# Matrix of a linear transformation that associates a matrix to its transpose

Find the eigenvalues and eigenvectors of the transformation $T: \mathbb{R^{2x2}} \rightarrow \mathbb{R^{2x2}}$, which associates, to each $A \in \mathbb{R^{2x2}}$, its transpose, that is, $T(A)=A^T$.

So, I know how to find the eigenvalues and eigenvalues of a matrix, but I'm confused about this question because I'm not sure how the matrix of $T$ looks like. This is what I thought:

if $$A= \left( \begin{array}{cc} a & b\\ c & d \end{array}\right)$$ then $$T(A)=T\left( \begin{array}{cc} a & b\\ c & d \end{array}\right)=\left( \begin{array}{cc} a & c\\ b & d \end{array}\right)$$

Is $A^T=\left( \begin{array}{cc} a & c\\ b & d \end{array}\right)$ the matrix of the transformation $T$?

• No, that is not the matrix of transformation of $T$. $A$ here is an element of the vector space that $T$ acts upon. – Omnomnomnom Dec 9 '15 at 17:43
• The eigenvectors of $T$ are precisely the matrices $A$ for which $A^T$ is a multiple of $A$. You can find these, in fact, without finding the matrix of $T$. – Omnomnomnom Dec 9 '15 at 17:44
• Why are those the eigenvectors? – Encontro Regional Sul Dec 9 '15 at 18:54
• That's the definition of an eigenvector, where the $A$'s are your "vectors". – Omnomnomnom Dec 9 '15 at 22:52

The transformation $T$ maps from the space $\mathbb R^{2\times 2}$ to itself, meaning it maps a $4$ dimensional space to a $4$ dimensional space.

This means the matrix will be a $4\times 4$ matrix. Think about where the transformation $T$ maps the basis of the space $\mathbb R^{2\times 2}$.

Think at $A$ as an element of the vector space $\mathbb{R}^4$: $$A=\begin {bmatrix} a&b\\c&d \end{bmatrix}= a\begin {bmatrix} 1&0\\0&0 \end{bmatrix} + b\begin {bmatrix} 0&1\\0&0 \end{bmatrix}+ c\begin {bmatrix} 0&0\\1&0 \end{bmatrix} + d\begin {bmatrix} 0&0\\0&1 \end{bmatrix}= \begin {bmatrix} a\\b\\c\\d \end{bmatrix}$$ where the matrices at the RHS are the canonical basis for $M_2(\mathbb{R})$.

Now your transformation is: $$T(A)=T\begin {bmatrix} a\\b\\c\\d \end{bmatrix}=\begin {bmatrix} a\\c\\b\\d \end{bmatrix}$$ can you see the $4\times4$ matrix that represent this transformation?

Since $(A^T)^T = A$, you have $T^2 = I$. Therefore the only eigenvalues are $\pm 1$. Eigenvectors for $+1$ are symmetric matrices, those for $-1$ are antisymmetric matrices.

• Doesn't $I$ have only one eigenvalue equal to 1? – Encontro Regional Sul Dec 9 '15 at 18:48
• If $T$ has eigenvalue $\lambda$, $T^2 - I$ has eigenvalue $\lambda^2 - 1$. So you must have $\lambda^2 -1 = 0$, which has solutions $\pm 1$. – Robert Israel Dec 9 '15 at 19:53