From c381ddd1275fe155bfc65ca2f721e92efd4aa85d Mon Sep 17 00:00:00 2001 From: jkostiuk Date: Wed, 12 Mar 2025 19:28:46 +0000 Subject: [PATCH 1/5] AT3 Incorporate Vector Equations --- source/linear-algebra/source/03-AT/03.ptx | 67 ++++++++++++++--------- 1 file changed, 40 insertions(+), 27 deletions(-) diff --git a/source/linear-algebra/source/03-AT/03.ptx b/source/linear-algebra/source/03-AT/03.ptx index 26e2dd151..e1bf03f67 100644 --- a/source/linear-algebra/source/03-AT/03.ptx +++ b/source/linear-algebra/source/03-AT/03.ptx @@ -82,12 +82,12 @@ the set of all vectors that transform into \vec 0?

-Let T: V \rightarrow W be a linear transformation, and let \vec{z} be the additive -identity (the zero vector) of W. The kernelkernel of T +Let T: V \rightarrow W be a linear transformation, and let \vec{0} be the zero vector of W. +The kernelkernel of T (also known as the null spacenull space of T) is an important subspace of V defined by -\ker T = \left\{ \vec{v} \in V\ \big|\ T(\vec{v})=\vec{z}\right\} +\ker T = \left\{ \vec{x} \in V\ \big|\ T(\vec{x})=\vec{0}\right\}

@@ -161,25 +161,26 @@ the set of all vectors that transform into \vec 0?

Let T: \IR^3 \rightarrow \IR^2 be the linear transformation given by the standard matrix -T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right]\right) = \left[\begin{array}{c} 3x+4y-z \\ x+2y+z \end{array}\right] + + A=\left[\begin{array}{ccc} 3 & 4 & -1 \\ 1 & 2 & 1 \end{array}\right]. + +

- -

-Set - - T\left(\left[\begin{array}{c}x\\y\\z\end{array}\right]\right) - = - \left[\begin{array}{c}0\\0\end{array}\right] - to find a linear system of equations whose solution set is the kernel. -

- - -

-Use \RREF(A) to solve this homogeneous system of equations and find a basis -for the kernel of T. -

- + + +

+ If \vec{x}=\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]=x_1\vec{e}_1+x_2\vec{e}_2+x_3\vec{e}_3\in\IR^3, write down a vector equation that is equivalent to the equation T(\vec{x})=\vec{0}. +

+ + + + +

+ Solve this vector equation and, in doing so, find a basis for \ker T. +

+ + @@ -369,6 +370,13 @@ the set of all vectors that are the result of using T to transform \Im T?

+ + +

+ Write down a vector equation equivalent to the equation T(\vec{x})=\left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right]. +

+ +

@@ -377,6 +385,13 @@ the set of all vectors that are the result of using T to transform

+ + +

+ Write down a vector equation equivalent to the equation T(\vec{x})=\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]. +

+ +

@@ -415,7 +430,7 @@ An arbitrary vector \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{ar - +

Let T: \IR^4 \rightarrow \IR^3 be the linear transformation given by the standard matrix @@ -464,16 +479,14 @@ Let T:\IR^n\to\IR^m be a linear transformation with standard matrix A

  • - The kernel of T is the solution set of the homogeneous system given -by the augmented matrix \left[\begin{array}{c|c}A&\vec 0\end{array}\right]. -Use the coefficients of its free variables to get a basis for the kernel (as in ). + The kernel of T is the solution set of the equation T(\vec{x})=\vec{0} or, equivalently, the linear system given by the augmented matrix \left[\begin{array}{c|c}A&\vec 0\end{array}\right]. + Use the coefficients of its free variables to get a basis for the kernel (as in ).

  • - The image of T is the span of the columns of A. Remove -the vectors creating non-pivot columns in \RREF A to get a basis -for the image (as in ). + The image of T is the set of vectors \vec{b}\in \IR^m for which T(\vec{x})=\vec{b} is consistent or, equivalently (by ), the column space of A. + Remove the vectors creating non-pivot columns in \RREF A to get a basis for the image (as in ).

