Finding an ordered basis to diagonalize Transpose matrix.

Question

We define $T : M_{n \times n}R \to M_{n\times n}R$ by $T(A) = A^t$.

We can write the matrix representation of this transformation as:

$[T]_\beta^\beta = \begin{pmatrix} 1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1 \end{pmatrix}$.

However, we also want to find an ordered basis $\beta$ for the space $M_{2 \times 2}$ such that this matrix representation $[T]_\beta^\beta$ is diagonal.

How could we do this? Would we somehow use the eigenvectors and eigenvalues of this matrix? I know the eigenvalues are $\pm 1$ and the eigenvectors are $A$ itself (the symmetric matrix) and $-A$ (the matrix with entries that are opposite of A).

Would we somehow use the diagonalization formula $A = QDQ^{-1}$?

Thank you!

Answer 1

You could definitely get a valid answer using the diagonalization formula. However, I find that for questions like these, it's easier to find the eigenvectors with an "educated guess".

In particular, try the basis $$ \left\{ \pmatrix{1&0\\0&0}, \pmatrix{0&0\\0&1}, \pmatrix{0&1\\1&0}, \pmatrix{0&-1\\1&0} \right\} $$

Answer 2

Another point of view: $T$ is a permutation s.t. $T^2=I$. Then $T$ decomposes in a product of cycles of order $2$. To use this fact, we choose the canonical basis with the following order: $E_{1,1},\cdots,E_{n,n}$ and for any $i<j$ the couple $E_{i,j},E_{j,i}$ (no matter about the order). Finally, the associated matrix is $diag(I_n,U_1,\cdots,U_{\binom{n}{2}})$ where $U_k=\begin{pmatrix}0&1\\1&0\end{pmatrix}$.