When is a matrix integral rank-deficient?

[Kreizhn]

Homework Statement

Let $X,A:\mathbb R \to M_n(\mathbb C)$ be two smooth functions mapping an interval $[t_0,t_1]$ to the set of nxn complex matrices. Assuming that X has a well defined inverse, determine when
$$\int_{t_0}^{t_1} (X^{-1}AX)(t) \ dt$$
is rank deficient.

The Attempt at a Solution

Since X and A are both square matrices of the same dimension, the resulting matrix will also be square and hence the matrix will be rank deficient precisely when it's singular. So normally I would do something fun like take a determinant in which case $\det(X^{-1}AX) = \det(X^{-1})\det(A)\det(X) = \det(A)$. However, the presence of this integral concerns me.

For example, I can imagine it's possible that the product matrix $X^{-1} A X$ could have components that are say, never all simultaneously zero (I know this isn't enough for singularity but bear with me), but that the integral could result in a zero matrix (which is certainly singular)!

So it seems to me that I really must take the integral into question. Does anybody have any ideas?

 

[Dick]

Take the case n=1. So you've got 1x1 matrices. Then 'rank deficient' just means zero. And the integral turns into the integral of A(t) from t1 to t2. If you set A(t)=e^(it) and integrate around the unit circle, you get zero even though A(t) is never zero. I'm not sure what kind of condition you would to put on A or X or what sort of conclusion you would be looking for?

 

[Kreizhn]

Yeah, I'm not sure what sort of conclusion I'm looking for either. It's possible that nothing more than this can be said, so I'm really just looking for anything.

 

[Dick]

Yeah, I'm not sure what sort of conclusion I'm looking for either. It's possible that nothing more than this can be said, so I'm really just looking for anything.

Well, you could try to prove if A is singular along the curve then the integral is singular. But I'm 99.9% sure that's not true either without even trying to make a counterexample. What are you really trying to prove?