From 90f778e1acc9934e9e5762b942ee3d3c83b11ff3 Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Fri, 14 Mar 2025 16:07:25 +0000 Subject: [PATCH 2/5] quick change to try out in 8 minutes --- source/linear-algebra/source/03-AT/03.ptx | 67 ++++++++++++++++++++--- 1 file changed, 60 insertions(+), 7 deletions(-) diff --git a/source/linear-algebra/source/03-AT/03.ptx b/source/linear-algebra/source/03-AT/03.ptx index e1bf03f67..0ae31f8cb 100644 --- a/source/linear-algebra/source/03-AT/03.ptx +++ b/source/linear-algebra/source/03-AT/03.ptx @@ -50,8 +50,11 @@ Let T: \IR^2 \rightarrow \IR^3 be given by \text{with standard matrix } \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right] -Which of these subspaces of \IR^2 describes -the set of all vectors that transform into \vec 0? +Which of these sets contain all vectors \left[\begin{array}{c}x \\ y \end{array}\right] +that transform into T\left(\left[\begin{array}{c}x \\ y \end{array}\right] \right) + = + \left[\begin{array}{c} x \\ y \\ 0 \end{array}\right]= + \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]?

  1. @@ -128,8 +131,13 @@ Let T: \IR^3 \rightarrow \IR^2 be given by \text{with standard matrix } \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] -Which of these subspaces of \IR^3 describes \ker T, -the set of all vectors that transform into \vec 0? +Which of these subsets of \IR^3 is equal to +\ker T=\left\{ \vec{x} \in \mathbb R^3\ \big|\ T(\vec{x})=\vec{0}\right\}, +the set of all vectors \left[\begin{array}{c}x \\ y \\ z\end{array}\right] +that transform into T\left(\left[\begin{array}{c}x \\ y \\ z \end{array}\right] \right) + = + \left[\begin{array}{c} x \\ y\end{array}\right]= + \left[\begin{array}{c} 0 \\ 0 \end{array}\right]?

    1. @@ -162,7 +170,8 @@ the set of all vectors that transform into \vec 0? Let T: \IR^3 \rightarrow \IR^2 be the linear transformation given by the standard matrix - A=\left[\begin{array}{ccc} 3 & 4 & -1 \\ 1 & 2 & 1 \end{array}\right]. + A=\left[\begin{array}{ccc} 3 & 4 & -1 \\ 1 & 2 & 1 \end{array}\right] + =\left[\begin{array}{ccc} T(\vec e_1) & T(\vec e_2) & T(\vec e_3)\end{array}\right].

      @@ -170,15 +179,59 @@ standard matrix

      - If \vec{x}=\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]=x_1\vec{e}_1+x_2\vec{e}_2+x_3\vec{e}_3\in\IR^3, write down a vector equation that is equivalent to the equation T(\vec{x})=\vec{0}. +Which of these is the most appropriate method to determine whether a \mathbb R^3 vector +\vec{x}=\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]=x_1\vec{e}_1+x_2\vec{e}_2+x_3\vec{e}_3, +belongs to \ker T=\left\{ \vec{x} \in \mathbb R^3\ \big|\ T(\vec{x})=\vec{0}\right\}? +

        +
      1. +

        + Determine if it belongs to the solution set of the vector equation + T(\vec x)=T\left(\left[\begin{array}{c} x_1 \\ x_2\\ x_3\end{array}\right]\right)= + x_1T(\vec e_1)+x_2T(\vec e_2)+x_3T(\vec e_3)= + \left[\begin{array}{c} 3 \\ 1\end{array}\right]. +

        +
        2. +
      2. +

        + Determine if it belongs to the solution set of the vector equation + T(\vec x)=T\left(\left[\begin{array}{c} x_1 \\ x_2\\ x_3\end{array}\right]\right)= + x_1T(\vec e_1)+x_2T(\vec e_2)+x_3T(\vec e_3)= + \left[\begin{array}{c} 0 \\ 0\end{array}\right]. +

        +
        3. +
      3. +

        + Determine if it belongs to the span of the vectors + \left\{T(\vec e_1), T(\vec e_2), T(\vec e_3)\right\}. +

        +
        4. +
      4. +

        + Determine if the set of vectors + \left\{T(\vec e_1), T(\vec e_2), T(\vec e_3), \vec x\right\} + is linearly independent. +

        +
        5. +

      + +

      + B. This is exactly the same as finding the solution space for the + homoegeneous vector equation T(\vec x)=\vec 0. +

      +

      - Solve this vector equation and, in doing so, find a basis for \ker T. + Use this method to find the kernel of T. +

      + +

      + B.

