4
$\begingroup$

We know that symmetric matrices are orthogonally diagonalizable and have real eigenvalues. Is the converse true? Does a matrix with real eigenvalues have to be symmetric?

A class of symmetric matrices, the positive definite matrices, have positive real eigenvalues. Is the converse true? Does a matrix with positive real eigenvalues have to be symmetric, positive-definite?

I think the answer to all this is "no", but I just wanted to confirm.

Thanks,

$\endgroup$
2
  • 2
    $\begingroup$ Consider [ 1 1; 0 1] $\endgroup$ Jul 2, 2015 at 3:34
  • $\begingroup$ no, try to find example on upper trianguler matrix $\endgroup$ Jul 2, 2015 at 3:34

4 Answers 4

5
$\begingroup$

Is the matrix

$$\begin{pmatrix}1&1\\0&1\end{pmatrix}$$

symmetric? It has only one positive eigenvalue of multiplicity two.

$\endgroup$
7
  • 1
    $\begingroup$ got it - thanks @Timbuc. $\endgroup$
    – User001
    Jul 2, 2015 at 3:45
  • $\begingroup$ Hi @Timbuc, if we have a symmetric matrix, then it is positive-definite (or positive-semidefinite) if and only if its eigenvalues are positive (or non-negative). I guess this is somewhat useful? Perhaps computing eigenvalues to prove a symmetric matrix is PSD may be easier than showing that $x^tAx$ >0 for all x. What do you think? Thanks, $\endgroup$
    – User001
    Jul 2, 2015 at 22:23
  • 1
    $\begingroup$ What is "somewhat useful"? To have positive definite matrices, or to know they are such by their eigenvalues? Both things are pretty important in certain things, dpeneding on your knowledge and7or to what you want to apply this stuff. $\endgroup$
    – Timbuc
    Jul 2, 2015 at 22:26
  • 1
    $\begingroup$ Either you calculate eigenvalues ... or ... you can also show all the main minors are positive (or the odd minors are negative in case of NSD matrices). $\endgroup$
    – Timbuc
    Jul 2, 2015 at 22:28
  • 1
    $\begingroup$ A quadratic form has to fulfill some basic requirements, and if the matrix isn't symmetric it doesn't. Try it with low dimensions, it's easy. $\endgroup$
    – Timbuc
    Jul 2, 2015 at 22:43
3
$\begingroup$

consider on this $$ \begin{bmatrix} 1 & 2 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 3 \\ \end{bmatrix} $$

$\endgroup$
1
  • $\begingroup$ You are welcome :) $\endgroup$ Jul 2, 2015 at 3:47
3
$\begingroup$

The first question seems to be handled quite well by other people but I just want to add one more thing.

Positive-definiteness always starts from symmetry. So the converse can never be true.

That being said, it's only when we have the guarantee that the matrix is symmetric that we can conclude a matrix is positive-definite.

$\endgroup$
1
  • 1
    $\begingroup$ Ok, awesome - thanks so much for adding this answer, @daeyounglim. So, positive eigenvalues does not imply positive-definiteness. $\endgroup$
    – User001
    Jul 2, 2015 at 17:06
2
$\begingroup$

No.

The simplest example is any upper triangular matrix whose diagonal entries are real:

$$\begin{pmatrix} 1 & 1-i & 2 \\ 0 & 2 & 3i \\ 0 & 0 & 3 \end{pmatrix}$$

has complex entries in the non-diagonal spots but its eigenvalues are $1,2,3$ and it is certainly not symmetric!

$\endgroup$
1
  • $\begingroup$ ok got it - thanks, @xoque55. $\endgroup$
    – User001
    Jul 2, 2015 at 3:46

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.