +\begin{bmatrix} -5 -(-1) & 2 \\ 2 & -2-(-1) \end{bmatrix} \vv{x}= \begin{bmatrix} -4 & 2 \\ 2 & -1 \end{bmatrix} \vv{x} =0 \\ \begin{bmatrix} -4 & 2 & 0 \\ 2 & -1 & 0 \end{bmatrix} \to \begin{bmatrix} -4 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to \begin{bmatrix} 1 & -1/2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to x_1 -\frac{x_2}{2} =0 \\ x_2 free @@ -47,18 +47,18 @@ x_2 free This solution is not unique, which makes sense from the equation. We may then choose $x_1$ or $x_2$ so that the eigenvector has a nice norm or values. e.g., \begin{align} -x_1 =1, x_2 = 2 \to \underline{x}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} +x_1 =1, x_2 = 2 \to \vv{x}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} \end{align} Repeat for $\lambda = -6$: \begin{align} -\begin{bmatrix} -5 -(-6) & 2 \\ 2 & -2-(-6) \end{bmatrix} \underline{x} = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}\underline{x} =0 +\begin{bmatrix} -5 -(-6) & 2 \\ 2 & -2-(-6) \end{bmatrix} \vv{x} = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}\vv{x} =0 \end{align} Row 2 is 2x Row 1, so we can just use Row 1!(This will be a pattern for 2x2 matricies). \begin{align} -x_1 +2x_2 =0 \to \underline{x}=\begin{bmatrix} 2 \\ -1 \end{bmatrix} +x_1 +2x_2 =0 \to \vv{x}=\begin{bmatrix} 2 \\ -1 \end{bmatrix} \end{align} Hence, the eigenvalues and eigenvectors are: From cffdd05aa1379e6f36592e5814a6532ebf528c19 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Fri, 9 May 2025 11:36:42 -0500 Subject: [PATCH 35/37] issue49 final draft all underlined vectors replaces with \vv and all matrix quantities with \vv should be good now --- linear-algebra/eigenvalue-problem.md | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 5ed2f1c01..51f067b82 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -41,13 +41,13 @@ $\lambda=-1$: \begin{align} \begin{bmatrix} -5 -(-1) & 2 \\ 2 & -2-(-1) \end{bmatrix} \vv{x}= \begin{bmatrix} -4 & 2 \\ 2 & -1 \end{bmatrix} \vv{x} =0 \\ -\begin{bmatrix} -4 & 2 & 0 \\ 2 & -1 & 0 \end{bmatrix} \to \begin{bmatrix} -4 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to \begin{bmatrix} 1 & -1/2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to x_1 -\frac{x_2}{2} =0 \\ -x_2 free +\begin{bmatrix} -4 & 2 & 0 \\ 2 & -1 & 0 \end{bmatrix} \to \begin{bmatrix} -4 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to \begin{bmatrix} 1 & -1/2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to \vv{x_1} -\frac{\vv{x_2}}{2} =0 \\ +\vv{x_2} free \end{align} This solution is not unique, which makes sense from the equation. We may then choose $x_1$ or $x_2$ so that the eigenvector has a nice norm or values. e.g., \begin{align} -x_1 =1, x_2 = 2 \to \vv{x}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} +\vv{x_1} =1, \vv{x_2} = 2 \to \vv{x}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} \end{align} Repeat for $\lambda = -6$: @@ -58,12 +58,12 @@ Repeat for $\lambda = -6$: Row 2 is 2x Row 1, so we can just use Row 1!(This will be a pattern for 2x2 matricies). \begin{align} -x_1 +2x_2 =0 \to \vv{x}=\begin{bmatrix} 2 \\ -1 \end{bmatrix} +\vv{x_1} +2\vv{x_2} =0 \to \vv{x}=\begin{bmatrix} 2 \\ -1 \end{bmatrix} \end{align} Hence, the eigenvalues and eigenvectors are: \begin{align} -\lambda_1 = -1 , \underline{x_1}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} \quad \text{and} \quad \lambda_2 = -6, \underline{x_2}= \begin{bmatrix} 2 \\ -1 \end{bmatrix} +\lambda_1 = -1 , \vv{x_1}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} \quad \text{and} \quad \lambda_2 = -6, \vv{x_2}= \begin{bmatrix} 2 \\ -1 \end{bmatrix} \end{align} From 35c040589eeec8113506277eb718806ce60dfbe5 Mon Sep 17 00:00:00 2001 From: oscarearhart2005 Date: Fri, 9 May 2025 11:51:45 -0500 Subject: [PATCH 36/37] issue49 final fix on missed vector in text --- linear-algebra/eigenvalue-problem.