# How to find "eigenmatrices "

On the space of 2 by 2 matrices, let $T$ be the transformation that transposes every matrix. Find the eigenvalues and "eigenmatrices" for

$A^T=eA$ where $e$ denotes the eigenvalue for the matrix.

• "The" eigenvalue is ill-defined here. Are you missing an hypothesis? Aug 4 '16 at 20:03
• I suggest beginning with a 2 by 2 matrix written as the usual $a,b;c,d$ but then writing that as a column vector with four entries, in order $a,b,c,d.$ Precisely what does the transpose operation on the matrix do to that vector with four entries? What 4 by 4 matrix accomplishes that, meaning $W$ is the 4 by 4 matrix, say $X$ is the column vector with four entries, and I am asking what $W$ needs to be to get $WX$ the desired outcome column vector? Aug 4 '16 at 20:05
• Since the transpose of the transpose is the original matrix, $T^2 = I$ restricts the possible eigenvalues considerably. Aug 4 '16 at 20:25

Hint: the transpose operator squares to $1$. That means its eigenvalues must be $\pm 1$. Symmetric and antisymmetric matrices.

Hint:

Representing a matrix $$A= \begin{bmatrix} a&b\\c&d \end{bmatrix}$$ as a vector (in the standard basis) $$A= \begin{bmatrix} a\\b\\c\\d \end{bmatrix}$$ The transpose operator acts as $$T(A)= \begin{bmatrix} a\\c\\b\\d \end{bmatrix}$$

and it is represented by the matrix:

$$T= \begin{bmatrix} 1&0&0&0\\ 0&0&1&0\\ 0&1&0&0\\ 0&0&0&1 \end{bmatrix}$$ Now you can find the eigenvalues and eigenvectors of this matrix.

Hint: Assume $A$ is an eigenmatrix with eigenvalue $e$. Now look at the identity $(A^T)^T=A$. What information can we get about $e$ from this?

$$A = \frac{1}{2} (A+A^T) + \frac{1}{2} (A - A^T)$$ gives precisely the projection of a matrix $A$ onto the space of symmetric and anti-symmetric matrices having evals 1 and -1, respectively (so there is nothing else, as already mentioned elsewhere). Irrespective of what norm you put on the space.