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The teacher has posted a solution for the following:

Let $A \in \mathbb{K}^{n \times n}$ and $\Phi_A: \mathbb{K}^{n \times n} \to \mathbb{K}^{n \times n}$ defined by $\Phi_A(X) := AX$. Show that $\chi_{\Phi_A} = (\chi_A)^n.$

but it is too confusing. What is strange to me is that the matrix representation of the linear map shown in the question (we need it because the characteristic polynomial for it is $\det(M_\phi(X) - \lambda I)$ has dimension $n^2$. Does anyone know why?

Original image.

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    $\begingroup$ Please take the time to enter key parts of your question as text instead of linking to an image of it, especially when the content of the image is only one sentence. Images are neither searchable nor accessible to screen readers. You can find a quick reference to formatting mathematical expressions in MathJax here. $\endgroup$
    – amd
    Aug 22, 2019 at 0:26

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I think that was a mistake. If $f:V\to W$ is a linear map between two finite dimensional vector spaces, the corresponding matrix representation has dimension $(\dim W)\times (\dim V)$ right?

In your case, $\dim \mathbb K^{n\times n}=n^2$, so indeed the dimension of the matrix representation of $\Phi_A$ is $n^2\times n^2$.

By the way, there's no dimension issue when looking at the desired result. Indeed $\deg\chi_{\Phi_A}=\dim\mathbb K^{n\times n}=n^2$ while $\deg(\chi_A)^n=n\deg\chi_A=n\dim\mathbb K^n=n^2$.

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  • $\begingroup$ okay this makes much more sense, thank you so much! $\endgroup$ Aug 22, 2019 at 1:34
  • $\begingroup$ You're welcome. $\endgroup$ Aug 22, 2019 at 8:54

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