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Course Notes for Linear Algebra

Remkes Kooistra

Section 7.5 Week 7 Activity

Subsection 7.5.1 Inversion

All row reductions in these activity were done with Wolfram Alpha.

Activity 7.5.1.

Calculate the inverse of this matrix. Check by matrix multiplication that the product of the matrix and its inverse (in both orders!) is the identity.
\begin{equation*} \left( \begin{matrix} 7 \amp 0 \\ -5 \amp -4 \end{matrix} \right) \end{equation*}
Solution.
I write the extended matrix with the original matrix and the identity.
\begin{equation*} \left( \begin{array}{cc|cc} 7 \amp 0 \amp 1 \amp 0 \\ -5 \amp -4 \amp 0 \amp 1 \\ \end{array} \right) \end{equation*}
I row reduce this extended matrix.
\begin{equation*} \left( \begin{array}{cc|cc} 1 \amp 0 \amp \frac{4}{28} \amp 0 \\ 0 \amp 1 \amp \frac{-5}{28} \amp \frac{-1}{4} \\ \end{array} \right) \end{equation*}
The left side of the extended matrix reduces to the identity, so the right side is the inverse matrix. Then I check the two products.
\begin{align*} \frac{1}{28} \left( \begin{matrix} 4 \amp 0 \\ -5 \amp -7 \end{matrix} \right) \left( \begin{matrix} 7 \amp 0 \\ -5 \amp -4 \end{matrix} \right) = \left( \begin{matrix} 1 \amp 0 \\ 0 \amp 1 \\ \end{matrix} \right)\\ \left( \begin{matrix} 7 \amp 0 \\ -5 \amp -4 \end{matrix} \right) \frac{1}{28} \left( \begin{matrix} 4 \amp 0 \\ -5 \amp -7 \end{matrix} \right) = \left( \begin{matrix} 1 \amp 0 \\ 0 \amp 1 \\ \end{matrix} \right) \end{align*}
Both products are the identity matrix, so this is the correct inverse.

Activity 7.5.2.

Calculate the inverse of this matrix. Check by matrix multiplication that the product of the matrix and its inverse (in both orders!) is the identity.
\begin{equation*} \left( \begin{matrix} 3 \amp 0 \amp 1 \\ -3 \amp -1 \amp 0 \\ 2 \amp 1 \amp -4 \end{matrix} \right) \end{equation*}
Solution.
I write the extended matrix with the original matrix and the identity.
\begin{equation*} \left( \begin{array}{ccc|ccc} 3 \amp 0 \amp 1 \amp 1 \amp 0 \amp 0 \\ -3 \amp -1 \amp 0 \amp 0 \amp 1 \amp 0 \\ 2 \amp 1 \amp -4 \amp 0 \amp 0 \amp 1 \end{array} \right) \end{equation*}
I row reduce this extended matrix.
\begin{equation*} \left( \begin{array}{ccc|ccc} 1 \amp 0 \amp 0 \amp \frac{4}{11} \amp \frac{1}{11} \amp \frac{1}{11} \\ 0 \amp 1 \amp 0 \amp \frac{-12}{11} \amp \frac{-14}{11} \amp \frac{-3}{11} \\ 0 \amp 0 \amp 1 \amp \frac{-1}{11} \amp \frac{-3}{11} \amp \frac{-3}{11} \end{array} \right) \end{equation*}
The left side reduces to the identity matrix, so the right side is the inverse matrix. Then I check the two products.
\begin{align*} \frac{1}{11} \left( \begin{matrix} 4 \amp 1 \amp 1 \\ -12 \amp -14 \amp -3 \\ -1 \amp -3 \amp -3 \end{matrix} \right) \left( \begin{matrix} 3 \amp 0 \amp 1 \\ -3 \amp -1 \amp 0 \\ 2 \amp 1 \amp -4 \end{matrix} \right) = \left( \begin{matrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \\ \end{matrix} \right)\\ \left( \begin{matrix} 3 \amp 0 \amp 1 \\ -3 \amp -1 \amp 0 \\ 2 \amp 1 \amp -4 \end{matrix} \right) \frac{1}{11} \left( \begin{matrix} 4 \amp 1 \amp 1 \\ -14 \amp -14 \amp -3 \\ -1 \amp -3 \amp -3 \end{matrix} \right) = \left( \begin{matrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \\ \end{matrix} \right) \end{align*}
Both products are the identity matrix, so this is the correct inverse.

