Eigenvalue of 3x3 matrix with real unknown constants how do i find the eigenvectors of this matrix which has unknown real constants a and b ?
$$
\begin{bmatrix}
a & -\sqrt{2}b & 0 \\
-\sqrt{2}b & 0 & -\sqrt{2}b \\
0 & -\sqrt{2}b & a \\
\end{bmatrix}
$$
Eigenvalues of the matrix are $\lambda_{1}=a, \lambda_{2}=\frac{a-\sqrt{a^2+16b^2}}{2}, \lambda_{3}=\frac{a+\sqrt{a^2+16b^2}}{2}$
When I tried to solve I am getting $\Big[ 1,0,-1 \Big], \Big[1,\tfrac{{4}\sqrt{2}b}{-a+\sqrt{a^2+16b^2}},1\Big], \Big[ 1,\tfrac{4\sqrt{2}b}{-a-\sqrt{a^2+16b^2}},1 \Big]$.
But in mathematica it gives $\Big[ 1,0,-1 \Big], \Big[1,\tfrac{a+\sqrt{a^2+16b^2}}{{2}\sqrt{2}b},1\Big], \Big[ 1,\tfrac{a-\sqrt{a^2+16b^2}}{2\sqrt{2}b},1 \Big]$
I do not understand what is wrong with the calculation

You have that $\frac{4\sqrt2 b}{-a+\sqrt{a^2+16b^2}}=\frac{a+\sqrt{a^2+16b^2}}{2\sqrt2 b}$ and $\frac{4\sqrt2 b}{-a-\sqrt{a^2+16b^2}}=\frac{a-\sqrt{a^2+16b^2}}{2\sqrt2 b}$. So your answer and the output from Mathematica are the same.
