3
$\begingroup$

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 $\mathbb R^n$.

$\endgroup$
  • 1
    $\begingroup$ In general, the maximum "stretch factor" is the largest singular value of $A$ $\endgroup$ – Omnomnomnom Apr 25 at 23:48
0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $r$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.