# to show $g$ attains maxima and minima

Let $A$ be a symmetric $n\times n$ real matrix and define $G:\mathbb{R}^n\rightarrow \mathbb{R}$ by $G(t)=\langle At,t\rangle$; let $g:S^{n-1}\rightarrow \mathbb{R}$ be the restriction of $G$ to the unit sphere $S^{n-1}$; Could you help me to how to show "If" $g$ attains a maximum or minimum value at a point $t$ then $t$ is an eigen vector for $A$ i.e $\exists \lambda\in\mathbb{R}\text{such that } At=\lambda t$; I have no idea how to proceed for this problem.any light will be helpful

-
Are you trying to show it attains a maximum and a minimum, or to show the maximum and minimum are eigenvectors, or both? Diagonalization is a general technique for matrix problems; have you tried it yet? – Hurkyl Apr 29 '12 at 13:09
thank you, I have edited. – La Belle Noiseuse Apr 29 '12 at 13:12
Do you mean $n\times1$? – nbubis Apr 29 '12 at 13:17
I changed $<At,t>$ to $\langle At,t\rangle$. That is standard. – Michael Hardy Apr 29 '12 at 18:46

Since $A$ is symmetric and real, we can find an orthogonal matrix $P$ such that $X=P^tDP$, where $D$ is diagonal. Since $P$ is orthogonal, $x\in S^{n-1}$ if and only if $Px\in S^{n-1}$, and $$g_A(x)=\langle Ax,x\rangle=\langle P^tDPx,x\rangle=\langle DPx,Px\rangle=g_D(Px).$$ So we have to deal with the case $D$ diagonal, namely $D=\operatorname{Diag}(\lambda_1,\ldots,\lambda_n)$, where $\lambda_1\leq \lambda_2\leq \ldots\leq \lambda_n$. We have, if $g_D$ reaches a maximum at $x$, that $$\max_j\lambda_j\leq g_D(x)=\sum_{j=1}^n\lambda_jx_j^2\leq \max_{j}\lambda_j$$ (the maximum is greater than $g_d(v_j)$, where $v_j$ is an eigenvector for $\lambda_j$) so $g_D(x)=\lambda_n$. Denoting $(v)_k$ the $k$-th component of the vector $v$, we have $(Dx)_k=\lambda_kx_k\leq \lambda_nx_k$, so we have $x_k=0$ if $\lambda_k<\lambda_n$ and $x$ is an eigenvector for $D$. We do the same for the minimum considering $-D$.
could you explain from here: We have, if $g_D$ reaches a maximum at $x$, that $$\max_j\lambda_j\leq g_D(x)=\sum_{j=1}^n\lambda_jx_j^2\leq \max_{j}\lambda_j,$$ so $g_D(x)=\lambda_n$. $(Dx)_k=\lambda_kx_k\leq \lambda_nx_k$, so we have $x_k=0$ if $\lambda_k<\lambda_n$ and $x$ is an eigenvector for $D$. – La Belle Noiseuse Apr 29 '12 at 13:35
what is $(Dx)_k=\lambda_k x_k$ and how $x_k=0$? – La Belle Noiseuse Apr 29 '12 at 13:48
If $x_k\neq 0$ for a $k$ such that $\lambda_k<\lambda_n$ then $g_D(x)$ won't be $\lambda_k$. – Davide Giraudo Apr 29 '12 at 13:52
is $x_k$ is a kth collumn vector ? – La Belle Noiseuse Apr 29 '12 at 14:00