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The matrix $$\pmatrix{100\sqrt{2}&x&0\\-x&0&-x\\0&x&100\sqrt{2}},\quad x>0$$ have two equal eigenvalues. How can I find $x$? What I tried is this. If $\lambda_1$ is doubly degenerate and $\lambda_2$ the third eigenvalue, then the characteristic equation is $(\lambda-\lambda_1)^2(\lambda-\lambda_2)=0$. Also $2\lambda_1+\lambda_2=200\sqrt{2},\quad \lambda_1^2\lambda_2=200\sqrt{2}x^2$. I do not know how to proceed from here.

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The eigenvalues are $100\sqrt{2}$ and $\sqrt{2}\left[50\pm\sqrt{2500-x^2}\right]$. There are double eigenvalues if $x\in\{-50,0,50\}$.

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Just calculate the characteristic polynomial of the matrix:

\begin{align}\begin{vmatrix} 100\sqrt{2} - \lambda&x&0\\ -x&-\lambda&-x\\ 0&x&100\sqrt{2}-\lambda \end{vmatrix} &= (100\sqrt{2} - \lambda)\begin{vmatrix} -\lambda&-x\\ x&100\sqrt{2}-\lambda \end{vmatrix} +x\begin{vmatrix} x&0\\ x&100\sqrt{2}-\lambda \end{vmatrix}\\ &= (100\sqrt{2}-\lambda)(\lambda^2-100\sqrt{2}\lambda + x^2)+2x^2(100\sqrt{2} - \lambda)\\ &= \lambda(100\sqrt{2} - \lambda)^2 + 2x^2(100\sqrt{2} - \lambda)\\ &= \lambda(100\sqrt{2} - \lambda)^2 + 2x^2(100\sqrt{2} - \lambda)\\ &= (100\sqrt{2} - \lambda)(\lambda(100\sqrt{2} - \lambda) + 2x^2)\\ &= (100\sqrt{2} - \lambda)(-\lambda^2 +100\sqrt{2}\lambda + 2x^2)\\ \end{align}

Thus, $100\sqrt{2}$ is certainly a root.

The solutions of $-\lambda^2 +100\sqrt{2}\lambda + 2x^2 = 0$ are:

$$\lambda \in \left\{50\sqrt{2}-\sqrt{2}\sqrt{2500-x^2}, 50\sqrt{2}+\sqrt{2}\sqrt{2500-x^2}\right\}$$

So, for the characteristic polynomial to have a double root, it must be either $2500-x^2 = 0$, or $50\sqrt{2}\pm\sqrt{2}\sqrt{2500-x^2} = 100\sqrt{2}$.

Since $\sqrt{2500} = 50$, the first condition gives $x = \pm 50$ and the second one gives $x = 0$.

Thus, $x \in \{0, -50, 50\}$.

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