# Maximum “stretch factor” of linear map

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 $$\mathbb R^n$$.

• In general, the maximum "stretch factor" is the largest singular value of $A$ – 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 $$r$$.