Activity 7.5.3.

Calculate the inverse of this matrix. Check by matrix multiplication that the product of the matrix and its inverse (in both orders!) is the identity.
\begin{equation*} \left( \begin{matrix} 0 \amp -5 \amp 2 \\ 3 \amp 0 \amp 1 \\ 0 \amp -4 \amp 1 \end{matrix} \right) \end{equation*}
Solution.
I write the extended matrix with the original matrix and the identity.
\begin{equation*} \left( \begin{array}{ccc|ccc} 0 \amp -5 \amp 2 \amp 1 \amp 0 \amp 0 \\ 3 \amp 0 \amp 1 \amp 0 \amp 1 \amp 0 \\ 0 \amp -4 \amp 1 \amp 0 \amp 0 \amp 1 \end{array} \right) \end{equation*}
I row reduce this extended matrix.
\begin{equation*} \left( \begin{array}{ccc|ccc} 1 \amp 0 \amp 0 \amp \frac{-4}{9} \amp \frac{3}{9} \amp \frac{5}{9} \\ 0 \amp 1 \amp 0 \amp \frac{3}{9} \amp 0 \amp \frac{-6}{9} \\ 0 \amp 0 \amp 1 \amp \frac{12}{9} \amp 0 \amp \frac{-15}{9} \end{array} \right) \end{equation*}
The left side reduces to the identity matrix, so the right side is the inverse matrix. Then I check the two products.
\begin{align*} \frac{1}{9} \left( \begin{matrix} -4 \amp 3 \amp 5 \\ 3 \amp 0 \amp -6 \\ 12 \amp 0 \amp -15 \end{matrix} \right) \left( \begin{matrix} 0 \amp -5 \amp 2 \\ 3 \amp 0 \amp 1 \\ 0 \amp -4 \amp 1 \end{matrix} \right) = \left( \begin{matrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \\ \end{matrix} \right)\\ \left( \begin{matrix} 0 \amp -5 \amp 2 \\ 3 \amp 0 \amp 1 \\ 0 \amp -4 \amp 1 \end{matrix} \right) \frac{1}{9} \left( \begin{matrix} -4 \amp 3 \amp 5 \\ 3 \amp 0 \amp -6 \\ 12 \amp 0 \amp -15 \end{matrix} \right) = \left( \begin{matrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \\ \end{matrix} \right) \end{align*}
Both products are the identity matrix, so this is the correct inverse.

Activity 7.5.4.

Calculate the inverse of this matrix. Check by matrix multiplication that the product of the matrix and its inverse (in both orders!) is the identity.
\begin{equation*} \left( \begin{matrix} 1 \amp -6 \amp 2 \\ -3 \amp 4 \amp 1 \\ -5 \amp 2 \amp 4 \end{matrix} \right) \end{equation*}
Solution.
I write the extended matrix with the original matrix and the identity.
\begin{equation*} \left( \begin{array}{ccc|ccc} 1 \amp -6 \amp 2 \amp 1 \amp 0 \amp 0 \\ -3 \amp 4 \amp 1 \amp 0 \amp 1 \amp 0 \\ -5 \amp 2 \amp 4 \amp 0 \amp 0 \amp 1 \end{array} \right) \end{equation*}
I row reduce this extended matrix.
\begin{equation*} \left( \begin{array}{ccc|ccc} 1 \amp 0 \amp -1 \amp 0 \amp \frac{2}{14} \amp \frac{-4}{14} \\ 0 \amp 1 \amp \frac{-1}{2} \amp 0 \amp \frac{5}{14} \amp \frac{-3}{14} \\ 0 \amp 0 \amp 0 \amp \frac{14}{14} \amp \frac{28}{14} \amp \frac{-14}{14} \end{array} \right) \end{equation*}
The right side does not reduce to the identity. Therefore, this matrix is not invertible.

