Polynomial of matrix

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#1 h Mar 23, 2025, 4:59 PM

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Let A, B are two square matrices with same size.
p(x) and q(x) are real coefficients polynomial
prove:

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#2 h Mar 23, 2025, 8:23 PM

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First assume that $p(A)$ and $q(B)$ are invertible. Then $p(A)^{-1}$ and $q(B)^{-1}$ are polynomials in $A$ and $B$ respectively, and so
they commute with $q(A)$ and $p(B)$ respectively, thus
$p(B)q(B)^{-1}+p(A)^{-1}q(A) = q(B)^{-1}p(B)+q(A)p(A)^{-1}$ and so
$|p(B)q(B)^{-1}+p(A)^{-1}q(A)| = |q(B)^{-1}p(B)+q(A)p(A)^{-1}|,$ $|p(A)|\cdot|p(B)q(B)^{-1}+p(A)^{-1}q(A)|\cdot|q(B)| = |q(B)|\cdot|q(B)^{-1}p(B)+q(A)p(A)^{-1}|\cdot|p(A)|,$ $|p(A)p(B)+q(A)q(B)| = |p(B)p(A)+q(B)q(A)|.$
If $p(A)$ and/or $q(B)$ are not invertible, then replace $p$ and/or $q$ with $p+\varepsilon$ and/or $q+\varepsilon$ for $\varepsilon$ close enough to $0$ and take limit $\varepsilon\to0$ .

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#3 h Monday at 5:09 PM

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GreenKeeper wrote:

First assume that $p(A)$ and $q(B)$ are invertible. Then $p(A)^{-1}$ and $q(B)^{-1}$ are polynomials in $A$ and $B$ respectively

is this really true? I know it's true in linear case (for example $(I+A)^{-1} = \frac{1}{I+A} = ...$ (Taylor series) but idk if it's true for a polynomial in general case

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#4 h Monday at 8:15 PM • 1 Y

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Yes, for any invertible matrix $X$ the inverse $X^{-1}$ is a polynomial in $X$ , this follows from the Cayley-Hamilton theorem. So $p(A)^{-1}$ is a polynomial in $p(A)$ which is (by definition) a polynomial in $A$ . Same for the other one.

Also, the Taylor series won't be of much help since it gives you an infinite power series in $A$ that doesn't need to converge (the commutativity argument would still work for infinite series but the convergence is the problem).

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#5 h Yesterday at 12:23 AM • 2 Y

Y by MihaiT, Mathloops

A stronger result is that if $X,Y,Z,T$ are square matrices of the same size over $\mathbb{C},$ then
$\det \begin{pmatrix}X & Y \\ Z & T\end{pmatrix} =\det(XT-ZY), \ \ \ \ \ \text{if} \ \ XZ=ZX,$ and
$\det \begin{pmatrix}X & Y \\ Z & T\end{pmatrix} =\det(TX-YZ), \ \ \ \ \ \text{if} \ \ YT=TY;$ see here for a proof.

So if we have both $XZ=ZX$ and $YT=TY,$ then $\det(XT-ZY)=\det(TX-YZ).$ Changing $Z$ to $-Z$ gives
$\det(XT+ZY)=\det(TX+YZ).$

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