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Let $x \in \mathbb{R}^{n \times1 }, A \in \mathbb{R}^{n \times n}$. We are looking for maximal and minimal value of $f(x) = x^TAx$ with constraint $g(x)=x^Tx$. We get that $\nabla g= 2x, \nabla f = Ax + A^Tx$ so we have to solve a system $$\left\{\begin{array}{l}(A+A^T)x = 2\lambda x \\ x^Tx = 1 \end{array} \right.$$

If $A$ was symmetric then we would get eigenvectors, but what with this more general case?

Rodrigo de Azevedo
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Naah
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  • The general case does not matter at all, because only the symmetric part of $A$ contributes to the quadratic form. Hence, you should assume $A$ is symmetric. – Rodrigo de Azevedo Jan 26 '20 at 18:24
  • Can you elaborate on why only the symmetric part of $A$ contribute to the quadratic form? – Naah Jan 26 '20 at 18:37
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    Take a look at this. Note that $$\begin{bmatrix} x_1\ x_2\end{bmatrix}^\top \begin{bmatrix} 0 & -1\ 1 & 0\end{bmatrix} \begin{bmatrix} x_1\ x_2\end{bmatrix} = - x_1 x_2 + x_2 x_1 = 0$$ – Rodrigo de Azevedo Jan 26 '20 at 18:39
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    That answers my question. Thank you. – Naah Jan 26 '20 at 18:42

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