Subsection 7.5.2 Transformations of Spans

Activity 7.5.5.

Calculate the image of the given span under the given matrix.
\begin{align*} \amp \Span \left\{ \left( \begin{matrix} 3 \\ -1 \end{matrix} \right) \right\} \amp \amp \left( \begin{matrix} 1 \amp -4 \\ -2 \amp 4 \end{matrix} \right) \end{align*}
Solution.
To calculate the image of the span, I apply the matrix to the vector.
\begin{align*} \left( \begin{matrix} 1 \amp -4 \\ -2 \amp 4 \end{matrix} \right) \left( \begin{matrix} 3 \\ -1 \end{matrix} \right) = \left( \begin{matrix} 7 \\ -10 \end{matrix} \right) \end{align*}
Then the image of the span is the span of the new vector.
\begin{equation*} \left( \begin{matrix} 1 \amp -4 \\ -2 \amp 4 \end{matrix} \right) \left[ \Span \left\{ \left( \begin{matrix} 3 \\ -1 \end{matrix} \right) \right\} \right] = \Span \left\{ \left( \begin{matrix} 7 \\ -10 \end{matrix} \right) \right\} \end{equation*}

Activity 7.5.6.

Calculate the image of the given span under the given matrix.
\begin{align*} \amp \Span \left\{ \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \right\} \amp \amp \left( \begin{matrix} -1 \amp -1 \amp 4 \\ 0 \amp 2 \amp -4 \\ -3 \amp -5 \amp 1 \end{matrix} \right) \end{align*}
Solution.
To calculate the image of the span, I apply the matrix to the vector.
\begin{align*} \left( \begin{matrix} -1 \amp -1 \amp 4 \\ 0 \amp 2 \amp -4 \\ -3 \amp -5 \amp 1 \end{matrix} \right) \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 2 \\ -2 \\ -7 \end{matrix} \right) \end{align*}
Then the image of the span is the span of the new vector.
\begin{equation*} \left( \begin{matrix} -1 \amp -1 \amp 4 \\ 0 \amp 2 \amp -4 \\ -3 \amp -5 \amp 1 \end{matrix} \right) \left[ \Span \left\{ \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \right\} \right] = \Span \left\{ \left( \begin{matrix} 2 \\ -2 \\ -7 \end{matrix} \right) \right\} \end{equation*}

Activity 7.5.7.

Calculate the image of the given span under the given matrix.
\begin{align*} \amp \Span \left\{ \left( \begin{matrix} 0 \\ -4 \\ 0 \end{matrix} \right), \left( \begin{matrix} -1 \\ -1 \\ 2 \end{matrix} \right) \right\} \amp \amp \left( \begin{matrix} -2 \amp 4 \amp 7 \\ 1 \amp 0 \amp 1 \\ 1 \amp 6 \amp -3 \end{matrix} \right) \end{align*}
Solution.
To calculate the image of the span, I apply the matrix to both vectors.
\begin{align*} \left( \begin{matrix} -2 \amp 4 \amp 7 \\ 1 \amp 0 \amp 1 \\ 1 \amp 6 \amp -3 \end{matrix} \right) \left( \begin{matrix} 0 \\ -4 \\ 0 \end{matrix} \right) = \left( \begin{matrix} -16 \\ 0 \\ -24 \end{matrix} \right) \\ \left( \begin{matrix} -2 \amp 4 \amp 7 \\ 1 \amp 0 \amp 1 \\ 1 \amp 6 \amp -3 \end{matrix} \right) \left( \begin{matrix} -1 \\ -1 \\ 2 \end{matrix} \right) = \left( \begin{matrix} 12 \\ 1 \\ -13 \end{matrix} \right) \end{align*}
Then the image of the span is the span of the new vectors.
\begin{equation*} \left( \begin{matrix} -2 \amp 4 \amp 7 \\ 1 \amp 0 \amp 1 \\ 1 \amp 6 \amp -3 \end{matrix} \right) \left[ \Span \left\{ \left( \begin{matrix} 0 \\ -4 \\ 0 \end{matrix} \right), \left( \begin{matrix} -1 \\ -1 \\ 2 \end{matrix} \right) \right\} \right] = \Span \left\{ \left( \begin{matrix} -16 \\0 \\ -24 \end{matrix} \right), \left( \begin{matrix} 12 \\ 1 \\ -13 \end{matrix} \right) \right\} \end{equation*}

