Orthonormal diagonalizable Let: 
$$ A = \begin{pmatrix} 5 &2 & -1 \\ 2 & 2 & 2 \\ -1 & 2 & 5 \end{pmatrix} \in Mat_3(\mathbb{R})$$
1) Show that $0$ and $6$ are eigenvalues for $A$ and find the basis for the corresponding eigenspace. 
2) Explain why $A$ is orthonormal diagonalizable and find an orthonormal basis for $\mathbb{R^3}$ consisting of eigenvectors for $A$ 

1) By solving the characteristic polynomial of $A$ it is possible to show that 0 and 6 are eigenvalues for A. 
$det(A-\lambda \cdot I) = det( \begin{pmatrix} 5-t &2 & -1 \\ 2 & 2-t & 2 \\ -1 & 2 & 5-t \end{pmatrix} = (5-t)(t^2-7t+6)-2(-2t+12)-1(6-t)=-t^3+12t^2-36t = -t(t^2-12t+36) = -t((t-6)(t-6))=-t(t-6)^2$
To solve $-t(t-6)^2 = 0$ we either have $0$ or $6$, which means that 0 and 6 are eigenvalues for $A$. 
The basis for the eigenspace can be found by calculating the null space and we get that: $E_A(0) = N(A) = span(\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix})$ and $E_A(6) = N(A-6I) = span(\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix})$
2) We see that the geometric multiplicity and the algebraic multiplicity are equal to each other, which means that A is diagonalizable. 
How do I go on from here? I know that I have an invertible matrix $P$ consisting of the eigenvectors $P=\begin{pmatrix} 1 & 2 & -1 \\ -2 & 1 & 0 \\ 1 & 0 & -1 \end{pmatrix}$ such that $D = P^{-1}AP$. But this only means that it is diagonalizable and not orthogonal diagonalizable. 

It should be obvious from inspection that the two eigenspaces are orthogonal, so you just need an orthogonal basis for each one. $E_A(0)$ trivially has an orthogonal basis, and one iteration of the Gram-Schmidt process will get you one for $E_A(6)$.
