Eigenvectors of $3 \times 3$ symmetric matrix For the following symmetric matrix, I want to compute the square root.
$$A=\begin{bmatrix}
D  & 0  & 0 \\  
0  & D/\sin^2(\theta)  & -2D \cos(\theta)/\sin^2(\theta)\\
 0  & -2D \cos(\theta)/\sin^2(\theta)  & D/\sin^2(\theta)
\end{bmatrix}$$
where $D > 0$ and $\theta \in[0,\pi]$. To compute the square root, I need to find its eigenvalues and eigenvectors. The eigenvalues I found are 
$$\lambda_1=D \qquad \qquad \lambda_2=D/\sin^2(\theta)$$ 
but confused on how to obtain the eigenvectors. Any help would be appreciated.
 A: You should go to get the basis vectors for the nullspaces of the matrices $A-\lambda I$, as the eigenvectors lay on those subspaces. This happens due to the fact that $(A-\lambda I)\nu = 0,\forall\lambda$. Using this fact, for the first eigenvalue:
$A-\lambda I =\left( \begin{array}{ccc}
0 & 0 & 0 \\
0 & \frac{D}{\sin^2\theta}-D & -2D\frac{\cos\theta}{\sin^2\theta} \\
0 & -2D\frac{\cos\theta}{\sin^2\theta} & \frac{D}{\sin^2\theta}-D \end{array} \right)$, so now you should just compute a basis for the nullspace of this matrix, and that basis will form a group of the eignevectors of the matrix A (then you should compute the values for the others to get the complete eigenbasis. For this $\lambda_1$, it's easy to see that $\nu_1=(1,0,0)$ is the basis of the nullspace for that matrix, so $\nu_1$ is an eigenvector for A.
A: Solving the characteristic equation:
$$
0=det|B-\lambda I|=\begin{vmatrix}
D-\lambda & 0 & 0\\
0 & \frac{D}{\sin^2\theta}-\lambda & -2D\frac{\cos\theta}{\sin^2\theta} \\
0 & -2D\frac{\cos\theta}{\sin^2\theta} & \frac{D}{\sin^2\theta}-\lambda
\end{vmatrix}=(D-\lambda)\Big[\big(\frac{D}{\sin^2\theta}-\lambda\big)^2 -4D^2\frac{\cos^2\theta}{\sin^4\theta}\Big].
$$
So, the eigenvalues are:
$$
\lambda_1=D, \quad \lambda_2=\frac{D(1+2\cos\theta)}{\sin^2\theta}, \quad \lambda_3=\frac{D(1-2\cos\theta)}{\sin^2\theta}.
$$
and the eigenvectors are:
$$
\mathbf{v_1}=(1,0,0). \quad \mathbf{v_2}=(0,1,-1), \quad \mathbf{v_3}=(0,1,1).
$$
