Let $A \in M_n (\mathbb R)$ be diagonalizable matrix, let $\lambda_1, \dots, \lambda_n$ be its eigenvalues. I want to know if the maximum "stretch factor" of $A$ is the maximum of its unsigned eigenvalues i.e. $$\sup_{\Vert X \Vert = 1} \Vert AX\Vert = \max_{1\leq i\leq n} |\lambda_i|$$ Any help would be appreciated.

$\Vert \cdot \Vert$ will be the euclidean norm of $\mathbb R^n$.

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    $\begingroup$ In general, the maximum "stretch factor" is the largest singular value of $A$ $\endgroup$ – Omnomnomnom Apr 25 at 23:48

Yes, as long as the matrix is symmetric (or Hermitian in the complex case).

For a counter-example in the non-symmetric case, take the matrix $A=\begin{bmatrix}1&1\\0&2\end{bmatrix}$ is diagonalizable: $$\begin{bmatrix}1&-1\\0&2\end{bmatrix}=P\begin{bmatrix}1&0\\0&2\end{bmatrix}P^{-1}$$ where $P=\begin{bmatrix}1&1\\0&1\end{bmatrix}$, $P^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}$. The eigenvalues of $A$ are $1$ and $2$, but $2$ is not the maximum "stretch factor": the vector $\begin{bmatrix}0\\1\end{bmatrix}$ is taken to $\begin{bmatrix}1\\2\end{bmatrix}$, which has norm $\sqrt{5}\simeq 2.236$.

On the other hand, suppose that $A$ is a symmetric $n\times n$ matrix. Let $\lambda_1,\ldots,\lambda_n$ be the its eigenvalues (counting multiplicity). Then there exists a orthonormal basis $v_1,\ldots,v_n$ of $\mathbb{R}^n$, where $v_i$ is an eigenvector associated to $\lambda_i$.

Any vector $x$ of $\mathbb{R}^n$ may be written uniquelly as $x=\sum x_iv_i$. Then $Ax=\sum x_i\lambda_i v_i$, so since the $v_i$ are orthonormal, $$\Vert Ax\Vert^2=\sum |x_i\lambda_i|^2\leq\max_i|\lambda_i|^2\sum_i|x_i|^2$$ i.e., $\Vert Ax\Vert\leq\max|\lambda_i|\Vert x\Vert$, so the "stretch factor" of $A$ is at most $\max|\lambda_i|$. It is attained at the eigenvector $v_i$ associated to the eigenvalue of largest modulus.


In general, the maximum stretch factor is the largest singular value $\sigma_1$ of $A$, i.e. the square root of the largest eigenvalue $\lambda_1$ of $A^TA$. Note that if you assume $A$ is symmetric, then $\sigma_1$ is exactly $\sqrt{\lambda^2} = |\lambda|$ where $\lambda$ is the largest eigenvalue of $A$.

This comes from our knowledge of quadratic forms. Since maximizing $\|Ax\|$ is equivalently maximizing the square root of $\|Ax\|^2 = (Ax)\cdot(Ax) = x^TA^TAx$, we’re really maximizing the square root of the quadratic form given by $A^TA$. The restriction $\|x\|=1$ tells us that this is exactly the largest singular value of $A$, $\sigma_1$.

Note that if $x$ is the eigenvector for $A^TA$ corresponding to $\lambda_1$, then

$$\sqrt{x^TA^TAx} = \sqrt{x^T(\lambda_1x)} = \sqrt{\lambda_1\|x\|^2} = \sigma_1\|x\|.$$

Thus, if you want to maximize the length of vectors on a sphere of any fixed radius, i.e. constraining yourself to $\|x\| = r$ for a fixed $r$, you just scale by $r$.


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