Test du plugin WP QuickLaTeX

Voici une équation

(X\cdot I_{n}-A)\cdot S=\chi_{A}\cdot\mathrm{Id}



and N(t) be two polynomials in t with matric coefficients

Let us use evaluation with coefficients on the right, that is M(P)\,:={\displaystyle \sum_{i}P\cdot M_{i}}

Then, if P commutes with all the coefficients M_{i}, (M\cdot N)(P)={\displaystyle M(P)\cdot N(P)}

The proof of this lemma is quite obvious: it is based on the equalities
P^{i+j}\cdot M_{i}\cdot N_{j}=P^{i}\cdot M_{i}\cdot P^{j}\cdot N_{j}

when P commutes with all the M_{j}

One can apply this lemma to the identity used for the proof, {\displaystyle (tI_{n}-A)\cdot B=p(t)\cdot I_{n}}

Obiously, A commutes with the left factor (tI_{n}-A)

It is then legitimate to replace t by A in this identity, and we are done.

It is true that A commutes also with the matrix coefficients in B, but this is not necessary for the proof.

I let it to the author to double-check the reasoning above and to simplify the demonstration acccordingly.

With this simplification, this proof using polynomials with matrix coefficients seems to me the nicest and simplest proof.

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