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I have a matrix $A \in \mathbb{R}^{d \times 2}$ such that it has full column rank, i.e, $a_1 \in \mathbb{R}^{d} $ and $a_2 \in \mathbb{R}^{d} $ are linearly independent where $A=\begin{bmatrix} a_1 & a_2\end{bmatrix}$. Since $A$ is of rank $2$, I am wondering if there are any closed form expressions for the two singular values of $A$ in terms of $a_1$ and $a_2$.

My attempt: The largest singular value can be characterized by $$ \sigma_1(A)^2=\sup_{\|x\|_2=1} \|A x\|_2^2, $$ and hence can be written as \begin{align*} \sigma_1(A)^2 &= \sup_{\theta} \| a_1 \cdot \cos(\theta) + a_2 \cdot \sin(\theta) \|_2^2,\\ &=\sup_{\theta} \| a_1 \|^2 \cdot \cos^2\theta + \|a_2\|^2 \cdot \sin^2 \theta \ + 2 \langle a_1, a_2 \rangle \sin \theta \cos \theta \\ &=\frac{\|a_1\|^2+\|a_2\|^2}{2} + \sup_\theta \left[ \cos 2\theta \left(\frac{\|a_1\|^2 -\|a_2\|^2}{2} \right) + \sin 2\theta \cdot \langle a_1, a_2 \rangle \right] \\ &=\frac{\|a_1\|^2+\|a_2\|^2}{2} + \sqrt{ \left(\frac{\|a_1\|^2 -\|a_2\|^2}{2} \right)^2 + \langle a_1, a_2 \rangle^2 }. \end{align*} Can a similar derivation be done for $\sigma_2(A)^2$?

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Start with a matrix of column vectors $$ \mathbf{A} = \left[ \begin{array}{c} v_{1} & v_{2} \end{array} \right] \in\mathbb{C}^{m\times 2}. $$ Form the product matrix $$ \begin{align} \mathbf{A}^{*} \mathbf{A} &= % \left[ \begin{array}{c} v_{1}\bar{v}_{1} & v_{1}\bar{v}_{2} \\ v_{2}\bar{v}_{1} & v_{2}\bar{v}_{2} \\ \end{array} \right] \in\mathbb{C}^{2\times 2} \\[3pt] % &= % \left[ \begin{array}{c} m & v_{1}\bar{v}_{2} \\ v_{2}\bar{v}_{1} & m \\ \end{array} \right] % \end{align} $$ The eigenvalues of this matrix are $$ \lambda \left( \mathbf{A}^{*} \mathbf{A} \right) = % \frac{1}{2} \left( m \pm \lvert v_{1}v_{2}^{*} \rvert \right) $$ The singular values are $$ \sigma_{\pm} = \sqrt{\lambda\left( \mathbf{A}^{*} \mathbf{A} \right)} = \sqrt{ \frac{1}{2} \left( m \pm \lvert v_{1}v_{2}^{*} \rvert \right) } $$

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