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The Question: Let $A$ be an $n \times n$ matrix with distinct eigenvalues $\lambda_{1},...,\lambda_{n}$. Show that ad$_{A}$ acting on the space of all $n \times n$ complex matrices has $n^{2}$ eigenvalues $\lambda_{i}-\lambda_{j}$, $1 \leq i,j \leq n$. Then, find the corresponding eigenvectors ("eigenmatrices").

The Attempt: So we can think of ad$_{A}$ as a matrix of dimension $n^{2} \times n^{2}$, with $n^{2}$ eigenvalues. I can diagonalize A to get the eigenvalues in the diagonal. We have distinct eigenvalues. I think this has to do with weights but I am unsure.

any hints would be appreciated.

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Is $\text{ad}_A(Y) = [A,Y]$? –  Juan S May 27 '12 at 7:29

1 Answer 1

up vote 4 down vote accepted

Let us denote by $u_1,\ldots,u_n$ and $v_1,\ldots,v_n$ the distinct eigenvectors of $A$ and $A^T$ respectively corresponding to the eigenvalues $\lambda_1,\ldots,\lambda_n.$

If we introduce $X_{ij}=u_i\otimes v_j,$ where $\otimes$ denote the Kronecker tensor product, then we get $$AX_{ij}=\lambda_iX_{ij},\quad X_{ij}A=\lambda_jX_{ij}.$$

So $\textrm{ad}_AX_{ij}\equiv[A,X_{ij}]\equiv AX_{ij}-X_{ij}A=(\lambda_i-\lambda_j)X_{ij}.$

I hope it helps.

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Yes, this was very consise thank you. This morning I had something similar using the Kronecker tensor product, but I thought I was going in the wrong direction, so I scrapped it. Thanks so much! –  Dylan Sabulsky May 27 '12 at 16:44

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