Show that there is a polynomial $p(z)$ of degree at most $k$ such that $e^A = p(A)$.

Question

Let $\dim V<\infty$, and let $A\in L(V)$ be diagonalizable with distinct eigenvalues $\lambda_0, \lambda_1,\ldots,\lambda_k \in \mathbb{N}_0$. Show that there is a polynomial $p(z)$ of degree at most $k$ such that $e^A = p(A)$. (Hint: You may use any information you find about the Vandermonde matrix.)

This is for a Linear Algebra course and I'm not really sure how to start this. I've researched a little bit about the Vandermonde matrix, but for it being my first time seeing what it's about, I really don't know how to apply it to this problem.

It would be helpful if I could be directed in the right direction.

Answer 1

Find a polynomial $p$ of degree (at most) $k$ such that $p(\lambda_i) = e^{\lambda_i}$, for $i=1,...,k$. This is where the Vandermonde matrix comes in.

If $\Lambda $ is diagonal with entries from $\lambda_1,...,\lambda_k$, then $p(\Lambda) = e^\Lambda$.

Note that if $V$ is invertible then for any polynomial $p$ we have $p(V^{-1} AV) = V^{-1} p(A) V$ and $e^{V^{-1} AV} = V^{-1} e^A V $.

Answer 2

This is true more generally for any linear operator. Hint: the set of polynomials in $A$ is a linear subspace of $\operatorname{End}(V)$, hence closed (for the Euclidean topology).