# Supremum equal Max

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

Hence

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

• 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. – stochasticboy321 Feb 10 '16 at 22:53
• I only get $$\sup_{\|\mathbf{x}\|_A=1}\|p(A)\mathbf{x}\|_A=\max\limits_{i}{|p(\lambda_i)| \cdot | \lambda _i|}$$ – Svetoslav Feb 10 '16 at 23:15
• Why do you get an extra $\lambda_i$? How did you derive this, in detail? – Clement C. Feb 10 '16 at 23:22
• $$\sup_{\|\mathbf{x}\|_A=1}\|p(A)\mathbf{x}\|_A^2=\sup_{\sum_{i=1}^nc_i^2\lambda_i=1}(p(A)\sum_{i=1}^nc_i\mathbf{v}_i).(A\,p(A)\sum_{i=1}^nc_i\mathbf{v}_i)=\\ \sup\limits_{\sum_{i=1}^nc_i^2\lambda_i=1}{(\sum\limits_{i}{c_iP(\lambda_i)v_i})\cdot (\sum\limits_{i}{c_iP(\lambda_i)\lambda_iv_i})}=\\ \sup\limits_{\sum_{i=1}^nc_i^2\lambda_i=1}{\sum\limits_{i}{c_i^2P(\lambda_i)^2 \lambda_i}}$$ – Svetoslav Feb 10 '16 at 23:41
• Aha, I see. I am sleepy, so I had squared the $\lambda_i$ also. – 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 $p(x)=a_nx^n+\ldots a_1x+a_0$):

$$p(A)x=a_nA^nx+\ldots+a_1Ax+a_0x=a_n(\sum_i\lambda_i^nx_iv_i)+\ldots+a_1(\sum_i\lambda_ix_i)+a_0(\sum_ix_iv_i)=\sum_i(a_n\lambda_i^n+\ldots+a_1\lambda_i+a_0)x_iv_i=\sum_ip(\lambda_i)x_iv_i$$

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).