Singular Values of negative eigenvalue?

Question

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?

score 4 · Accepted Answer · answered Jun 03 '14 at 21:51

4

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 $\alpha$.

answered Jun 03 '14 at 21:51

Algebraic Pavel

22,928

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? – BlizzardAlpha Jun 03 '14 at 22:14
They cannot be. $A^A$ (if that's what $AA$ means) is positive semidefinite, $-2$ cannot be its eigenvalue, $-2$ is not nonnegative. Otherwise, yes. – Algebraic Pavel Jun 03 '14 at 22:23
Understood, if we had eigenvalues of $A$ then, can we find eigenvalues of $A*A$ to in return find the singular values? – BlizzardAlpha Jun 03 '14 at 22:24
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.). – Algebraic Pavel Jun 03 '14 at 22:26
Ok. Thanks again! – BlizzardAlpha Jun 03 '14 at 22:27
You're welcome. – Algebraic Pavel Jun 03 '14 at 22:28

ABC · Answer 2 · 2014-06-03T21:47:55.347

1

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

edited Jun 03 '14 at 21:47

answered Jun 03 '14 at 20:34

ABC

997

1

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 $A$". – Algebraic Pavel Jun 03 '14 at 20:53
OK, so to correct my answer it's right to delete the last sentence? – ABC Jun 03 '14 at 20:58
So, per your definition there should never have been a negative eigenvalue in the question? – BlizzardAlpha Jun 03 '14 at 21:39
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. – Algebraic Pavel Jun 03 '14 at 21:52
Oh, I know that, we're assuming here that some matrix M*M = A and the eigenvalues are found after that computation. – BlizzardAlpha Jun 03 '14 at 21:59

score 1 · Answer 3 · answered Jun 03 '14 at 21:54

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.

Understood, thank you! – BlizzardAlpha Jun 03 '14 at 21:56 — BlizzardAlpha, Jun 03 '14 at 21:56

score 0 · Answer 4 · answered Apr 03 '17 at 22:21

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

Singular Values of negative eigenvalue?

4 Answers4