Connection betweeen exponential of a matrix and Hermite polynomials While doing a course on differential equations I came across the following question: Assume A is a real or complex $n$ x $n$ matrix with characteristic polynomial $p(\lambda) = (\lambda - \lambda_{1})^{m_{1}}...(\lambda - \lambda_{k})^{m_{k}}$ with $\sum m_{i} = n$. I'm asked to prove that if Q(\lambda) is the Hermite interpolation polynomial of degree $n-1$ (i.e, such that $Q^{(j)}(\lambda_{i}) = e^{\lambda_{i}} \forall j = 0,...,m_{i}-1, i = 1,...,k$), then $e^A = Q(A)$. In case you don't know, the exponential of a matrix is defined as its "Taylor series". I was thinking maybe of applying Cayley-Hamilton theorem since it kind of relates the two concepts, but I'm not sure how to do it.
 A: We put the matrix ${\bf A} $ in Jordan normal form
$$
{\bf A} = {\bf V}\,{\bf J}\,{\bf V}^{\, - \,{\bf 1}} \quad  \to \quad \exp \left( {\bf A} \right) = {\bf V}\,\exp \left( {\,{\bf J}} \right)\,{\bf V}^{\, - \,{\bf 1}} 
$$
what we can do since the total algebraic multiplicity equal the dimension.
Then each block of ${\bf J} $  can be written as
$$
{\bf J}_{\,k}  = {\bf \Lambda }_{\,k}  + {\bf E} = \lambda _{\,k} {\bf I} + {\bf E}\quad  \to \quad \exp \left( {{\bf J}_{\,k} } \right) = \exp \left( {{\bf \Lambda }_{\,k}  + {\bf E}} \right) = \exp \left( {\bf E} \right)\,\exp \left( {{\bf \Lambda }_{\,k} } \right) = e^{\,\lambda _{\,k} } \exp \left( {\bf E} \right)
$$
where each matrix has dimension $m_k \times m_k$, and where $ {\bf E} $ is the shift matrix, i.e the matrix with the off-diagonal elements $=1$ and the remaining null.
From what we are given about $Q(x)$ follows that it can be developed around $x=\lambda_k$ as
$$
\left\{ \matrix{
  Q\left( {\lambda _{\,k} } \right) = e^{\,\lambda _{\,k} }  \hfill \cr 
  Q^{\left( 1 \right)} \left( {\lambda _{\,k} } \right) = e^{\,\lambda _{\,k} }  \hfill \cr 
  \quad  \vdots  \hfill \cr 
  Q^{\left( {\,m_{\,k}  - 1} \right)} \left( {\lambda _{\,k} } \right) = e^{\,\lambda _{\,k} }  \hfill \cr}  \right.\quad  \to \quad Q\left( x \right) = e^{\,\lambda _{\,k} } \left( {1 + {{\left( {x - \lambda _{\,k} } \right)} \over {1!}} +  \cdots  + {{\left( {x - \lambda _{\,k} } \right)\,^{m_{\,k}  - 1} } \over {\left( {\,m_{\,k}  - 1} \right)!}}} \right) + O\left( {\left( {x - \lambda _{\,k} } \right)\,^{m_{\,k} } } \right)
$$
So
$$
\eqalign{
  & Q\left( {{\bf J}_{\,k} } \right) = e^{\,\lambda _{\,k} } \left( {1 + {{\left( {{\bf J}_{\,k}  - \lambda _{\,k} {\bf I}} \right)} \over {1!}} +  \cdots  + {{\left( {{\bf J}_{\,k}  - \lambda _{\,k} {\bf I}} \right)\,^{m_{\,k}  - 1} } \over {\left( {\,m_{\,k}  - 1} \right)!}}} \right) + O\left( {\left( {{\bf J}_{\,k}  - \lambda _{\,k} } \right)\,^{m_{\,k} } } \right) =   \cr 
  &  = e^{\,\lambda _{\,k} } \left( {1 + {{\bf E} \over {1!}} +  \cdots  + {{{\bf E}\,^{m_{\,k}  - 1} } \over {\left( {\,m_{\,k}  - 1} \right)!}}} \right) + O\left( {{\bf E}\,^{m_{\,k} } } \right) =   \cr 
  &  = e^{\,\lambda _{\,k} } \exp \left( {\bf E} \right) = \exp \left( {{\bf J}_{\,k} } \right) \cr} 
$$
since $\bf E$ is $m_k$ nil-potent.
Finally, a block-matrix raised to any (integer) exponent, is a matrix with each block raised at that exponent.
So a polynomial applied to the whole matrix, turns into a matrix composed with the polynomial applied to each block.
$$
Q\left( {\bf J} \right) = \exp \left( {\bf J} \right)
$$
And the polynomial applied to similar matrices gives similar polynomials, thus
$$
Q({\bf A}) = {\bf V}\,Q\left( {\bf J} \right)\,{\bf V}^{\, - \,{\bf 1}}  = {\bf V}\,\exp \left( {\,{\bf J}} \right)\,{\bf V}^{\, - \,{\bf 1}}  = \exp \left( {\bf A} \right)
$$
Q.E.D.
