Let $A$ be a matrix with no repeated eigenvalues: $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}.$ Let $p(x)$ and $r(x)$ be two polynomials satisfying $p(\lambda_{i})=r(\lambda_{i}) \text{ for } i = 1, 2, \ldots, n.$ Show that $p(A)=r(A).$
Show matrix polynomials are equal
3 Answers
Hint: What are some of the roots of $p-r$? The Cayley-Hamilton theorem will also be useful.
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0You would have $(p-r)(\lambda_i)=p(\lambda_i)-r(\lambda_i)=0$ for all $i$. So by the remainder theorem, $\prod_{i=1}^n(x-\lambda_i)$ divides $p(x)-r(x)$. But that product is the characteristic polynomial $\chi_A(x)$ of the matrix $A$, and $A$ satisfies this polynomial. So $p(A)-r(A)=\chi_A(A)h(A)=0$ (where h is some other polynomial). That other result sounds similar, but I don't see immediately how to use it. – 2012-06-20
You need to show that $p(A) = r(A)$ or $p(A) - r(A) = 0.$ In other words, the eigenvalues of $A$ are roots of $p(x) - r(x).$ What are the values of $p(x) - r(x)$ at $x = \lambda_i,$ for $i = 1, 2, \ldots, n$?
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0This is (more or less) the same route taken [here](http://books.google.com/books?hl=en&id=S6gpNn1JmbgC&pg=$P$A5). – 2012-07-23
Note that $p(A)$ and $q(A)$ are two matrices which have the same set of eigenvectors as $A$ (say $\{v_{i}\}$) and eigenvalues ($\{p(\lambda_{i})\}$ and $\{q(\lambda_{i})\}$ respectively). As both of them have $N$ linearly independent eignevectors (follows as $A$ has N distinct eigenvalues), they admit a spectral decomposition as $ p(A) = \sum_{i=0}^{N-1}{p(\lambda_{i})v_{i}v_{i}^{H}} \\ q(A) = \sum_{i=0}^{N-1}{q(\lambda_{i})v_{i}v_{i}^{H}} $ It is therefore easy to see that they are equal.