# Expressing the singular values of a 2-by-2 real-valued matrix by the norm of the two columns and the angle between them

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!

• I'm afraid these expressions for $\sigma_1$ and $\sigma_2$ are incorrect, the correct expressions are a bit more lengthy (see the answer box, where I've worked it out). Commented Sep 19, 2021 at 11:49

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.