Can we have $A(t)\exp(\int_{t_0}^t A(s)\,ds )=\left(\exp(\int_{t_0}^t A(s)\,ds )\right)A(t)$, when $\,A(t)\,$ is symmetric?

Question

$A(t)$ is a symmetric matrix for $t\in [t_0,a]$. Can we say $$A(t)\exp\left(\int_{t_0}^t A(s)ds \right)=\exp\left(\int_{t_0}^t A(s)\,ds \right) A(t)$$

My attempt: it is easy to see that $B(s)=\int_{t_0}^t A(s)\,ds $ is symetric . symetric matrices are diagnosable and also $\exp(B(t))$ is diagnosable. We know that two diagnosable matrices are commutative iff they have same eigenvectors.

Can someone help me.

Answer 1

EDIT 1: unfortunately, I wrote too fast.

Let $B(t)=\int_{t_0}^tA(s)ds$. Clearly $BB'=B'B$ implies that $Ae^B=e^BA$, that is $B'e^B=e^BB'$ .

Is the converse true ? Consider $(Be^B-e^BB)'=0=B'e^B+B(e^B)'-(e^B)'B-e^BB'=[B,(e^B)']$. Indeed we do not know if $(e^B)'=e^BB'=B'e^B$. Then I cannot give a positive answer to my question.

Yet, if $A$ is symmetric, then, in general, the equality $B'e^B=e^BB'$ is not satisfied ; indeed it suffices to consider $B(t)=\begin{pmatrix}t^2&t\\t&t^4\end{pmatrix}$ and to show (by a numeric calculation) that $A(1)e^{B(1)}\not= e^{B(1)}A(1)$ (cf. the Omnomnomnom's comment).

EDIT 2. More precisely $\begin{pmatrix}2&1\\1&4\end{pmatrix}\exp(\begin{pmatrix}1&1\\1&1\end{pmatrix})-\exp(\begin{pmatrix}1&1\\1&1\end{pmatrix})\begin{pmatrix}2&1\\1&4\end{pmatrix}=\begin{pmatrix}0&-e^2+1\\e^2-1&0\end{pmatrix}$.

Answer 2

Observe that $$ \exp\left(\int_{t_0}^t A(s)\,ds\right)=\lim_{n\to\infty}\sum_{k=0}^n \frac{\left(\int_{t_0}^t A(s)\,ds\right)^k}{k!}. $$ Hence it suffices to show that $$ A(t)\left(\int_{t_0}^t A(s)\,ds\right)^k=\left(\int_{t_0}^t A(s)\,ds\right)^kA(t), $$ for all $k$. And clearly it suffices to show that $$ A(t)\left(\int_{t_0}^t A(s)\,ds\right)=\left(\int_{t_0}^t A(s)\,ds\right)A(t) $$ Next note that $$ \int_{t_0}^t A(s)\,ds=\lim_{n\to\infty}\frac{t-t_0}{n}\sum_{k=1}^n A\left(t_0+\frac{k}{n}(t-t_0)\right), $$ and hence it suffices to show that $$ A(t)A(s)=A(s)A(t), \tag{1}. $$

If we further assume that $$ A(t)\int_{\tau}^t A(s)\,ds=\left(\int_{\tau}^t A(s)\,ds\right)A(t), \tag{2} $$ for all $\tau,t\in[a,b]$, then differentiating with respect to $\tau$ we obtain that $$ A(t)A(\tau)=A(\tau)A(t), \quad\text{for all $t,\tau\in[a,b]$} \tag{3} $$ Thus $(2)$ and $(3)$ are equivalent.