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One may define the following map : $\begin{array}{l|rcl} f : & M_n(\mathbb R) & \longrightarrow & \mathbb R_n[X] \\ & A & \longmapsto & p_A \end{array}$

Why is $f$ continuous ? Which norms are convenient for this problem ? Note that $f$ isn't linear.

I tried to use the sequential way, without success...

Gabriel Romon
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  • see also http://math.stackexchange.com/questions/121831/why-is-the-determinant-m-n-mathbb-r-to-mathbb-r-continuous – ir7 Jan 12 '14 at 15:18

3 Answers3

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It's continuous because it is given by $\det(A-\lambda I)$. Now the determinant function is continuous because it is a multilinear map, all multilinear maps are continuous. $A-\lambda I$ is obviously continuous. Since the composition of continuous maps is continuous the map in question is continuous.

Wintermute
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  • -1 The question asked why the map from the matrix to the characteristic polynomial is continuous. This is entirely different from why the characteristic polynomial itself is continuous. (it's a polynomial, of course its continuous!) – Poseidaan Feb 27 '22 at 19:27
  • UPDATE: I just noticed I misread your proof and the answer is actually entirely correct. Do you know how I can get rid of my downvote? I get a notification that my vote is locked until the post is edited. – Poseidaan Feb 27 '22 at 19:44
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In the form you are trying:

If there is a convergent sequence $A_n\to A$ then $(A_n-X1\!\!1)\to (A-X1\!\!1)$. Now since $\det$ is continuous $\det(A_n-X1\!\!1)\to\det(A-X1\!\!1)$. So $f$ is continuous.

The arrow $\to$ means "converges".

janmarqz
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It is a polynomial map (each coefficient of the characteristic polynomial is a polynomial in the matrix entries). Polynomials are easily seen to be continuous.

Igor Rivin
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