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 51f067b82..52432a9fa 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -45,7 +45,7 @@ $\lambda=-1$: \vv{x_2} free \end{align} -This solution is not unique, which makes sense from the equation. We may then choose $x_1$ or $x_2$ so that the eigenvector has a nice norm or values. e.g., +This solution is not unique, which makes sense from the equation. We may then choose $\vv{x_1}$ or $\vv{x_2}$ so that the eigenvector has a nice norm or values. e.g., \begin{align} \vv{x_1} =1, \vv{x_2} = 2 \to \vv{x}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} \end{align} From 3fd85f09ba35336782fcd445596d8b6718801319 Mon Sep 17 00:00:00 2001 From: Michael Howard Date: Fri, 9 May 2025 13:29:44 -0500 Subject: [PATCH 37/37] Revise and reformat --- linear-algebra/eigenvalue-problem.md | 150 ++++++++++++++++++++------- 1 file changed, 111 insertions(+), 39 deletions(-) diff --git a/linear-algebra/eigenvalue-problem.md b/linear-algebra/eigenvalue-problem.md index 52432a9fa..a891cce80 100644 --- a/linear-algebra/eigenvalue-problem.md +++ b/linear-algebra/eigenvalue-problem.md @@ -1,71 +1,144 @@ # Eigenvalue problem +For an *n* x *n* square matrix **A**, we seek a scalar $\lambda$ and vector +**x** such that: -For an *n* x *n* square matrix $ \vv{A} $ , there is a scalar $\lambda$ -and vector $\vv{x}$ such that : +\begin{align} +\vv{A} \vv{x} = \lambda \vv{x} +\end{align} +We want a *nontrivial* solution ($\vv{x} \ne \vv{0}$). This pair represents a +vector **x** that, when multiplied into **A**, does not change its direction. +However, it may adopt a new magnitude $\lambda$. Borrowing from a German word +for "own", we call **x** an *eigenvector* and $\lambda$ an *eigenvalue* of +**A**. +To find the eigenvalues of **A**, rearrange: \begin{align} -\vv{A} \vv{x} =\lambda\vv{x} +\vv{A} \vv{x} - \lambda \vv{x} &= \vv{0} \\ +(\vv{A} - \lambda \vv{I}) \vv{x} &= \vv{0} \end{align} -We want $\underline{nontrivial}$ solutions $(\vv{x} \ne \underline{0})$. To find them: +If $\vv{A} - \lambda \vv{I}$ is invertible, we have only $\vv{x} = \vv{0}$ as a +solution. Hence, we require this matrix to be singular! This occurs when its +[determinant](./determinants.md) is zero: \begin{align} -\vv{A} \vv{x} - \lambda \vv{x} = \underline{0} \\ +|\vv{A}- \lambda \vv{I}| = 0 +\end{align} -\vv{A} \vv{x} - \lambda \vv{I} \vv{x} = \underline{0} \\ +This equation creates a characeristic polynomial of degree *n* for $\lambda$ +that can be solved. Then, the eigenvector **x** that corresponds to each root +$\lambda$ can be determined. For example, to find the eigenvalues of: -(\vv{A} - \lambda \vv{I} )\vv{x} = \underline{0} -\end{align} +\begin{equation} +\vv{A} = \begin{bmatrix} -5 & 2\\ 2 & -2 \end{bmatrix} +\end{equation} -If $\vv{A} - \lambda \vv{I}$ is invertible, we have only $\lambda = \underline{0}$ as a solution. Hence, we want this to be singular! This occurs when the determinant is zero: +First compute the determinant: \begin{align} -|\vv{A}- \lambda \vv{I}|= 0 +|\vv{A} -\lambda \vv{I}| +&= \begin{vmatrix} -5-\lambda & 2 \\ 2 & -2-\lambda \end{vmatrix} \\ +&= (-5 -\lambda)(-2-\lambda) - 2 \cdot 2 \\ +&= \lambda^2 +7\lambda + 6 \\ +&= (\lambda +1)(\lambda +6) = 0 \end{align} -This determinant creates a $\underline{ characteristic \, polynomial}$ for $\lambda$ that can be solved. Then, the $\vv{x}$ that corresponds to a given $\lambda$ can be determined. +Hence, the eigenvalues of **A** are $\lambda_1 = -1$ and $\lambda_2 = -6$. Note +that the particular ordering of the eigenvalues is not important, and we are +only labeling them to make it convenient to refer to a particular eigenvalue +later. -Ex: -\begin{align} -\vv{A} = \begin{bmatrix} -5 & 2\\ 2 & -2 \end{bmatrix} -\\ -| \vv{A} -\lambda \vv{I} | = \begin{bmatrix} -5-\lambda & 2 \\ 2 & -2-\lambda \end{bmatrix} = (-5 -\lambda)(-2-\lambda)-2\cdot2 =(\lambda + 5)(\lambda +2) -4 -\\ -=\lambda^2 +7\lambda + 6 = (\lambda +1)(\lambda +6)=0 \to \underline{\lambda = -1, -6} -\end{align} +Next, we seek the eigenvector $\vv{x}_1$ that corresponds to $\lambda_1$. This +vector must solve the system: -$\lambda=-1$: \begin{align} -\begin{bmatrix} -5 -(-1) & 2 \\ 2 & -2-(-1) \end{bmatrix} \vv{x}= \begin{bmatrix} -4 & 2 \\ 2 & -1 \end{bmatrix} \vv{x} =0 -\\ -\begin{bmatrix} -4 & 2 & 0 \\ 2 & -1 & 0 \end{bmatrix} \to \begin{bmatrix} -4 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to \begin{bmatrix} 1 & -1/2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \to \vv{x_1} -\frac{\vv{x_2}}{2} =0 \\ -\vv{x_2} free +(\vv{A} - \lambda_1 \vv{I}) \vv{x}_1 &= \vv{0} \\ +\begin{bmatrix} -4 & 2 \\ 2 & -1 \end{bmatrix} \vv{x}_1 &= \vv{0} \end{align} -This solution is not unique, which makes sense from the equation. We may then choose $\vv{x_1}$ or $\vv{x_2}$ so that the eigenvector has a nice norm or values. e.g., -\begin{align} -\vv{x_1} =1, \vv{x_2} = 2 \to \vv{x}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} -\end{align} +We can solve for $\vv{x}_1$ using row reduction. Since the last column of the +augmented matrix would be only zeros, it is not necessary to include it, and we +row reduce only the matrix itself: -Repeat for $\lambda = -6$: -\begin{align} -\begin{bmatrix} -5 -(-6) & 2 \\ 2 & -2-(-6) \end{bmatrix} \vv{x} = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}\vv{x} =0 -\end{align} +\begin{equation} +\begin{bmatrix} +-4 & 2 \\ +2 & -1 +\end{bmatrix} +\begin{matrix}\vphantom{R_1} \\ +R_2/2 \end{matrix} +\to \begin{bmatrix} +-4 & 2 \\ +0 & 0 +\end{bmatrix} +\begin{matrix} \div -2 \\ \vphantom{R_2} \end{matrix} +\to +\begin{bmatrix} +1 & -1/2 \\ +0 & 0 +\end{bmatrix} +\end{equation} -Row 2 is 2x Row 1, so we can just use Row 1!(This will be a pattern for 2x2 matricies). +Note that this reduced matrix is equivalent to $x_1 = x_2/2$ (using *x* to +represent elements of $\vv{x}_1$ for convenience) with $x_2$ being free. This +means that the eigenvector is not unique, which makes sense both from the +original equation we are trying to solve and us forcing the matrix to be +singular! Equivalently, eigenvectors can be scaled up or down by an arbitrary +nonzero constant. We may then choose $x_2$ so that the eigenvector has either +nice values (like like integers) or a nice norm (1 is conventional). For +example, choosing $x_2 = 2$ gives: -\begin{align} -\vv{x_1} +2\vv{x_2} =0 \to \vv{x}=\begin{bmatrix} 2 \\ -1 \end{bmatrix} -\end{align} +\begin{equation} +\vv{x}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} +\end{equation} + +```{tip} +For a 2x2 matrix, if the eigenvalue $\lambda$ has been found correctly, one row +of $\vv{A}-\lambda\vv{I}$ will always be a multiple of the other. This means +the eigenvector can be determined using only one row. Moreover, the solution +for **x** can be obtained by swapping the entries in a row and changing the sign +of one of them! For example, we had + +\begin{equation} +\begin{bmatrix} +-4 & 2 \\ +2 & -1 +\end{bmatrix} +\end{equation} + +so using the second row, we find: + +\begin{equation} +\vv{x}_1 = \begin{bmatrix}-(-1) \\ 2\end{bmatrix} = \begin{bmatrix}1 \\ 2\end{bmatrix} +\end{equation} + +However, we could also have negated the other entry: + +\begin{equation} +\vv{x}_1 = \begin{bmatrix}-1\\ -2\end{bmatrix} +\end{equation} + +or used the first row: + +\begin{equation} +\vv{x}_1 = \begin{bmatrix}2 \\ 4\end{bmatrix} +\end{equation} + +Both of these are also valid eigenvectors because they are multiples of the one +we found already. +``` + +Let's use the same strategy to get the eigenvector for $\lambda_2 = -6$. We'll +jump straight to the matrix we need: -Hence, the eigenvalues and eigenvectors are: \begin{align} -\lambda_1 = -1 , \vv{x_1}= \begin{bmatrix} 1 \\ 2 \end{bmatrix} \quad \text{and} \quad \lambda_2 = -6, \vv{x_2}= \begin{bmatrix} 2 \\ -1 \end{bmatrix} +(\vv{A} - \lambda_2 \vv{I}) = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} +\to \vv{x}_2 = \begin{bmatrix}2 \\ -1 \end{bmatrix} \end{align} +using the swapping trick on the first row. ## Multiple eigenvalues @@ -312,7 +385,6 @@ with $x_2$ free. Choosing $x_2 = -i$ gives - If **A** is skew-symmetric ($\vv{A}^{\rm T} = -\vv{A}$), its eigenvalues are purely imaginary. - ## Skill builder problems Find the eigenvalues and eigenvectors