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!