I'm looking for an elegant way to show the following claim.


**Claim:** Let $m\_1, m\_2 \in \mathbb{R}^2$ be the two columns of matrix $M \in \mathbb{R}^{(2 \times 2)}$. The singular values of the matrix are $\sigma\_1 = \sqrt{\|m\_1\|\_2 + \left|\cos{\measuredangle \left( m\_1, m\_2 \right)}\right| \|m\_2\|\_2}$ and $\sigma\_2 = \sqrt{\sin{\measuredangle \left( m\_1, m\_2 \right)} \|m\_2\|\_2}$ so that $\sigma\_1 > \sigma\_2$.


What I have so far is only pretty messy first and second derivations of $max\_{a:\|a\|\_2=1} \|Ma\|\_2$ and $min\_{a:\|a\|\_2=1} \|Ma\|\_2$ w.r.t. the first coordinate of $a$, denoted by $a\_1$, after replacing the second coordinate of $a$, which we denote by $a\_2$, with $\sqrt{1-a\_1^2}$.


Another try that I had is to start by claiming that $\exists P$ s.t. $PM = \begin{bmatrix} \|m\_1\|\_2 & \|m\_2\|\_2 \cos a \\ 0 & \|m\_2\|\_2 \sin a\end{bmatrix}$, where $a := \angle(m\_1, m\_2) = \frac{m\_1^T m\_2}{\|m\_1\|\_2 \|m\_2\|\_2}$. Then, that $PM$ and $M$ have the same singular values, so we could just compute them for $PM$. Unfortunately, I'm getting an unexpected expression so I guess I've missed something...


Tnx!



Expressing the singular values of a 2-by-2 real-valued matrix by the norm of the two columns and the angle between them
Since $\sigma\_1^2=\lambda\_+$ and $\sigma\_2^2=\lambda\_-$ are the two eigenvalues of the symmetric matrix product $MM^t$, we have $\lambda\_++\lambda\_-={\rm tr}\,MM^t=\|m\_1\|^2+\|m\_2\|^2$. Hence we may write WLOG
$$\lambda\_\pm=\tfrac{1}{2}\left(\|m\_1\|^2+\|m\_2\|^2\right)\pm\Delta.$$
To determine $\Delta$ we equate $$\lambda\_+\lambda\_-={\rm det}\,MM^t=(m\_1\times m\_2)^2=\|m\_1\|^2\|m\_2\|^2\sin^2\measuredangle \left( m\_1, m\_2 \right),$$
hence
$$\lambda\_\pm=\tfrac{1}{2}\left(\|m\_1\|^2+\|m\_2\|^2\right)\pm\sqrt{\tfrac{1}{4}\left(\|m\_1\|^2+\|m\_2\|^2\right)^2-\|m\_1\|^2\|m\_2\|^2\sin^2\measuredangle \left( m\_1, m\_2 \right)}.$$
If the two vectors $m\_1$ and $m\_2$ have the same norm $\|m\|$, this simplifies to
$$\lambda\_\pm=\|m\|^2\bigl(1\pm\cos \measuredangle \left( m\_1, m\_2 \right)\bigr),\;\;\text{if}\;\;\|m\_1\|=\|m\_2\|\equiv\|m\|.$$


This differs from the result in the OP. Let me check, as an example, 
$$M=\begin{pmatrix}
1&1\\
0&1
\end{pmatrix},\;\;\sigma\_1^2=\tfrac{3}{2}+\tfrac{1}{2}\sqrt 5,\;\;\sigma\_2^2=\tfrac{3}{2}-\tfrac{1}{2}\sqrt 5.$$
Since the angle between the vectors $m\_1={1\choose 0}$ and $m\_2={1\choose 1}$ is $\pi/4$, the formula in the OP would give $\sigma\_1^2=1+\sqrt 2$ and $\sigma\_2^2=\sqrt 2$, which is incorrect.
