Let $x\in \mathbb{R}$ and the Matrix $M_x$ = $\begin{pmatrix} 1 & x+1 & 0 & 0 \\ 0 & x & x-1 & 0 \\ 0 & x-1 & x & 0 \\ 1 & 0 & x+1 & x \end{pmatrix}$ determine x such that M is diagonalizable. I know that the eigenvalues of this matrix are $\lambda_1 = 1 , \lambda_2 = x , \lambda_3 =2x-1$. But how does that help now?
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$\begingroup$ Is there one which has guaranteed algebraic multiplicity $\ge 2$? That helps cutting off some cases. $\endgroup$ – user228113 Feb 9 '17 at 23:03
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$\begingroup$ You have the eigenvalues. If you have proven these are the only eigenvalues, compute (brute force) the dimension of the null space of $M_x - \lambda_i$ for each $i$. What does the sum of the dimensions of the eigenspaces have to be in order for a matrix to be diagonalizable? What values of x would make that sum what you want it to be? $\endgroup$ – Christian Feb 9 '17 at 23:04
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Hint:
You have to determine which one is a double root (this may depend on the value of $x$). The matrix is diagonalisable if and only if the corresponding eigenspace has dimension $2$.
