Singular Values of negative eigenvalue? 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?
 A: 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 $\alpha$.
A: The singular values of a matrix $A$ are in fact the square roots of the eigenvalues of $AA^*$ and $A^*A$.
A: 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.  
A: @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:
$$
\mathbf{A}\left( 1, -1 \right) = 
\left[
\begin{array}{cc}
 \frac{1}{2} \left(1+\sqrt{5}\right) & 1 \\
 -1 & \frac{1}{2} \left(1-\sqrt{5}\right) \\
\end{array}
\right]
$$
