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I have a quick question.

Let $A,B$ be square matrix. And let $p(t)=a_kt^k+\cdots+a_1t+a_0)$ be a polynomial.

Prove that if $A$ and $B$ are similar then $p(A)$ and $p(B)$ are similar.

Proof: If $A$ and $B$ are similar there exists a non singular matrix $P$ such that $B=P^{-1}AP$

We then place the matrix A in the polynomial equation then

$P(A)=a_kA^k+\cdots+a_1A^1+a_0I$ So then we must show for a nonsingular matrix $P$

$P^{-1}P(A), \quad P= P(B)$

So from the left

\begin{align} & =P^{-1}(a_kA^k+\cdots+a_2A^2+A_1A+a_0I)P \\[01pt] & =a_kP^{-1}A^{k}P+\cdots+a_1P^{-1}A^1P+a_0P^{-1}IP \end{align}

and we for any natural number $K$ then$B^K=(P^{-1}AP)^k$

so then

\begin{align} & =a_kB^k+\cdots+a_1 B+a_0 I \\[10pt] & =P(B) \end{align}

so $P(A)$ and $P(B)$ are similar.

Michael Hardy
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Fernando Martinez
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    A caveat: similar polynomials are equal, since polynomial multiplication is commutative – Asinomás Sep 04 '16 at 17:09
  • @CarryonSmiling The title's use of polynomial is a misnomer: it doesn't refer to the polynomials being similar, but the images of the two matrices under the same polynomial. – Erick Wong Sep 04 '16 at 17:13
  • Since $A$ and $B$ are similar, $B=P^{-1}AP$ for some nonsingular $P$, by definition. Moreover, $$ B^{n}=\left(P^{-1}AP\right)^{n}=\underbrace{P^{-1}AP\cdots P^{-1}AP}{n\text{ times}}=P^{-1}A^{n}P. $$ Therefore, $$ p(B)=\sum{n=0}^{k}a_{n}B^{n}=\sum_{n=0}^{k}a_{n}P^{-1}A^{n}P=P^{-1}\left(\sum_{n=0}^{k}a_{n}A^{n}\right)P=P^{-1} p(A) P. $$ – parsiad Sep 04 '16 at 17:16

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Yes, that seems about right. The key point to note is that $(P^{-1}AP)^k =P^{-1}A^kP $.

Personally I find it easier to write the other way round. We have $B=P^{-1}AP $, so $p (B) = a_kB^k +\dots +a_1 B + a_0 I = a_k(P^{-1}AP)^k +\dots +a_1 P^{-1}AP + a_0 I = a_kP^{-1}A^kP + \dots +a_1 P^{-1}AP + a_0 P^{-1}IP = P^{-1}(a_kA^k + \dots + a_1A + a_0I )P = P^{-1}p(A)P.$

Michael Hardy
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