Easiest way to compute singular values of matrix
Let $A\in GL_2(\mathbb{R})$ be an invertible matrix. I know $A$ has a singular value decomposition $A=U\Sigma V^T$ where $U$ and $V$ are orthogonal matrices and $\Sigma$ is diagonal. I call "singular values of $A$" the entries of $\Sigma$. Which is the easiest (meaning less computations where the entries of $A$ are complicated) to compute the singular values of $A$? To me the obvious one is to compute the eigenvalues of $AA^T$, but I was wondering if there is an easier one.
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$\begingroup$You indeed need to compute the eigenvalues of $A^TA$. However, we can compute these eigenvalues without computing $A^TA$.
In particular, it suffices to note that $$ \lambda_1 + \lambda_2 = \operatorname{trace}(A^TA) = \sum_{i,j=1}^2 a_{ij}^2\\ \lambda_1 \lambda_2 = \det(A^TA) = \det(A)^2 $$ If we call the first number $a$ and the second number $b$, then the eigenvalues are the solutions $\lambda$ to $$ \lambda^2 - a\lambda + b = 0 $$ By the quadratic formula, we find that $$ \lambda_1 = s_1^2 = \frac{a + \sqrt{a^2 - 4b}}{2}, \qquad \lambda_2 = s_2^2 = \frac{a - \sqrt{a^2 - 4b}}{2} $$
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