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Suppose $M$ is an $n \times n$ matrix. Then the family $T(t) := \exp (tM)$ solves the initial value problem

$$T'(t) = M \cdot T(t) \textrm{ and } T(0)=I.$$

Here is my question. Suppose instead of a single matrix $M$ we start with a continuous path $t \mapsto M(t)$ of matrices. Is there always a corresponding path $t \mapsto T(t)$ of invertible matrices such that

$$T'(t) = M(t) \cdot T(t) \textrm{ and } T(0)=I?$$

If the answer to this is yes, here is another question. For just a single matrix $M$, if $M$ is skew-symmetric then the corresponding maps $T(t)$ are unitary. Is this still true if we start with a path $M(t)$ of skew symmetric matrices?

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Yes to both questions. The matrix equation $T'(t)=M(t)T(t)$ is a linear system of differential equations with continuous coefficients ($n\times n$ equations, and as many unknown functions, namely the entries of $T$). By the standard existence-uniqueness theorem for linear ODE, there exists a unique global solution of the initial value problem $T'(t)=M(t)T(t)$, $T(0)=I$.

Now suppose that $M(t)$ is skew symmetric for all $t$. The transpose $T^*$ satisfies $$(T^*)'=(T')^*=(MT)^*=T^*M^*=-T^*M$$ (I omit the argument $t$, but it's understood to be there.) By the product rule (which works for bilinear operations such as matrix multiplications), $$ (T^* T)'=(T^*)'T+T^* T'=-T^*MT+T^*MT=0 $$ Hence $T^*T \equiv T^*(0)T(0)=I$, which means $T$ is unitary at all times.

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