      +       From 8098e692b262dac225e2a02dda075dd7604a5a8f Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Fri, 14 Mar 2025 16:10:21 +0000 Subject: [PATCH 3/5] on more tweak --- source/linear-algebra/source/03-AT/03.ptx | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/source/linear-algebra/source/03-AT/03.ptx b/source/linear-algebra/source/03-AT/03.ptx index 0ae31f8cb..40f5df793 100644 --- a/source/linear-algebra/source/03-AT/03.ptx +++ b/source/linear-algebra/source/03-AT/03.ptx @@ -186,9 +186,9 @@ belongs to \ker T=\left\{ \vec{x} \in \mathbb R^3\ \big|\ T(\vec{x})=\vec{0}\

    2. Determine if it belongs to the solution set of the vector equation - T(\vec x)=T\left(\left[\begin{array}{c} x_1 \\ x_2\\ x_3\end{array}\right]\right)= + T(\vec x)= x_1T(\vec e_1)+x_2T(\vec e_2)+x_3T(\vec e_3)= - \left[\begin{array}{c} 3 \\ 1\end{array}\right]. + \left[\begin{array}{c} 3+4-1 \\ 1+2+1\end{array}\right].

    3. From bc1cf8242c2caf0d39a981d97b73fa89b7748e72 Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Fri, 14 Mar 2025 18:40:36 +0000 Subject: [PATCH 4/5] tweak kernel discovery --- source/linear-algebra/source/03-AT/03.ptx | 45 ++++++++++++++++------- 1 file changed, 31 insertions(+), 14 deletions(-) diff --git a/source/linear-algebra/source/03-AT/03.ptx b/source/linear-algebra/source/03-AT/03.ptx index 40f5df793..9d7cb09db 100644 --- a/source/linear-algebra/source/03-AT/03.ptx +++ b/source/linear-algebra/source/03-AT/03.ptx @@ -88,7 +88,7 @@ that transform into T\left(\left[\begin{array}{c}x \\ y \end{array}\right] \r Let T: V \rightarrow W be a linear transformation, and let \vec{0} be the zero vector of W. The kernelkernel of T (also known as the null spacenull space of T) -is an important subspace of V defined by +is an important subset of V defined by \ker T = \left\{ \vec{x} \in V\ \big|\ T(\vec{x})=\vec{0}\right\} @@ -185,31 +185,34 @@ belongs to \ker T=\left\{ \vec{x} \in \mathbb R^3\ \big|\ T(\vec{x})=\vec{0}\

      1. - Determine if it belongs to the solution set of the vector equation - T(\vec x)= - x_1T(\vec e_1)+x_2T(\vec e_2)+x_3T(\vec e_3)= - \left[\begin{array}{c} 3+4-1 \\ 1+2+1\end{array}\right]. + Determine if the set of vectors + \left\{T(\vec e_1), T(\vec e_2), T(\vec e_3), T(\vec x)\right\} + spans \IR^3.

      2. - Determine if it belongs to the solution set of the vector equation - T(\vec x)=T\left(\left[\begin{array}{c} x_1 \\ x_2\\ x_3\end{array}\right]\right)= - x_1T(\vec e_1)+x_2T(\vec e_2)+x_3T(\vec e_3)= - \left[\begin{array}{c} 0 \\ 0\end{array}\right]. + Determine if the set of vectors + \left\{T(\vec e_1), T(\vec e_2), T(\vec e_3), T(\vec x)\right\} + is linearly dependent.

      3. - Determine if it belongs to the span of the vectors - \left\{T(\vec e_1), T(\vec e_2), T(\vec e_3)\right\}. + Determine if + \vec x=\left[\begin{array}{c} x_1 \\ x_2\\ x_3\end{array}\right] + belongs to the solution set of the vector equation + T(\vec x)=T\left(\left[\begin{array}{c} x_1 \\ x_2\\ x_3\end{array}\right]\right)= + x_1T(\vec e_1)+x_2T(\vec e_2)+x_3T(\vec e_3)= + \left[\begin{array}{c} 0 \\ 0\end{array}\right].

      4. - Determine if the set of vectors - \left\{T(\vec e_1), T(\vec e_2), T(\vec e_3), \vec x\right\} - is linearly independent. + Determine if the equation + T(\vec 0)=T\left(\left[\begin{array}{c} 0 \\ 0\\ 0\end{array}\right]\right)= + 0T(\vec e_1)+0T(\vec e_2)+0T(\vec e_3)= + T(\vec x) is consistent.