Activity 7.5.8.

Calculate the image of the given offset span under the given matrix.
\begin{align*} \left( \begin{matrix} 4 \\ 0 \\ -1 \end{matrix} \right) + \amp \Span \left\{ \left( \begin{matrix} 2 \\ 1 \\ -7 \end{matrix} \right), \left( \begin{matrix} 1 \\ -4 \\ 1 \end{matrix} \right) \right\} \amp \amp \left( \begin{matrix} 0 \amp -4 \amp -4 \\ -2 \amp 6 \amp 1 \\ -5 \amp 3 \amp 1 \end{matrix} \right) \end{align*}
Solution.
To calculate the image of the offset span, I apply the matrix to all three vectors.
\begin{align*} \left( \begin{matrix} 0 \amp -4 \amp -4 \\ -2 \amp 6 \amp 1 \\ -5 \amp 3 \amp 1 \end{matrix} \right) \left( \begin{matrix} 4 \\ 0 \\ -1 \end{matrix} \right) = \left( \begin{matrix} 4 \\ -9 \\ -21 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp -4 \amp -4 \\ -2 \amp 6 \amp 1 \\ -5 \amp 3 \amp 1 \end{matrix} \right) \left( \begin{matrix} 2 \\ 1 \\ -7 \end{matrix} \right) = \left( \begin{matrix} 24 \\ -5 \\ -14 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp -4 \amp -4 \\ -2 \amp 6 \amp 1 \\ -5 \amp 3 \amp 1 \end{matrix} \right) \left( \begin{matrix} 1 \\ -4 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 12 \\ -25 \\ -16 \end{matrix} \right) \end{align*}
Then the image of the offset span is the offset span of the new vectors.
\begin{align*} \amp \left( \begin{matrix} 0 \amp -4 \amp -4 \\ -2 \amp 6 \amp 1 \\ -5 \amp 3 \amp 1 \end{matrix} \right) \left[ \left( \begin{matrix} 4 \\ 0 \\ - 1 \end{matrix} \right) + \Span \left\{ \left( \begin{matrix} 2 \\ 1 \\ -7 \end{matrix} \right), \left( \begin{matrix} 1 \\ -4 \\ 1 \end{matrix} \right) \right\} \right] \\ \amp = \left( \begin{matrix} 4 \\ -9 \\ -21 \end{matrix} \right) + \Span \left\{ \left( \begin{matrix} 24 \\ -5 \\ -14 \end{matrix} \right), \left( \begin{matrix} 12 \\ -25 \\ -16 \end{matrix} \right) \right\} \end{align*}

Activity 7.5.9.

