If $A$ is a positive semi-definite matrix and $B$ has a arbitary rank, how to determine the the positve definiteness of $B^TAB$ ? ($B^T$ is the transpose matrix of $B$)
I found that this product $B^TAB$ can be positive definite or positive semi-definite. So I think the positve definiteness may depend on the rank of $B$, but I cannot give a rigorous derivation.
Best Answer
Hint:
You know that the matrix $|M$ is positive definite if, for all $x\neq 0$, you have $$x^T Mx > 0$$ and it is positive semidefinite if you replace "$>$" above with "$\geq$".
Now, taking $M=B^TAB$, you can rewrite $x^TMx$ as
$$x^T(B^TAB)x = (x^TB^T) A (Bx) = (Bx)^T A (Bx).$$
Can you continue from here?