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While reading an old book, I came across this theorem:

enter image description here

Neither name nor proof was given, can somebody provide some further information about this throrem? Thanks.

JSCB
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  • Looks like Lagrange interpolation. If you diagonalize $A$ (possible since it has $n$ distinct eigenvalues), it looks like it's a rather direct variant of that. – Daniel Fischer Sep 05 '13 at 09:01
  • Divide $F$ by the characteristic polynomial of $A$. Then $F(A)$ equals the remainder of the division evaluated on $A$. We have the value of the remainder at the points $\lambda_i$, because those are $F(\lambda_i)$. Now interpolate the remainder. – OR. Sep 05 '13 at 09:11

2 Answers2

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This formula is known as Sylvester's formula in the matrix theory literature.

Samrat Mukhopadhyay
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Let $P$ be the characteristic polynomial of $A$. Divide $F$ by $P$ to get $$F=PQ+R$$

where $R$ is a polynomial of degree smaller than that of $P$. By Hamilton-Cayley $P(A)=0$. Then $$F(A)=R(A).$$

All we need is to compute $R$. But we also have that $P(\lambda_i)=0$ for the $n$ different eigenvalues $\lambda_i$. Therefore $R(\lambda_i)=F(\lambda_i)$. Using lagrange interpolation we get that $$R(x)=\sum_{i=1}^{n}F(\lambda_i)G_i(x),$$

where $$G_i(x)=\frac{1}{\prod_{j\neq i}(\lambda_j-\lambda_i)}\prod_{j\neq i}(\lambda_j-x).$$ Evaluating in $A$ the formula for $R$ you get the formula for $F(A)$.

OR.
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