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Suppose $A \in \mathbb{R}^{n \times n}$ is an arbitrary matrix (not necessarily a square matrix), I want to solve the following optimization problem: $$ \min_{x \in \mathbb{R}^{n}} \frac{\langle x,Ax \rangle}{\| x \|_2 \| Ax \|_2}, $$ namely, I want to find a direction $x$ so that the matrix $A$ could rotate the direction to the maximal angle.

A similar question is asked in Maximum angle between a vector $x$ and its linear transformation $A x$, in which, however, the matrix $A$ is symmetric. In this special case, the maximal angle depends on the condition number of $A$. I also wonder whether this conclusion holds in the general case that $A$ is non-symmetric.

  • A potential first step: let $A = U\Sigma V^T$ be a singular value decomposition. Note that $$ \langle x, Ax \rangle = \langle x, U\Sigma V^T x \rangle = \langle U^Tx, A\Sigma V^TU(U^Tx)\rangle. $$ So, setting $W = V^TU$ and $y = U^Tx$, we can rewrite this minimum as $$ \min_{y \in \Bbb R^n} \frac{\langle y, \Sigma W y \rangle}{|y|\cdot |\Sigma W y|} = \min_{y \in \Bbb R^n} \frac{\langle \Sigma^T y, W y \rangle}{|y|\cdot |\Sigma W y|} $$ – Ben Grossmann May 14 '21 at 14:26
  • Cool, but I have tried some similar approaches but got stuck somewhere. For example, we can write the inner-product and vector norm as the form of matrix multiplication: $$ \frac{\langle x,Ax \rangle}{| x |_2 | Ax |_2} = \sqrt{\frac{x^{\top} A^{\top} x x^{\top} A x}{x^{\top} x x^{\top} A^{\top} A x}} = \sqrt{\frac{{\rm tr} (x^{\top} A^{\top} x x^{\top} A x)}{{\rm tr} (x^{\top} x x^{\top} A^{\top} A x)}} = \sqrt{\frac{{\rm tr} (x x^{\top} A^{\top} x x^{\top} A)}{{\rm tr} (x x^{\top} x x^{\top} A^{\top} A)}}. $$ – Zhang Xinyu May 15 '21 at 06:48
  • So the problem can be re-formulated as \begin{align} &\min_{X \in \mathbb{R}^{n \times n}} \frac{{\rm tr} (X A^{\top} X A)}{{\rm tr} (X^2 A^{\top} A)}\ &\mbox{s.t. rank}(X) = 1\ &\qquad \quad~, X \succcurlyeq O. \end{align} But I cannot figure out what conclusion this re-formulation can lead to. – Zhang Xinyu May 15 '21 at 06:55
  • I think you meant "max" not "min" ? – K.defaoite May 27 '21 at 00:48

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