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I'm having a hard time putting this into a proof. It's posed as a question in Section 6.2 of Hoffman and Kunze, asking whether it is true or not. It seems obviously true, since if T is left multiplication by A, isn't A just the matrix representation of the transformation? So they would clearly have the same characteristic values? I seem to be missing something. From there I wanted to prove that if A is diagonalizable, this implies L is diagonalizable since they have the same characteristic values.

Thanks sincerely for any clarification you can provide.

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up vote 1 down vote accepted

Careful here! $T$ here is a map of matrices not of vectors.
To clarify, suppose that $A$ is a $n \times n$ matrix of real numbers. Then $A$ is the matrix representation of a linear map $\mathbb R^n \rightarrow \mathbb R^n$. However, $T$ here acts on
$ T : \mathbb R^{n \times n} \rightarrow \mathbb R^{n \times n}$,
the space of $n \times n$ real matrices, which is $n^2$-dimensional not $n$-dimensional. Hence the matrix representation of $T$ will be a $n^2 \times n^2$ matrix.

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