Is there a general form for the matrix representation of the transpose operator?

Question

The problem I'm doing requires me to find the eigenvectors, and prove the eigenvalues are 1 and -1, of a linear transformation that takes the transpose of an nxn Matrix, and I'm wondering if there's a general form for the function's matrix representation.

Answer 1

There's a neat representation of the transpose operator using Kronecker products.

In particular: let $\{e_1,\dots,e_n\}$ denote the canonical basis of $\Bbb R^n$. We note that the transpose operator can be defined by $$ (e_ie_j^T)^T = e_je_i^T $$ With the standard basis $$ \mathcal B = \{e_1e_1^T, e_1e_2^T,\dots,e_1e_n^T,e_2e_1^T,e_2e_2^T,\dots\dots e_ne_n^T\} $$ we end up with the coordinate vectors $$ [e_i e_j^T]_{\mathcal B} = e_i \otimes e_j $$ So, we have $$ [T]_{\mathcal B} (e_i \otimes e_j) = e_j \otimes e_i $$

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As far as directly answering the question you've been assigned goes, however, the matrix representation is irrelevant.

In order to prove the eigenvalues, all you need to know is that transposing twice puts you where you started (that is, $(A^T)^T = A$). In order to find the eigenvectors: which matrices satisfy $A^T = A$? Which satisfy $A^T = -A$? That's all there is to it.