$P(A) = Q(A)$ if $P(\lambda) = Q(\lambda)$ for all eigenvalues $\lambda$ of $A$

Assuming that characteristic polynomial $\chi_A$ has roots $\lambda_1, \cdots, \lambda_n$ only of odd power, prove that if $P(\lambda_i) = Q(\lambda_i)$ for all $i \in [1,\cdots,n]$ then $P(A) = Q(A)$.

(This is happening in a vector space over an algebraically closed field).

I don't really have much ideas but I think I have to use Jordan normal form or Caley-Hamilton theorem, I am not sure how to use them tho.

Can you please give me a hint on how to start proving this?

EDIT: it looks like my conjecture is false, sorry for the inconvenience.

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2 Answers

Seems to be false. Try $$A=\pmatrix{2&1&1\cr0&4&2\cr0&-2&0\cr}$$ I believe the only eigenvalue is 2, of multipicity 3. Let $P(x)=x$, $Q(x)=2x-2$. Then $P$ and $Q$ agree at 2, but $P(A)\ne Q(A)$.

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An easier example would be $A$ a $3 \times 3$ nilpotent matrix and $P=0$, $Q=x$. – Dan Petersen Nov 21 '11 at 7:30

I think what you mean is this. Suppose $A$ is an $n \times n$ matrix whose characteristic polynomial $\chi_A(x)$ has $n$ distinct roots $\lambda_i$ (which are thus eigenvalues of $A$). If $P$ and $Q$ are polynomials with $P(\lambda_i) = Q(\lambda_i)$ for all $i$, then $P(A) = Q(A)$. This is because $P(x) - Q(x)$ is divisible by $\prod_{i=1}^n (x - \lambda_i) = \chi_A(x)$, and Cayley-Hamilton says $\chi_A(A) = 0$.

In the more general case where root $\lambda_i$ has multiplicity $m_i$, what you want in order to conclude $P(A) = Q(A)$ is not just $P(\lambda_i) = Q(\lambda_i)$ but $P^{(j)}(\lambda_i) = Q^{(j)}(\lambda_i)$ for all nonnegative integers $j < m_i$.

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Except that for that implication you also need characteristic 0. Think about the companion matrix of $X^3$ over a field of characteristic 2, with $P=0$, $Q=X^2$. – Marc van Leeuwen Nov 21 '11 at 10:48
True. OK, the criterion in general is simply that $P(x) - Q(x)$ is divisible (as a polynomial over the field) by $\chi_A(x)$. – Robert Israel Nov 22 '11 at 4:29