1
$\begingroup$

Consider a $4 \times 4$ matrix with eigenvalues $0,1,-2,4$. What are the singular values of this matrix?

I am aware that singular values are just square root of the eigenvalues, but what will be the answer for $-2$ eigenvalue? Is it undefined?

$\endgroup$
2
$\begingroup$

Eigenvalues and singular values of a given square matrix are not related so easily!

We know, e.g., that the maximal singular value bounds from above the magnitude of all eigenvalues. We also know, that since the number of nonzero eigenvalues bounds from below the rank, the number of zero singular values is bounded from above by the number of zero eigenvalues.

This implies (from the given data) that one singular value is $0$ and the maximal singular value is at least $4$.

There are more fancy eigenvalue vs singular value bounds but in no way we can say that there is a particular relation between them without further information. Ideally, if the matrix is normal ($AA^*=A^*A$) then the singular values are simply the absolute value of the eigenvalues.

To illustrate the fact that singular values and eigenvalues can be completely different, you can consult the following example:

Consider $\alpha\neq 0$ and $$ A=\begin{bmatrix}0&\alpha\\1/\alpha&0\end{bmatrix}. $$ It is easy to see that the singular values of $A$ are $\alpha$ and $1/\alpha$ (and thus can be arbitrarily large/small), while the eigenvalues are $-1$ and $1$ (and thus "nice") independently of $\alpha$.

$\endgroup$
  • $\begingroup$ Thank you, what if these eigenvalues were of the resulting matrix $A*A$, does that provides us with a simpler relation to singular values of being the square-root of eigenvalues? $\endgroup$ – BlizzardAlpha Jun 3 '14 at 22:14
  • $\begingroup$ They cannot be. $A^*A$ (if that's what $A*A$ means) is positive semidefinite, $-2$ cannot be its eigenvalue, $-2$ is not nonnegative. Otherwise, yes. $\endgroup$ – Algebraic Pavel Jun 3 '14 at 22:23
  • $\begingroup$ Understood, if we had eigenvalues of $A$ then, can we find eigenvalues of $A*A$ to in return find the singular values? $\endgroup$ – BlizzardAlpha Jun 3 '14 at 22:24
  • $\begingroup$ No. If we know only the eigenvalues of $A$, there's no way to construct the eigenvalues of $A^*A$ unless that $A$ is a priori assumed to be normal (e.g., Hermitian, unitary, skew-Hermitian, etc.). $\endgroup$ – Algebraic Pavel Jun 3 '14 at 22:26
  • $\begingroup$ Ok. Thanks again! $\endgroup$ – BlizzardAlpha Jun 3 '14 at 22:27
1
$\begingroup$

The singular values of a matrix $A$ are in fact the square roots of the eigenvalues of $AA^*$ and $A^*A$.

$\endgroup$
  • 1
    $\begingroup$ But the question does not mention anything about $AA^*$ or $A^*A$. In fact, the given numbers are not eigenvalues of any matrix like that as $AA^*$ and $A^*A$ are both positive semidefinite (and hence they cannot have negative eigenvalues). So, the question is rather about "how the eigenvalues of $A$ are related to singular values of $A$". $\endgroup$ – Algebraic Pavel Jun 3 '14 at 20:53
  • $\begingroup$ OK, so to correct my answer it's right to delete the last sentence? $\endgroup$ – ABC Jun 3 '14 at 20:58
  • $\begingroup$ So, per your definition there should never have been a negative eigenvalue in the question? $\endgroup$ – BlizzardAlpha Jun 3 '14 at 21:39
  • $\begingroup$ I'm not certain now what "definition" are you talking about. I just said that $A^*A$ and $AA^*$ have nonnegative eigenvalues and apparently $-2$ is not nonnegative. $\endgroup$ – Algebraic Pavel Jun 3 '14 at 21:52
  • $\begingroup$ Oh, I know that, we're assuming here that some matrix M*M = A and the eigenvalues are found after that computation. $\endgroup$ – BlizzardAlpha Jun 3 '14 at 21:59
1
$\begingroup$

If the matrix is symmetric then the singular values just absolute values of your eigen values. Because your matrix has a representation

$A=U \Lambda U^T$ , where $\Lambda$ is diagonal of eigen values and $U$ are eigen vectors. Defining singular left signular vector $v=-u$ for negative eigen values, you get the singular value decomposition $A=V |\Lambda| U^T$ However if matrix is not symmetric - there is no direct connection between eigenvalues and singular values.

$\endgroup$
  • $\begingroup$ Understood, thank you! $\endgroup$ – BlizzardAlpha Jun 3 '14 at 21:56
0
$\begingroup$

@Alexander Vigodner provides the punchline $$ \sigma \left( \mathbf{A} \right) = \sqrt{\lambda \left( \mathbf{A} \right)} $$ only when $\mathbf{A} = \mathbf{A}^{*}$.

We can provide a bit more insight with a smaller matrix. What are the $2\times 2$ matrices with the eigenvalue spectrum $\left( 1, 0 \right)$?

The characteristic polynomial is $$ p(\lambda) = \lambda^{2} - \lambda. $$

Symmetric matrices

The symmetric matrices have the form $$ \mathbf{A} = \left[ \begin{array}{cc} a & b \\ b & c \\ \end{array} \right] $$ with the characteristic equation: $$ p(\lambda) = \lambda^{2} - \lambda \, \text{tr } \mathbf{A} + \det \mathbf{A} $$ The matrices can be expressed in terms of one parameter, $b$: $$ \mathbf{A}(b) = \left[ \begin{array}{cc} \frac{1}{2} \left(1 \pm \sqrt{1-4 b^2} \right) & b \\ b & 1 - \frac{1}{2} \left(1 \pm \sqrt{1-4 b^2} \right) \\ \end{array} \right] $$ Example: $$ \mathbf{A}\left( 1 \right) = \left[ \begin{array}{cc} \frac{1}{2} \left(i \sqrt{3}+1\right) & 1 \\ 1 & \frac{1}{2} \left(1-i \sqrt{3}\right) \\ \end{array} \right] $$

Asymmetric matrices

Relax the symmetry requirement $$ \mathbf{A} = \left[ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right] $$

$$ \mathbf{A}(b,c) = \left[ \begin{array}{cc} \frac{1}{2} \left(1 \pm \sqrt{1-4 b c} \right) & b \\ c & 1 - \frac{1}{2} \left(1 \pm \sqrt{1-4 b c} \right) \\ \end{array} \right] $$

Example: $$ \mathbf{A}\left( 1, -1 \right) = \left[ \begin{array}{cc} \frac{1}{2} \left(1+\sqrt{5}\right) & 1 \\ -1 & \frac{1}{2} \left(1-\sqrt{5}\right) \\ \end{array} \right] $$

$\endgroup$

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.