0
$\begingroup$

Theorem: Prove that any two regular symmetric $\;2\times 2\;$ real matrices are either simultaneously diagonalizable or else they are congruent to each other.

Now, my work: If $\;A,B\in Sym_2(\Bbb R\;$) are regular then each of them either has one single non-zero eigenvalue of (algebraic) multiplicity two, or else each has two different non-zero eigenvalues.

I also know that if two matrices are are diagonalizable, then they're simultaneously diagonalizable iff they commute.

But I also know two square matrices are congruent iff they have the same rank and the same signature, so I have this problem: say we have

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

Both matrices above are regular and symmetric and thus diagonalizable, but

1) They're not simultaneously diagonalizable since $\;AB\neq BA\;$ , and also

2) They are not congruent since certainly $\;\text{rk}\, A= \text{rk}\, B =2\;$, yet their signature is different: $\;\sigma(A)=(2,0)\;,\;\;\sigma(B)=(1,1)\;$ (this I got after I calculated both matrices' eigenvalues: the first one has two positive ones whereas the second one has one positive and one negative)

So either the theorem above is wrong, or else I'm misunderstanding something badly. Any help will be appreciated.

$\endgroup$

1 Answer 1

1
$\begingroup$

I think by "simultaneously diagonalizable", it means "simultaneously diagonalizable by congruence", not "simultaneously diagonalizable by similarity transform". If $A$ and $B$ are not congruent to each other, they have different signatures. Since they are invertible $2\times2$ real symmetric matrices, it follows that one of them must be either positive definite or negative definite. Hence they are simultaneously diagonalizable by congruence.

$\endgroup$
4
  • $\begingroup$ Ah, this makes sense, but: how is it then possible to find the invertible $\;P\;$ s.t. $\;P^tAP\;,\;\;P^tBP\;$ are both diagonal? Say, in my example? This could really help me. Thanks a lot.+1 $\endgroup$
    – Timbuc
    May 20, 2014 at 9:17
  • 1
    $\begingroup$ @Timbuc Assume that $A$ is positive (or negative) definite. First, find an invertible $P$ such that $P^TAP=I$ (or $-I$). At this point, $P^TBP$ is not yet diagonal, but it is a real symmetric matrix. So ... $\endgroup$
    – user1551
    May 20, 2014 at 10:06
  • $\begingroup$ Yes, thanks @user1551: positive/negative definiteness is a necessary condition, but if both quadratic forms aren't positive/negative definite how can be proved/disproved they're simultaneously diagonalizable? $\endgroup$
    – Timbuc
    May 20, 2014 at 14:14
  • 1
    $\begingroup$ @Timbuc They are not. In this case both of them are indefinite and hence they have the same signatures (because they are nonsingular and 2x2). Recall that there're two "either-or" clauses in the question. $\endgroup$
    – user1551
    May 20, 2014 at 22:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .