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Let $A\in\mathbb R^{n\times n}$ be positive-definite and $\langle Ax,x\rangle=1$ be the equation of an ellipsoid $M\subset\mathbb R^n$. Use Lagrange multipliers to prove that the greatest distance of a point in $M$ to the origin is the smallest eigenvalue of $A$ raised to $-1/2$.

$A$ is too general, so I tried using that there is an invertible matrix $P$ such that $D\triangleq PAP^{-1}$ is diagonal to solve the problem with $D$ and then go back to the original problem. For simplicity, I'm maximizing the squared distance, so, with $\tilde x$ being the critical distance, the system of equations is $$\left\{\begin{align} \nabla\|x\|^2&=\lambda\nabla\langle Dx,x\rangle\Leftarrow x=\tilde x\\ \langle D\tilde x,\tilde x\rangle&=1 \end{align}\right.\quad.$$ This is equivalent to $$\left\{\begin{align} \nabla\left(\sum_{i=1}^n x_i^2\right)&=\lambda\nabla\left(\sum_{i=1}^n\mu_i x_i^2\right)\Leftarrow x=\tilde x\\ \sum_{i=1}^n\mu_i\tilde x_i^2&=1 \end{align}\right.\quad,$$ where $\mu_1,\ldots,\mu_n$ are the diagonal entries of $D$, which are the eigenvalues of $A$. The first equation gives $$2\tilde x_i=2\lambda\mu_i\tilde x_i : i=1,\ldots,n\quad.$$ If every $\tilde x_i$ is zero, the restriction is impossible, because every $\mu_i$ is positive, thus I get the very questionable conclusion that every $\mu_i$ equals $1/\lambda$. What is going on?

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Since the $\tilde{x}_i$ aren't linearly independent on $M$, you only get that $$\sum_{i=1}^n (1-\lambda \mu_i)x_i = 0.$$ –  Branimir Ćaćić Apr 22 '13 at 19:05

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