Entries of exp(tA) given A's polynomials

[undone hazel]

If we're given the characteristic and minimal polynomial of a (real) matrix A, how can we use that to show what form the entries of exp(tA) has?

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[undone hazel]

The question is a specific example of this, and we're given the entries should be in a specific span, and the generators of the span are that of a linear ODE's solution space with the same characteristic polynomial, but idk how to actually prove it is such

My idea was to consider the system dx/dt = Ax and the linear ODE L[y]=0 where L=p_A(d/dt), cause then if they give the same solutions (?) then it's done

[undone hazel]

Bumping my question smile

[kindred ore]

If you know the matrix $A$, the minimal polynomial $m$ of $A$, its distinct eigenvalues $\lambda_1,\cdots,\lambda_s$ and the indices of nilpotence $(n_i)$ of the $(\lambda_i)_i$, then you can calculate the Hermite interpolating polynomial $p$ defined by:
$degree(m)=n_1+\cdots+n_s$, and $p^{(j)}(\lambda_i)=\exp(\lambda_i),j=0,\cdots, n_i-1; i=1,\cdots,s$.
Then $\exp(A)=p(A)$.
Unfortunately, in general when $n\geq 5$, we don't know how to calculate the $(\lambda_i)$. See for example
$A=\begin{pmatrix} 0& 0& 0&0& -1\ 1& 0& 0&0& 1\ 0& 1&0& 0&0\ 0& 0& 1& 0&0\ 0& 0& 0& 1& 0\end{pmatrix}$.
In other words, one doesn't know how to calculate the exponential of a matrix of dimension $\geq 5$ except if we build it so that this is possible !!

For a general $A$, the only thing you can say is that the entries of $e^{tA}$ are linear combinations of $t^{a_i}\exp(\lambda_i t)$, where $a_i\leq n_i-1;i=1\cdots s$.

[opal chasm]

arohi

[undone hazel]

We do in general know how to find exp(tA) provided the char polynomial splits into linear, it's a consequence of SN-decomp. The problem is the char poly given doesnt split into linear, it has irred. quadratics

But your very last sentence is precisely what I was asking about how to do

how do we know the entries are such?

[kindred ore]

Your first sentence is wrong. If you randomly choose a matrix -of dimension $\geq 5$-, then its characteristic polynomial is irreducible -then its roots are simple- and its Galois group is $S_n$; then it is non-solvable by radicals.
About your question -line 4-, it suffices to calculate $\exp(t(\lambda I+J))$, where $J$ is a nilpotent Jordan block.

[opal chasm]

arohi

[undone hazel]

exp(t(S+N))=exp(tS)exp(tN) by SN decomp, and each factor is as easy to compute as diagonalizing $S$

[opal chasm]

Omegabet_

[undone hazel]

No clue what else you're saying as you're assuming knowledge

If you need the explicit example instead of the general idea, here you go

[undone hazel]

+close