I have the matrix equation $$ M'(t) = A(t) M(t)$$ with the initial condition $M(0) = I$, with $I$ the identity matrix, and where both $M$ and $A$ are $3\times 3$ matrices, $A(t)$ is real and can be partitioned as $A(t)=B(t)+D(t)$ where B(t) is a skew-symmetric matrix and $D(t)$ is a diagonal matrix.
It seems that $M(t=T)$ has eigenvalues all with modulus less than one. How can I prove this?
It is not true that $M$ has all eigenvalues smaller than one. Take, for example, $D=0$ and $A = B$ a constant anti-symmetric matrix. Then $$ M(t)=e^{tA}. $$ If all eigenvalues have modulus smaller than one, the determinant must have modulus smaller than one as well, but $$ \det{M}=\det{e^{tA}}=e^{t\textbf{trace}(A)}=e^0=1. $$ If $D$ is a multiple of the identity matrix, the determinant can be made even larger than one.