# 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.

• Apply the Gram Schmidt process to your eigenvectors to end up with an orthonormal eigenbasis. – Ben Grossmann May 31 '18 at 12:06
• Or, if you prefer to work by inspection, note that $E_A(6)$ is spanned by the orthogonal vectors $(-1,0,1)$ and $(1,1,1)$ – Ben Grossmann May 31 '18 at 12:08
• Thanks! I think I will just use the Gram Schmidt process – Mads Jeppesen May 31 '18 at 12:39

Note that $A$ is symetric hence by the spectral theorem it's orthogonal diagonalizable.
• The spectral theorem in my textbook states that: "Let $L: V \to V$ be a self-adjoint operator, then there exists a orthonormal basis for $V$ consisting of eigenvectors for $L$. In particular $L$ is orthonormal diagonalizable." So because $A$ is symmetric, then the corresponding linear operator $L_A$ is self-adjoint and therefore it is possible to apply the spectral theorem? – Mads Jeppesen May 31 '18 at 12:38
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 $E_A(6)$.
• Why is it trivially that $E_A(0)$ ha an orthogonal basis? Couldn't I just use the Gram Schmidt process on all three eigenvectors to form an orthogonal basis? And would I not need to conclude beforehand that $A$ is orthonormal diagonalizable? – Simbörg Jun 1 '18 at 11:11
• @Simbörg $E_A(0)$ is one-dimensional. Any basis for it is vacuously orthogonal. – amd Jun 1 '18 at 17:58