# Matrix representation of basis with $n\times n$ dimension

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?

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– amd
Aug 22, 2019 at 0:26

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 $$\deg(\chi_A)^n=n\deg\chi_A=n\dim\mathbb K^n=n^2$$.