## Linear Algebra and Its Applications, Exercise 2.6.16

Exercise 2.6.16. Consider the space of 2 by 2 matrices. Any such matrix can be represented as the linear combination of the matrices $\begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \qquad \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix} \qquad \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix} \qquad \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}$

that serve as a basis for the space. Find a matrix $A$ corresponding to the linear transformation of transposing 2 by 2 matrices, and explain why $A^2 = I$.

Answer: Each of the elementary 2 by 2 matrices can be considered as equivalent to one of the elementary vectors in $\mathbb{R}^4$ $e_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \qquad e_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} \qquad e_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} \qquad e_4 = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$

(Simply read off the entries in the 2 by 2 matrices starting with the first row left to right and then the second row left to right.)

We want $A$ to transpose each of the elementary 2 by 2 matrices into the following matrices: $\begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \qquad \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix} \qquad \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix} \qquad \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}$

corresponding to the following vectors in $\mathbb{R}^4$ $\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$

In other words we must have $Ae_1 = e_1$, $Ae_2 = e_3$, $Ae_3 = e_2$, and $Ae_4 = e_4$.

Since $Ae_1 = e_1$ when multiplying the first row of $A$ by the entries of $e_1$ (which produces the first entry in the resulting vector) we must produce 1, and when multiplying the second through fourth row of $A$ by the entries of $e_1$ we must produce 0. This means $A$ must look as follows: $A = \begin{bmatrix} 1&?&?&? \\ 0&?&?&? \\ 0&?&?&? \\ 0&?&?&? \end{bmatrix}$

Note that we don’t care what the other entries in $A$ are, since when multiplying $e_1$ those entries will end up multiplying zeroes: $Ae_1 = \begin{bmatrix} 1&?&?&? \\ 0&?&?&? \\ 0&?&?&? \\ 0&?&?&? \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} = e_1$

Since $Ae_2 = e_3$ when multiplying the first row of $A$ by the entries of $e_2$ we must produce 0,  and similarly when multiplying the second and fourth rows of $A$ by the entries of $e_2$. However when multiplying the third row of $A$ by the entries of $e_2$ (which produces the third entry in the resulting vector) we must produce 1. This means $A$ must look as follows: $A = \begin{bmatrix} ?&0&?&? \\ ?&0&?&? \\ ?&1&?&? \\ ?&0&?&? \end{bmatrix}$

Note again that we don’t care what the other entries in $A$ are, since when multiplying $e_2$ those entries will end up multiplying zeroes: $Ae_2 = \begin{bmatrix} ?&0&?&? \\ ?&0&?&? \\ ?&1&?&? \\ ?&0&?&? \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} = e_3$

We next have $Ae_3 = e_2$ so when multiplying the second row of $A$ by the entries of $e_3$ we must produce 1,  and when multiplying the other rows of $A$ by the entries of $e_3$ must produce 0. This means $A$ must look as follows: $A = \begin{bmatrix} ?&?&0&? \\ ?&?&1&? \\ ?&?&0&? \\ ?&?&0&? \end{bmatrix}$

so that $Ae_3 = \begin{bmatrix} ?&?&0&? \\ ?&?&1&? \\ ?&?&0&? \\ ?&?&0&? \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} = e_2$

Finally since $Ae_4 = e_4$ when multiplying the first, second, and third rows of $A$ by the entries of $e_4$ we must produce 0, and when multiplying the fourth row of $A$ by the entries of $e_4$ we must produce 1. This means $A$ must look as follows: $A = \begin{bmatrix} ?&?&?&0 \\ ?&?&?&0 \\ ?&?&?&0 \\ ?&?&?&1 \end{bmatrix}$

with $Ae_4 = \begin{bmatrix} ?&?&?&0 \\ ?&?&?&0 \\ ?&?&?&0 \\ ?&?&?&1 \end{bmatrix}\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} = e_4$

Combining the above results we see that $A = \begin{bmatrix} 1&0&0&0 \\ 0&0&1&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{bmatrix}$

Note that we could have achieved the same result by having each column of $A$ be the vector into which the corresponding elementary vector should be transformed. In other words, the first column of $A$ should be $e_1$ (since $A$ transforms $e_1$ into $e_1$), the second column should be $e_3$ (since $A$ transforms $e_2$ into $e_3$), and similarly the third and fourth columns should be $e_2$ and $e_4$ respectively.

When multiplying $A$ by itself we see that $A^2 = \begin{bmatrix} 1&0&0&0 \\ 0&0&1&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} 1&0&0&0 \\ 0&0&1&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{bmatrix} = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} = I$

In other words, applying $A$ twice to a given vector leaves that vector unchanged. This effect can be explained in multiple ways.

First, when applied to vectors in $\mathbb{R}^4$ the matrix $A$ permutes the second and third entries of the vectors but leaves the first and fourth entries unchanged: $A = \begin{bmatrix} 1&0&0&0 \\ 0&0&1&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix} = \begin{bmatrix} v_1 \\ v_3 \\ v_2 \\ v_4 \end{bmatrix}$

Applying $A$ again reverses the effect of the permutation of the second and third entries, again leaving the first and fourth entries unchanged, and thus restores the original vector.

Second, when applied to our original space of 2 by 2 matrices $A$ has the effect of transposing a matrix, transforming $\begin{bmatrix} v_1&v_2 \\ v_3&v_4 \end{bmatrix}$

into $\begin{bmatrix} v_1&v_3 \\ v_2&v_4 \end{bmatrix}$

Applying $A$ twice has the effect of taking the transpose of the transpose. Since $(B^T)^T = B$ for any matrix $B$ this has the effect of restoring the original 2 by 2 matrix.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

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