formulas for exact values of singular values in low dimension? Are there formulas for the singular values of a real matrix in low dimension, i.e. for a $2 \times 2$ matrix or a $2 \times 3$ matrix? 
Any comment is welcome.
 A: While there surely are a handful of formulas which would fit this question (e.g. here, here or here), they should all be based on the following nice "closed" form: Given any complex $2\times 2$ matrix $A$ its singular values are
$$
\boxed{\frac{ \sqrt{ \|A\|_2^2+2|\det(A)| }\pm\sqrt{ \|A\|_2^2-2|\det(A)| } }2}\tag{1}
$$
where $\|A\|_2^2=a_{11}^2+a_{12}^2+a_{21}^2+a_{22}^2$ is the (square of the) Frobenius norm of $A$. One can see this as follows:
If $A=U\Sigma V$ is any singular value decomposition of $A$, that is, $U,V\in\mathbb C^{2\times 2}$ are unitary and $\Sigma=\operatorname{diag}(s_1,s_2)$ with $s_1,s_2$ the singular values of $A$, then
$$
\|A\|_2^2=\|U\Sigma V\|_2^2=\|\Sigma\|_2^2=s_1^2+s_2^2
$$
and
$$
|\det(A)|=|\det(U\Sigma V)|=|\det(U)||\det(\Sigma)||\det(V)|=|\det(\Sigma)|=s_1s_2\,.
$$
Here we used that the Frobenius norm is unitarily equivalent and that all unitary matrices have eigenvalues on the unit circle (so the absolute value of their determinant is $1$). Now all that is left is to re-write $s_1\pm s_2$ as
$$
\sqrt{(s_1\pm s_2)^2}=\sqrt{s_1^2+s_2^2\pm 2s_1s_2}=\sqrt{\|A\|_2^2\pm 2|\det(A)|}
$$
and take their sum (resp. difference) to get $2s_1,2s_2$ which reproduces (1). This has two immediate consequences:

*

*The trace norm of $A$ gets a nice "closed" form:
$$
\|A\|_1^2=(s_1+s_2)^2=(\sqrt{ \|A\|_2^2+2|\det(A)| })^2= \|A\|_2^2+2|\det(A)| 
$$

*If $A$ is a $2\times 3$ complex matrix then (1) implies the, admittedly not as elegant, formula
$$
\frac{ \sqrt{ \|A\|_2^2+2\sqrt{|\det(AA^*)|} }\pm\sqrt{ \|A\|_2^2-2\sqrt{|\det(AA^*)|} } }2\tag{2}
$$
for its singular values. This follows from the fact that given any complex matrix $A$ its singular values coïncide with the singular values of $\sqrt{AA^*},\sqrt{A^*A}$ (readily verified, again by means of the singular value decomposition). If $A\in\mathbb C^{3\times 2}$ then (2) holds, but with $A^*A$ instead of $AA^*$.