      @@ -238,6 +241,20 @@ belongs to \ker T=\left\{ \vec{x} \in \mathbb R^3\ \big|\ T(\vec{x})=\vec{0}\ + + +

      + The kernel of a transformation T + is exactly the solution subspace of + the homogeneous equation T(\vec{x})=\vec{0}. +

      +

      + In particular, the kernel has a basis which may be found as in + . +

      +       +       +

      From 190b03e7f9c71cdad319b7a0cbee1700f4bb860a Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Fri, 14 Mar 2025 19:18:27 +0000 Subject: [PATCH 5/5] more edits based on ad hoc adjustments made today --- source/linear-algebra/source/03-AT/03.ptx | 127 +++++++++++++++------- 1 file changed, 88 insertions(+), 39 deletions(-) diff --git a/source/linear-algebra/source/03-AT/03.ptx b/source/linear-algebra/source/03-AT/03.ptx index 9d7cb09db..3b4fe43d2 100644 --- a/source/linear-algebra/source/03-AT/03.ptx +++ b/source/linear-algebra/source/03-AT/03.ptx @@ -232,7 +232,8 @@ belongs to \ker T=\left\{ \vec{x} \in \mathbb R^3\ \big|\ T(\vec{x})=\vec{0}\

      - B. + \ker T=\left\{\left[\begin{array}{c}3a\\-2a\\a\end{array}\right]\middle| + a\in\IR\right\}

      @@ -241,19 +242,28 @@ belongs to \ker T=\left\{ \vec{x} \in \mathbb R^3\ \big|\ T(\vec{x})=\vec{0}\ - +

      The kernel of a transformation T - is exactly the solution subspace of + is exactly the solution space of the homogeneous equation T(\vec{x})=\vec{0}. + If its standard matrix is A, then we may write + A\vec x=\vec 0 and use \RREF[A\,|\,\vec 0] to + find this kernel.

      - In particular, the kernel has a basis which may be found as in - . + In particular, the kernel is a subspace of the transformation's + domain, and has a basis which may be found as in + : + + \ker T=\left\{\left[\begin{array}{c}3a\\-2a\\a\end{array}\right]\middle| + a\in\IR\right\} \hspace{2em} + \text{Basis for }\ker T=\left\{\left[\begin{array}{c}3\\-2\\1\end{array}\right]\right\}. +

      -       + @@ -316,9 +326,9 @@ Which of these subspaces of \IR^3 describes the set of all vectors that a

      Let T: V \rightarrow W be a linear transformation. The imageimage of T -is an important subspace of W defined by +is an important subset of W defined by -\Im T = \left\{ \vec{w} \in W\ \big|\ \text{there is some }\vec v\in V \text{ with } T(\vec{v})=\vec{w}\right\} +\Im T = \left\{ T(\vec v) \in W\ \big| \vec v\in V \right\}

      @@ -394,7 +404,7 @@ Let T: \IR^3 \rightarrow \IR^2 be given by \text{with standard matrix } \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] -Which of these subspaces of \IR^2 describes \Im T, +Which of these subsets of \IR^2 describes \Im T = \left\{ T(\vec v) \in \IR^2\ \big| \vec v\in \IR^3 \right\}, the set of all vectors that are the result of using T to transform \IR^3 vectors?

      @@ -437,13 +447,27 @@ the set of all vectors that are the result of using T to transform

      Consider the question: Which vectors \vec{w} in \IR^3 belong to - \Im T? + \Im T=\left\{ T(\vec v) \in \IR^3\ \big| \vec v\in \IR^4 \right\}?

      - Write down a vector equation equivalent to the equation T(\vec{x})=\left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right]. + Recall that + + T\left(\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right] \right) = + x_1T(\vec e_1)+x_2T(\vec e_2)+x_3T(\vec e_3)+x_4T(\vec e_4) + . + Complete the following vector equation which must be consistent + in order for \left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right] + to belong to \Im T: + + x_1 \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]+ + x_2 \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]+ + x_3\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]+ + x_4\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]= + \left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right] +

      @@ -458,7 +482,9 @@ the set of all vectors that are the result of using T to transform

      - Write down a vector equation equivalent to the equation T(\vec{x})=\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]. + Write down the vector equation which must be consistent in + order for \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] + to belong to \Im T.

      @@ -475,12 +501,13 @@ the set of all vectors that are the result of using T to transform

      An arbitrary vector \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right] belongs to \Im T provided the equation -x_1 T(\vec{e}_1)+x_2 T(\vec{e}_2)+x_3T(\vec{e}_3)+x_4T(\vec{e}_4)=\vec{w} has... +x_1 T(\vec{e}_1)+x_2 T(\vec{e}_2)+x_3T(\vec{e}_3)+x_4T(\vec{e}_4)= +\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right] has...

