I have the differential equation $$\partial/\partial t A(t)=B(t)A(t)$$ on the interval $[0,1]$ with initial condittion $A(0)=A_0$. Here $A$ and $B$ are $n\times n$ real matrices depending smoothly on $t$.
I want to prove that the solution $A(t)$ has constant rank on $[0,1]$.
A result by Magnus* states that the solution has the expression $$A(t)=\exp(\Omega(t))A(0)$$
for a suitable $\Omega(t)$. The thing is that this experession holds only on a neighborhood of $0$. Some extra conditions on $B$ would gurantee that in fact such expression holds in $[0,1]$, but in my case $B$ is quite general.
Is there any other result that could help me in this case?
*Magnus, W. (1954), On the exponential solution of differential equations for a linear operator. Comm. Pure Appl. Math., 7: 649–673.
Use that $$ \det(A(t))=\exp\left(\int_0^t\text{trace}(B(s))ds)\right)·\det(A(0)) $$ to conclude that if $A(0)$ has full rank then also $A(t)$ has full rank.
Replace linearly dependent columns of $A(0)$ by independent ones until the new matrix $\tilde A(0)$ has full rank. $A(0)$ and $\tilde A(0)$ still coincide in $k=\text{rank}(A(0))$ columns, and as the columns integrate independently, these same columns still coincide in $A(t)$ and $\tilde A(t)$ and are thus linearly independent. It is easy to show that the dependent columns of $A(0)$ stay dependent with exactly the same linear combinations in $A(t)$.