Calculate the image of the given offset span under the given matrix.
\begin{align*} \left( \begin{matrix} 1 \\ 0 \\ 1 \\ 1 \end{matrix} \right) + \amp \Span \left\{ \left( \begin{matrix} -1 \\ 5 \\ 6 \\ -1 \end{matrix} \right), \left( \begin{matrix} 0 \\ -3 \\0 \\ -3 \end{matrix} \right), \left( \begin{matrix} -1 \\ 0 \\ 5 \\ 5 \end{matrix} \right) \right\} \amp \amp \left( \begin{matrix} 0 \amp -5 \amp 1 \amp 1 \\ -1 \amp -2 \amp 6 \amp -3 \\ 0 \amp 1 \amp 0 \amp -4 \end{matrix} \right) \end{align*}
Solution.
To calculate the image of the offset span, I apply the matrix to all four vectors.
\begin{align*} \left( \begin{matrix} 0 \amp -5 \amp 1 \amp 1 \\ -1 \amp -2 \amp 6 \amp -3 \\ 0 \amp 1 \amp 0 \amp -4 \end{matrix} \right) \left( \begin{matrix} 1 \\ 0 \\ 1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 2 \\ 2 \\ -4 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp -5 \amp 1 \amp 1 \\ -1 \amp -2 \amp 6 \amp -3 \\ 0 \amp 1 \amp 0 \amp -4 \end{matrix} \right) \left( \begin{matrix} -1 \\ 5 \\ 6 \\ -1 \end{matrix} \right) = \left( \begin{matrix} -20 \\ 30 \\ 9 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp -5 \amp 1 \amp 1 \\ -1 \amp -2 \amp 6 \amp -3 \\ 0 \amp 1 \amp 0 \amp -4 \end{matrix} \right) \left( \begin{matrix} 0 \\ -3 \\ 0 \\ -3 \end{matrix} \right) = \left( \begin{matrix} 12 \\ 15 \\ 9 \end{matrix} \right) \\ \left( \begin{matrix} 0 \amp -5 \amp 1 \amp 1 \\ -1 \amp -2 \amp 6 \amp -3 \\ 0 \amp 1 \amp 0 \amp -4 \end{matrix} \right) \left( \begin{matrix} -1 \\ 0 \\ 5 \\ 5 \end{matrix} \right) = \left( \begin{matrix} 10 \\ 16 \\ -20 \end{matrix} \right) \end{align*}
Then the image of the offset span is the offset span of the new vector.
\begin{align*} \amp \left( \begin{matrix} 0 \amp -5 \amp 1 \amp 1 \\ -1 \amp -2 \amp 6 \amp -3 \\ 0 \amp 1 \amp 0 \amp -4 \end{matrix} \right) \left[ \left( \begin{matrix} 1 \\ 0 \\ 1 \\ 1 \end{matrix} \right) + \Span \left\{ \left( \begin{matrix} -1 \\ 5 \\ 6 \\ -1 \end{matrix} \right), \left( \begin{matrix} 0 \\ -3 \\ 0 \\ -3 \end{matrix} \right), \left( \begin{matrix} -1 \\ 0 \\ 5 \\ 5 \end{matrix} \right) \right\} \right] \\ = \amp \left( \begin{matrix} 2 \\ 2 \\ -4 \end{matrix} \right) + \Span \left\{ \left( \begin{matrix} -20 \\ 30 \\ 9 \end{matrix} \right), \left( \begin{matrix} 12 \\ 15 \\ 9 \end{matrix} \right), \left( \begin{matrix} 10 \\ 16 \\ -20 \end{matrix} \right) \right\} \end{align*}

Subsection 7.5.3 Kernels and Images

Activity 7.5.10.

Calculate the kernel and image of this matrix.
\begin{equation*} \left( \begin{matrix} 2 \amp 3 \\ -3 \amp 2 \end{matrix} \right) \end{equation*}
Solution.
The kernel is the solution to the system with all constants as zero. Here is the matrix of that system.
\begin{equation*} \left( \begin{array}{cc|c} 2 \amp 3 \amp 0 \\ -3 \amp 2 \amp 0 \end{array} \right) \end{equation*}
I row reduce this system.
\begin{equation*} \left( \begin{array}{cc|c} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \end{array} \right) \end{equation*}
The only solution is the zero vector.
\begin{equation*} \Ker \left( \begin{matrix} 2 \amp 3 \\ -3 \amp 2 \end{matrix} \right) = \left\{ \left( \begin{matrix} 0 \\ 0 \end{matrix} \right) \right\} \end{equation*}
The image is the span of the columns. I put the columns as rows in a matrix and row reduce (without exchanging rows).
\begin{equation*} \left( \begin{matrix} 2 \amp -3 \\ 3 \amp 2 \end{matrix} \right) \end{equation*}
I row reduce this matrix.
\begin{equation*} \left( \begin{matrix} 1 \amp 0 \\ 0 \amp 1 \end{matrix} \right) \end{equation*}
I see that all of these are linearly independent. I write the image as the span of the linearly independent columns.
\begin{equation*} \Image \left( \begin{matrix} 2 \amp 3 \\ -3 \amp 2 \end{matrix} \right) = \Span \left\{ \left( \begin{matrix} 2 \\ -3 \end{matrix} \right), \left( \begin{matrix} 3 \\ 2 \end{matrix} \right) \right\} \end{equation*}
Alternatively, I could simply say that the image is all of \(\RR^2\text{,}\) since these two vectors span the whole space.

