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I was reading the answer of this post but I don't understand why ''the maximum value of $\langle Ax,x\rangle$ on the unit sphere is attained only for an eigenvector for value $\lambda_1$''. I mean, the equality $$\langle Ax,x\rangle = \sum_{i=1}^n\alpha_i^2\lambda_i$$ implies that $\langle Ax,x\rangle\leq \lambda_1$ on the unit sphere, and if $v$ is a unit vector such that $Av=\lambda_1 v$ then $$\langle Av,v\rangle = \langle \lambda_1 v,v\rangle = \lambda_1\langle v,v\rangle =\lambda_1\|v\|^2 = \lambda_1.$$ However, can I show that the only vectors such that $\langle Ax,x\rangle = \lambda_1$ on the unit sphere are eigenvectors for the value $\lambda_1$?

Thanks in advance.

Lord Vader
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  • See my answer here, especially equation $(*)$. And a slightly related post you may find helpful here. – peek-a-boo Aug 26 '23 at 02:05
  • Say we have $\lambda_1=\lambda_2>\lambda_3\ge\cdots\ge\lambda_n$. Recall that $|\alpha|=1$ because the matrix $(e_1\cdots e_n)$ is unitary and $|x|=1$ by definition. So, if there exists $i>2$ with $\alpha_i\neq 0$ we have $\langle Ax,x\rangle=\sum_i|\alpha_i|^2\lambda_i<\sum_i|\alpha_i|^2\lambda_1=\lambda_1$. Hence, the maximum can only be achieved if $\alpha_i=0$ for $i>2$. But then $x$ is a linear combination of the eigenvectors for $\lambda_1$, so $x$ is an eigenvector for $\lambda_1$. The general case for multiple maximizers $\lambda_i$ with arbitrary indices should now also be clear. – Matija Aug 28 '23 at 17:45
  • Remark: I think that in the answer in the linked post it should also be $|\alpha_i|^2$ since we consider complex $\alpha_i\in\mathbb C$, so $\langle\alpha,\alpha\rangle=\sum_i|\alpha_i|^2$ (using the complex conjugate for the first argument to the Hermitian form). – Matija Aug 28 '23 at 17:48

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