3
$\begingroup$

First off, the question: $A=QLQ'$ given that $Q$ is orthogonal, and $L$ is a real diagonal matrix, is the following true: $A$ is a real symmetric matrix.

My attempts

The easy part is to recognize $Q' = Q^T$. Therefore we have a similarity equation between A and L. This question basically boils down to whether all similar matrices to L is real symmetric. I know that L has all real eigenvalues, and so does A. However, could there exist not-real-symmetric matrices that also have all real eigenvalues?

So first I tried to think of easy contradictory examples but they all came out to be real symmetric. Take this 1 for example:

Take L to be $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$, I know that the characteristic polynomial must be the same, the rank must be the same, and the geometric multiplicities of its eigenvalues(consequently) is also the same.

The charpoly(L) is $(1-\lambda)(-1-\lambda)$, which is $\lambda^2-1$.

Take A to be $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. The charpoly(A) is $\lambda^2 - 1$ as well. Its rank is also 2.

But argh, this doesn't help! This is a real symmetric!

So I turn to math SO for help - am I missing an obvious example here? If this is actually true, then what's the proof behind it? Thanks!

$\endgroup$
3
  • 1
    $\begingroup$ Hint: use the property of transpose: $(AB)^T = B^T A^T$. It is much simpler than you think. $\endgroup$
    – Malkoun
    Sep 30, 2016 at 7:21
  • $\begingroup$ Let me make an educated guess here: Since $L^T = L$, we can transpose both sides of the equation, giving $A^T = (QLQ^{-1})^T$, which gives us $A^T = Q^TLQ$. We know that $Q^-1=Q^T$, then we can see that $QA^T = LQ$. We can see that $A^T$ is also similar to L, which we can now say $A^T$ is similar to $A$, which... makes it real symmetric? I think somewhere in there I went wrong. $\endgroup$
    – OneRaynyDay
    Sep 30, 2016 at 7:25
  • $\begingroup$ Oh wait I am stupid. $\endgroup$
    – OneRaynyDay
    Sep 30, 2016 at 7:31

1 Answer 1

Reset to default
4
$\begingroup$

Oops. Thanks to @Malkoun for giving me the gratuitous hint:

$A = QLQ^T$

$A^T = (QLQ^T)^T = (Q^T)^TL^TQ^T = QL^TQ^T$. We know $L^T=L$, therefore $A^T=QLQ^T$. Therefore $A^T=A$

$\endgroup$

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.