Question: What value in $d$ makes the matrix
$$ \begin{bmatrix} 0 & d & 0 \\ 1 & 0 & d \\ d & 1 & 0 \\ \end{bmatrix} $$ diagonalisable over the field $\mathbb{R}$?
My workings:
$=[(-1)(d)*(1- \lambda) (0- \lambda) - d^2]$
$=-d x( \lambda^2 - \lambda - d^2)$
$=(d)\lambda^2 + d(\lambda) +d^3$
After throwing it into $b^2-4ac$ :
$=d^2 - 4d^4$
The answer I got is $1/2$ as a root, but I am not sure if I am correct or if it's diagonalisable.
Edit : Apology for the terrible formatting, I am very new to stackexchange and I could not find a clue on how to type Lambda out. A clue on that as well would be awesome, thank you!
The characteristic polynomial of your matrix is $-\lambda^3+2\lambda d+d^3$, whose discriminant is $32d^3-27d^6$. So, the discriminant is $0$ if and only if $d=0$ or $d=\frac23\sqrt[3]4$. For every value of $d$ between those two numbers, your matrix will have $3$ distinct real eigenvalues and therefore will be diagonalizable over $\mathbb R$ if $d>\frac23\sqrt[3]4$ or $d<0$, the discriminant will be negative, and therefore your matrix will not be diagonalizable over $\mathbb R$. Now, check by hand the cases in which $d=0$ and $d=\frac23\sqrt[3]4$.