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Multiply a vector $\vec x$ by matrix $A$ from the left. $|\vec x|$ is the length of $\vec x$. $|A\vec x|$ is the length of $A\vec x$.

I try to simplify $\displaystyle \frac{|Ax|}{|x|}$, but failed. I can only make it to $\displaystyle \frac{\sqrt{x^TA^2x}}{\sqrt{x^Tx}}$.

If it can't be simplified when $A$ is a general matrix, what about it's symmetric?

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    For $x=e_j$, $Ax=A_j$, the $j^\text{th}$ column of $A$. For $|x|=1$, $x$ is a unit direction vector and $Ax$ is a projection of $A$. Hence $|Ax|$ lies between $0$ (inclusive if $A$ has nonempty kernel or nonzero nullity) and some maximum value. For $A$ square, this value will be the magnitude of the largest eigenvalue, $|\lambda_i|$, of $A$.2012-02-24

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Assume that $A$ is symmetric, and let $\lambda_1,...,\lambda_n$ its eigenvalues and $b_1,...,b_n$ be an orthonormal basis of eigenvectors. Then $x$ can be written as $x_1 b_1 + ... + x_n b_n$. Then $\left ( \frac{|Ax|}{|x|} \right ) ^2 = \frac{x_1^2 \lambda_1^2 + ... + x_n^2 \lambda_n^2}{x_1^2 + ... + x_n^2}$ so it is a weighted sum of squares of eigenvalues of the matrix $A$, where the weights are determined by the coordinates in the eigenvector basis.

If $A$ is not symmetric, $\lambda_1^2,...,\lambda_n^2$ have to be replaced by the eigenvalues of $A^t A$ (and the basis is the eigenvector base of $A^t A$).