Activity 7.5.11.

Calculate the kernel and image of this matrix.
\begin{equation*} \left( \begin{matrix} 2 \amp 5 \\ -4 \amp -10 \end{matrix} \right) \end{equation*}
Solution.
The kernel is the solution to the system with all constants as zero. Here is matrix of that system.
\begin{equation*} \left( \begin{array}{cc|c} 2 \amp 5 \amp 0 \\ -4 \amp -10 \amp 0 \end{array} \right) \end{equation*}
I row reduce this system.
\begin{equation*} \left( \begin{array}{cc|c} 1 \amp \frac{5}{2} \amp 0 \\ 0 \amp 0 \amp 0 \end{array} \right) \end{equation*}
I interpret the solution as a span. In the solution, \(x + \frac{5}{2} y = 0\text{.}\) If I take \(y = 2\) then \(x = -5\) gives us a vector to describe the span.
\begin{equation*} \Ker \left( \begin{matrix} 2 \amp 5 \\ -4 \amp -10 \end{matrix} \right) = \Span \left\{ \left( \begin{matrix} -5 \\ 2 \end{matrix} \right) \right\} \end{equation*}
The image is the span of the columns. I put the columns as rows in a matrix and row reduce (without exchanging rows).
\begin{equation*} \left( \begin{matrix} 2 \amp -4 \\ 5 \amp -10 \end{matrix} \right) \end{equation*}
I row reduce this matrix.
\begin{equation*} \left( \begin{matrix} 1 \amp -2 \\ 0 \amp 0 \end{matrix} \right) \end{equation*}
I see that only one of these is linearly independent. I write the image as the span of the linearly independent columns.
\begin{equation*} \Image \left( \begin{matrix} 2 \amp 5 \\ -4 \amp -10 \end{matrix} \right) = \Span \left\{ \left( \begin{matrix} 2 \\ -4 \end{matrix} \right) \right\} \end{equation*}

Activity 7.5.12.

Calculate the kernel and image of this matrix.
\begin{equation*} \left( \begin{matrix} 4 \amp 0 \amp -1 \\ -2 \amp -2 \amp -2 \\ -1 \amp 0 \amp 3 \end{matrix} \right) \end{equation*}
Solution.
The kernel is the solution to the system with all constants as zero. Here is matrix of that system.
\begin{equation*} \left( \begin{array}{ccc|c} 4 \amp 0 \amp -1 \amp 0 \\ -2 \amp -2 \amp -2 \amp 0 \\ -1 \amp 0 \amp 3 \amp 0 \end{array} \right) \end{equation*}
I row reduce this system.
\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ \end{array} \right) \end{equation*}
All components must be set to zero, so the kernel is only the zero vector.
\begin{equation*} \Ker \left( \begin{matrix} 4 \amp 0 \amp -1 \\ -2 \amp -2 \amp -2 \\ -1 \amp 0 \amp 3 \end{matrix} \right) = \left\{ \left( \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right) \right\} \end{equation*}
The image is the span of the columns. I put the columns as rows in a matrix and row reduce (without exchanging rows).
\begin{equation*} \left( \begin{matrix} 4 \amp -2 \amp -1 \\ 0 \amp -2 \amp 0 \\ -1 \amp -2 \amp 3 \end{matrix} \right) \end{equation*}
I row reduce this matrix.
\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \\ \end{matrix} \right) \end{equation*}
I see that all of these are linearly independent. I write the image as the span of the linearly independent columns.
\begin{equation*} \Image \left( \begin{matrix} 4 \amp 0 \amp -1 \\ -2 \amp -2 \amp -2 \\ -1 \amp 0 \amp 3 \end{matrix} \right) = \Span \left\{ \left( \begin{matrix} 4 \\ -2 \\ -1 \end{matrix} \right), \left( \begin{matrix} 0 \\ -2 \\ 0 \end{matrix} \right), \left( \begin{matrix} -1 \\ -2 \\ 3 \end{matrix} \right) \right\} \end{equation*}
Alternatively, I can just saw that the image is \(\RR^3\text{,}\) since these three vectors span the entire space.

