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!

  • $\begingroup$ 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

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:


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

  • $\begingroup$ 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

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