      1. no solutions.
        2. +
      2. at most one solution.
      3. exactly one solution.
      4. at least one solution.
        5. -
      5. infinitely-many solutions.

       @@ -488,12 +515,25 @@ An arbitrary vector \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{ar

      - Based on this, how do \Im T and \vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} relate to each other? + Based on this, what do we know?

        -
      1. The set \Im T contains \vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} but is not equal to it.
        2. -
      2. The set \vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} contains \Im T but is not equal to it.
        3. +
      3. + The set \left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} spans + \Im T whenever + \left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} + is linearly independent. +
        4. +
      4. + The set \Im T equals the codomain whenever + \left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} + is linearly independent. +
        5. +
      5. + The set \Im T is simply \{\vec 0\} whenever + \left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} + is linearly dependent. +
      6. The set \Im T and \vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} are equal to each other.
        7. -
      7. There is no relation between these two sets.

       @@ -502,32 +542,41 @@ An arbitrary vector \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{ar

      -Let T: \IR^4 \rightarrow \IR^3 be the linear transformation given by the +Let T: \IR^n \rightarrow \IR^m be the linear transformation given by the standard matrix A = - \left[\begin{array}{cccc} 3 & 4 & 7 & 1\\ -1 & 1 & 0 & 2 \\ 2 & 1 & 3 & -1 \end{array}\right] + \left[\begin{array}{ccc} T(\vec e_1) & \cdots & T(\vec e_n) \end{array}\right] .

      -Since the set - - \setList{ - \left[\begin{array}{c}3\\-1\\2\end{array}\right], - \left[\begin{array}{c}4\\1\\1\end{array}\right], - \left[\begin{array}{c}7\\0\\3\end{array}\right], - \left[\begin{array}{c}1\\2\\-1\end{array}\right] +Then we have + + \Im T=\vspan\setList{ + T(\vec e_1), + \cdots, + T(\vec e_n) } - -spans \Im T, we can obtain a basis for \Im T by finding - - \RREF A + +and in particular, see that \Im T is a subspace of the codomain, and thus +has a kernel which may be found as in : +given + + \RREF \left[\begin{array}{cccc} 3 & 4 & 7 & 1\\ -1 & 1 & 0 & 2 \\ 2 & 1 & 3 & -1 \end{array}\right] = - \left[\begin{array}{cccc} 1 & 0 & 1 & -1\\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] - -and only using the vectors corresponding to pivot columns: + \left[\begin{array}{cccc} \markedPivot{1} & 0 & 1 & -1\\ 0 & \markedPivot{1} & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] + +we see that we have +\Im T=\vspan\left\{ + \left[\begin{array}{c} 3 \\ -1 \\ 2 \end{array}\right], + \left[\begin{array}{c} 4 \\ 1 \\ 1 \end{array}\right], + \left[\begin{array}{c} 7\\ 0 \\ 3 \end{array}\right], + \left[\begin{array}{c} 1\\ 2 \\ -1 \end{array}\right], + \right\} + and +\text{Basis for }\Im T= \setList{ \left[\begin{array}{c}3\\-1\\2\end{array}\right], \left[\begin{array}{c}4\\1\\1\end{array}\right] @@ -535,9 +584,7 @@ and only using the vectors corresponding to pivot columns:

      - In general, the column spacecolumn space of a matrix M refers to the subspace obtained by considering the span of its column vectors. - Using this terminology, if the transformation T is represented by the matrix A, - then the image of T is the column spacecolumn space of A. +This justifies why the image of a transformation is also called the column spacecolumn space of its standard matrix.

      @@ -545,7 +592,8 @@ and only using the vectors corresponding to pivot columns:

      -Let T:\IR^n\to\IR^m be a linear transformation with standard matrix A. +To summarize, +let T:\IR^n\to\IR^m be a linear transformation with standard matrix A.

      • @@ -555,7 +603,8 @@ Let T:\IR^n\to\IR^m be a linear transformation with standard matrix A

      • - The image of T is the set of vectors \vec{b}\in \IR^m for which T(\vec{x})=\vec{b} is consistent or, equivalently (by ), the column space of A. + The image of T is the span of the columns of its standard matrix A, + a.k.a. its column space. Remove the vectors creating non-pivot columns in \RREF A to get a basis for the image (as in ).