Let $p$ be a polynomial and $\|.\|_A$ is a norm defined by $$\|\mathbf{x}\|_A:=\sqrt{\mathbf{x}.A\mathbf{x}},$$ for $\mathbf{x}\in\mathbb{R}^n$ and $A\in\mathbb{R}^{n\times n}$. Let $A$ be a symmetric positive definite matrix with eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_n$. How to prove that

$$\sup_{\|\mathbf{x}\|_A=1}\|p(A)\mathbf{x}\|_A=\max_i|p(\lambda_i)|$$ for any polynomial $p$? I tried this: since $A$ is symmetric then there exists orthonormal basis $\mathbf{v}_1,\mathbf{v}_2,\dots,\mathbf{v}_n$ so that $\mathbf{v}_i$ is eigenvector of $A$. Therefore for any $\mathbf{x}$ there exist scalars $c_1,c_2,\dots,c_n$ such that

$$\mathbf{x}=\sum_{i=1}^nc_i\mathbf{v}_i,$$ and $$\|\mathbf{x}\|_A^2=\sum_{i=1}^nc_i^2\lambda_i.$$


$$\sup_{\|\mathbf{x}\|_A=1}\|p(A)\mathbf{x}\|_A^2=\sup_{\sum_{i=1}^nc_i^2\lambda_i=1}(p(A)\sum_{i=1}^nc_i\mathbf{v}_i).(A\,p(A)\sum_{i=1}^nc_i\mathbf{v}_i).$$ I am stuck in here and don't know what to do next.

  • 3
    $\begingroup$ Hint: Write $p(A) = \sum_{j=0}^m \alpha_j A^j$ for some $m$. Now demonstrate that $p(A) v_i = p(\lambda_i) v_i$. The rest should follow. $\endgroup$ – stochasticboy321 Feb 10 '16 at 22:53
  • $\begingroup$ I only get $$\sup_{\|\mathbf{x}\|_A=1}\|p(A)\mathbf{x}\|_A=\max\limits_{i}{|p(\lambda_i)| \cdot | \lambda _i|}$$ $\endgroup$ – Svetoslav Feb 10 '16 at 23:15
  • $\begingroup$ Why do you get an extra $\lambda_i$? How did you derive this, in detail? $\endgroup$ – Clement C. Feb 10 '16 at 23:22
  • $\begingroup$ $$\sup_{\|\mathbf{x}\|_A=1}\|p(A)\mathbf{x}\|_A^2=\sup_{\sum_{i=1}^nc_i^2\lambda_i=1}(p(A)\sum_{i=1}^nc_i\mathbf{v}_i).(A\,p(A)\sum_{i=1}^nc_i\mathbf{v}_i)=\\ \sup\limits_{\sum_{i=1}^nc_i^2\lambda_i=1}{(\sum\limits_{i}{c_iP(\lambda_i)v_i})\cdot (\sum\limits_{i}{c_iP(\lambda_i)\lambda_iv_i})}=\\ \sup\limits_{\sum_{i=1}^nc_i^2\lambda_i=1}{\sum\limits_{i}{c_i^2P(\lambda_i)^2 \lambda_i}}$$ $\endgroup$ – Svetoslav Feb 10 '16 at 23:41
  • $\begingroup$ Aha, I see. I am sleepy, so I had squared the $\lambda_i$ also. $\endgroup$ – Svetoslav Feb 10 '16 at 23:42

First, notice that, once $A$ is symmetric and positive definite, $\langle x,y\rangle\stackrel{\textrm{def}}{=}x^TAy$ is a scalar product from which the norm you gave in the question is induced. Notice also that the usual euclidean scalar product on $\mathbb{R}^n$ is a special case of this scalar product, namely, with $A$ being the identity matrix.

Let $\mathcal{B}=\{v_1,\ldots,v_n\}$ be an orthonormal basis, not with respect to the preceding scalar product, but with respect to the usual euclidean scalar product, i.e, $\langle v_i,v_j\rangle=v_i^Tv_j=\delta_{ij}$. Now, with respect to this basis $\mathcal{B}$, the matrix $A$ is the diagonal matrix given by (we are considering that $\lambda_i$ is the eigenvalue associated with the eigenvector with same index):

$$A=\begin{pmatrix} \lambda_1 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \lambda_n \end{pmatrix}$$

Thus, writing $x=\sum_i x_iv_i$, we have that:

$$\lVert x\rVert_A^2=x^TAx=\sum_i\sum_jx_i\,\lambda_j x_j\,\langle v_i,v_j \rangle=\sum_ix^2_i\lambda_i$$

Also, we have that the vector $p(A)x$, written in the basis $\mathcal{B}$, is given by (if $p(x)=a_nx^n+\ldots a_1x+a_0$):


Now, combining the above two equalities, we find that:

$$\lVert p(A)x\rVert_A^2=\sum_ip(\lambda_i)^2x_i^2\lambda_i$$

It is easy to see that:

$$\lVert p(A)x\rVert_A^2\leq (\textrm{max}_i\;p(\lambda_i)^2 )\cdot(\sum_ix_i^2\lambda_i)=\textrm{max}_i\;p(\lambda_i)^2$$

where the equality holds by the supremum condition under $\lVert x\rVert_A$. Hence we have:

$$\textrm{sup}_{\lVert x\rVert_A=1}\;\lVert p(A)x\rVert_A^2\leq\textrm{max}_i\;p(\lambda_i)^2$$

For the other inequality, we take a particular $y$ such that $\lVert y\rVert_A=1$, namely, $y=(1/\sqrt{\lambda_k})v_k$ (remember that the eigenvalues are all strictly positve), where $k$ is such that $\textrm{max}_i\;p(\lambda_i)=\lambda_k$. Once we are taking the supremum, we have, for this particular $y$:

$$\textrm{sup}_{\lVert x\rVert_A=1}\;\lVert p(A)x\rVert_A^2\geq p(\lambda_k)^2=\textrm{max}_i\;p(\lambda_i)^2$$

And we are done (just take the square root on both sides).


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.