Activity 7.5.13.

Calculate the kernel and image of this matrix.
\begin{equation*} \left( \begin{matrix} 0 \amp -5 \amp 2 \\ 3 \amp -4 \amp 1 \\ -3 \amp -1 \amp 1 \end{matrix} \right) \end{equation*}
Solution.
The kernel is the solution to the system with all constants as zero. Here is matrix of that system.
\begin{equation*} \left( \begin{array}{ccc|c} 0 \amp -5 \amp 2 \amp 0 \\ 3 \amp -4 \amp 1 \amp 0 \\ -3 \amp -1 \amp 1 \amp 0 \end{array} \right) \end{equation*}
I row reduce this system.
\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp 0 \amp \frac{-1}{5} \amp 0 \\ 0 \amp 1 \amp \frac{-2}{5} \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \end{array} \right) \end{equation*}
I interpret the solution as a span. There is one free variable. If I set \(z = 5\) then \(x = 1\) and \(y=2\) gives a vector to describe the span.
\begin{equation*} \Ker \left( \begin{matrix} 0 \amp -5 \amp 2 \\ 3 \amp -4 \amp 1 \\ -3 \amp -1 \amp 1 \end{matrix} \right) = \Span \left\{ \left( \begin{matrix} 1 \\ 2 \\ 5 \end{matrix} \right) \right\} \end{equation*}
The image is the span of the columns. I put the columns as rows in a matrix and row reduce (without exchanging rows).
\begin{equation*} \left( \begin{matrix} 0 \amp 3 \amp -3 \\ -5 \amp -4 \amp -1 \\ 2 \amp 1 \amp 1 \end{matrix} \right) \end{equation*}
I row reduce this matrix.
\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp 1 \\ 0 \amp 1 \amp -1 \\ 0 \amp 0 \amp 0 \end{matrix} \right) \end{equation*}
I see that two of these are linearly independent. I write the image as the span of the linearly independent columns.
\begin{equation*} \Image \left( \begin{matrix} 0 \amp -5 \amp 2 \\ 3 \amp -4 \amp 1 \\ -3 \amp -1 \amp 1 \end{matrix} \right) = \Span \left\{ \left( \begin{matrix} 0 \\ 3 \\ -3 \end{matrix} \right) \left( \begin{matrix} -5 \\ -4 \\ -1 \end{matrix} \right) \right\} \end{equation*}

Activity 7.5.14.

