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I couldn't get very far with this at all, I was only able to find the trivial solutions.

$$ a=1/2 \mathbf{v^T}(A-\alpha)\mathbf{v} $$ if we differentiate this and by knowing that $a^T=a$ we get: $$ \frac{\partial a}{\partial v_i}=\begin{bmatrix} \delta_{1i} \\ \delta_{2i} \\ \vdots \\ \delta_{ni} \end{bmatrix} (A-\alpha)\mathbf{v}=\sum^n_k(A_{ik}-\alpha)v_k $$

However apart from the trivial solutions: $A=I\alpha$ and $\mathbf{v=0}$ I can't seem to solve it.

I'm quite sure I differentiated it correctly. Any help would be grateful. Thank you

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    $\begingroup$ You write $A-\alpha$, but one is a matrix and the other is a scalar, no? $\endgroup$ – Bobson Dugnutt May 29 '17 at 10:00
  • $\begingroup$ Yes, how would you want me to differentiate between them $\endgroup$ – Toby Peterken May 29 '17 at 14:41
  • $\begingroup$ It's not that, it is just that I'm not sure you can "factor out" like that, as the operation inside the parenthesis is not defined. $\endgroup$ – Bobson Dugnutt May 29 '17 at 15:07
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    $\begingroup$ I suppose that by $A-\alpha$ you mean $A-\alpha I$? $\endgroup$ – amd May 29 '17 at 19:01
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You’re close. Convert the system of equations ${\partial a\over\partial v_i}=0$ that you derived into matrix form: $$(\mathbf A-\alpha\mathbf I)\mathbf v=0.$$ It should now be obvious that this is satisfied by any $\mathbf v\in\ker(\mathbf A-\alpha\mathbf I)$. This kernel is nontrivial when $\alpha$ is an eigenvalue of $\mathbf A$.

This might’ve been easier to spot had you computed the differential of $a$ directly: $$a=\frac12(\mathbf v^T\mathbf A\mathbf v+\alpha|\mathbf v|^2)=\frac12(\mathbf v^T\mathbf A\mathbf v+\alpha\mathbf v^T\mathbf v)=\frac12\mathbf v^T(\mathbf A-\alpha\mathbf I)\mathbf v$$ so $$\operatorname{d}a=\mathbf v^T(\mathbf A-\alpha\mathbf I).$$ $\mathbf A-\alpha\mathbf I$ is symmetric, so the condition $\operatorname{d}a=0$ is equivalent to $(\mathbf A-\alpha\mathbf I)\mathbf v=0$.

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