Let $A\in M_3$ some arbitrary positive definite matrices and
B=$\begin{bmatrix}
1& 1& 0& 2\\
2& -1& 1& 2\\
1 & 0& 0& 1
\end{bmatrix}$.
Conclude connection between eigenvalue of matrices $B^{T}AB$, how much it has zero eigenvalues, how much it has positive and negative eigenvalues.
I find that $rankB=3$ then if I use some eigenvector $x\in \mathbb R^4$ such that $x\not=0$ then exist some eigenvalue $\lambda$ that $B^{T}ABx=\lambda x$, if I multiply both side with $x^T$ I get this $x^TB^{T}ABx=\lambda x^Tx$
then if $Bx=y$ and if I write $x^TB^{T}ABx=\lambda x^Tx$ like this $y^TAy=\lambda x^Tx$ then since $y^TAy>0$, $x^Tx>0$ so then $\lambda>0$ but If I pick some eigenvector from $kerB$ then $\lambda=0$ so it can be one zero and three positive eigenvalue, what you think?
Best Answer
From $A = Q \Lambda Q^\top$ with $\Lambda$ a diagonal matrix we know that $A$ is self-adjoint. Thus, we know that all eigenvalues are real. Moreover, $B^\top A B$ is self-adjoint: $$ (B^\top A B)^\top = B^\top (B^\top A)^\top = B^\top A^\top B = B^\top A B $$ Thus, the eigenvalues of $B^\top A B$ are also real. Moreover, since $A$ is positive definite, $B^\top A B$ is positive semidefinite and we know that the eigenvalues are all nonnegative, as you showed. This would be the formal argument.
Your remaining analysis is correct up to typos.