Calculate the kernel and image of this matrix.
\begin{equation*} \left( \begin{matrix} 1 \amp 3 \amp 0 \amp 1 \\ -1 \amp -2 \amp 0 \amp 1 \end{matrix} \right) \end{equation*}
Solution.
The kernel is the solution to the system with all constants as zero. Here is matrix of that system.
\begin{equation*} \left( \begin{array}{cccc|c} 1 \amp 3 \amp 0 \amp 1 \amp 0 \\ -1 \amp -2 \amp 0 \amp 1 \amp 0 \end{array} \right) \end{equation*}
I row reduce this system.
\begin{equation*} \left( \begin{array}{cccc|c} 1 \amp 0 \amp 0 \amp -5 \amp 0 \\ 0 \amp 1 \amp 0 \amp 2 \amp 0 \end{array} \right) \end{equation*}
I interpret the solution as a span. There are two free variables (I take the variables for \(\RR^4\) as \(w,x,y,z\) in that order). The other variables don’t depend at all on \(y\text{,}\) which gives one spanning vector. Otherwise, if I set \(y=0\) and \(z=1\) then \(w=5\) and \(x=-2\) produces another vector.
\begin{equation*} \Ker \left( \begin{matrix} 1 \amp 3 \amp 0 \amp 1 \\ -1 \amp -2 \amp 0 \amp 1 \end{matrix} \right) = \Span \left\{ \left( \begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix} \right) \left( \begin{matrix} 5 \\ -2 \\ 0 \\ 1 \end{matrix} \right) \right\} \end{equation*}
The image is the span of the columns. I put the columns as rows in a matrix and row reduce (without exchanging rows).
\begin{equation*} \left( \begin{matrix} 1 \amp -1 \\ 3 \amp -2 \\ 0 \amp 0 \\ 1 \amp 1 \end{matrix} \right) \end{equation*}
I row reduce this matrix.
\begin{equation*} \left( \begin{matrix} 1 \amp 0 \\ 0 \amp 1 \\ 0 \amp 0 \\ 0 \amp 0 \end{matrix} \right) \end{equation*}
I see that two of these are linearly independent. I write the image as the span of the linearly independent columns.
\begin{equation*} \Image \left( \begin{matrix} 1 \amp 3 \amp 0 \amp 1 \\ -1 \amp -2 \amp 0 \amp 1 \end{matrix} \right) = \Span \left\{ \left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) \left( \begin{matrix} 3 \\ -2 \end{matrix} \right) \right\} \end{equation*}

Subsection 7.5.4 Proof Questions

Activity 7.5.15.

Let \(M\) and \(N\) be \(n \times n\) invertible matrices. Prove that
\begin{equation*} (MN)^{-1} = N^{-1} M^{-1} \end{equation*}
Solution.
There are a number of ways to argue this. I could write out general matrices and argue algebraically, but that seems very difficult and cumbersome. Instead, I’d like to give a geometric argument. (Such an argument is not as formal but, depending on the context, could still be considered a proof.)
If I interpret \(MN\) as a transformation \(\RR^n \rightarrow \RR^n\text{,}\) I apply \(N\) first, and then I apply \(M\) (composition works from right to left). Since both are invertible matrices, I can reverse both processes. If I wanted to reverse the composition, I would have to start with the last thing I did, which is \(M\text{.}\) Therefore, the first step in the inverse would be \(M^{-1}\text{.}\) Then I would get back to the point after I applied \(N\text{,}\) so the next thing to do would be \(N^{-1}\text{.}\) The way to write the composition of \(M^{-1}\) and \(N^{-1}\) would be \(N^{-1}M^{-1}\text{.}\) Therefore, I have proven that the inverse of \(MN\) should be the composition \(N^{-1}M^{-1}\text{.}\)

Activity 7.5.16.

Let \(M\) be a \(3 \times 3\) matrix. Assume that the kernel of \(M\) is 2 dimensional. Prove that \(M\) can have at most one linearly independent column.
Solution.
I can use the proposition that the kernel and the image have dimensions that add to the dimension of the domain. I know the dimension of the kernel is \(2\text{,}\) and the dimension of the domain is \(3\text{.}\) Therefore, the dimension of the image must be \(1\text{.}\) But I also know that the image is the span of the columns. Since the image has dimension 1, the span of the columns has dimension 1. This means that there is only one linearly independent column.

Subsection 7.5.5 Conceptual Review Questions

  • What is an inverse matrix? What does it mean?
  • What does it mean to transform a linear or affine subspace under a linear transformation?
  • What is a kernel? What is an image? How do these two things give valuable information about a transformation.
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