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...


  • 2
    $\begingroup$ 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). $\endgroup$ Commented Sep 19, 2021 at 11:49

1 Answer 1


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.

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