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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?

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  • $\begingroup$ Yes, but I have no idea how to prove, $\endgroup$ – Marko Škorić Sep 4 '18 at 6:04
  • $\begingroup$ Do we have any self-adjoint assumption on A? $\endgroup$ – AlgebraicsAnonymous Sep 4 '18 at 6:07
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    $\begingroup$ i know that $A=Q\Lambda Q^T$ $\endgroup$ – Marko Škorić Sep 4 '18 at 6:09
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    $\begingroup$ yes with positive eigenvalues $\endgroup$ – Marko Škorić Sep 4 '18 at 6:11
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    $\begingroup$ Whan you write that $y^TAy>0$, that's true only if $y\ne0$. $\endgroup$ – Jean-Claude Arbaut Sep 4 '18 at 6:22
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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.

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