1
$\begingroup$

struggling with a question from homework and would appreciate some assistance.

Let $A, B \in M_2^{\mathbb{R}}$ be defined as follows:

$$A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}, B = \begin{pmatrix} -4 & 1 \\ 1 & 1 \end{pmatrix} $$

Find a regular $P \in M_2^\mathbb{R}$ such that $P^tAP=I$ and $P^tBP$ is diagonal.

I'm familiar with the theorem that states that a simultaneous diagonlization exists if one of the matrices is positive-definite (or negative-definite).

I've found an invertible P such that $P^tAP=I$: $$P=\begin{pmatrix}1\over\sqrt{2}&-1\over\sqrt{2}\\0&\sqrt{2}\end{pmatrix}$$

But $P^tBP$ is not diagonal.

Would appreciate any assistance / hints. Thanks!

$\endgroup$
2
  • $\begingroup$ Are you aware of the concept of congruent matrices? $\endgroup$ – Git Gud Apr 1 '13 at 9:05
  • $\begingroup$ I am, and also of the invariance of $\rho(A)$ across congruence, Sylvester's law of inertia and such. $\endgroup$ – iravid Apr 1 '13 at 9:19
2
$\begingroup$

Assuming you can find a matrix $S$ such that $S^TAS=I$ (such a matrix does exist because $A$ is positive definite - I think this is called Sylvester's theorem - or rather a consequence of it), consider the matrix $V=S^TBS$. You should be able to prove that $V$ is symmetric.

Since $V$ is symmetric, there exist an orthogonal matrix $Q$ and a diagonal matrix $D$ such that $Q^TVQ=D$.

Now let $P=SQ$.You should be able to prove this one works:

$P^TAP=Q^TS^TASQ=Q^TIQ=I \wedge P^TBP=Q^TS^TBSQ=Q^TVQ=D$

Note that this method works for any positive definite matrix $A$ and any symmetric matrix $B$.

$\endgroup$
2
  • $\begingroup$ Thanks! The proof that SQ satisfies the requirements is indeed trivial. I'll now try to find the matrix itself. $\endgroup$ – iravid Apr 1 '13 at 9:44
  • $\begingroup$ @iravid No problem. $\endgroup$ – Git Gud Apr 1 '13 at 9:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.