# 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. – Omnomnomnom 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 $(1,1,1)$ – Omnomnomnom 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
• @Simbörg In general, you can’t simply throw the entire eigenbasis into G-S because the output might no longer consist entirely of eigenvectors. You need to take care to keep the eigenspaces separate. In this case you can get away with it because the matrix is symmetric, so you know that the eigenspaces are orthogonal. – amd Jun 1 '18 at 18:03
• @Simbörg If eigenspaces are not orthogonal, you have no hope of finding an orthogonal eigenbasis for the entire space. If, on the other hand, all of the eigenspaces are orthogonal, then you can find an orthogonal eigenbasis by applying G-S individually to each eigenspace or to the entire mess as a whole, but why do the extra work that the latter entails? – amd Jun 2 '18 at 0:27
• @Simbörg Do you understand what it means for subspaces of a vector space to be orthogonal to each other? – amd Jun 2 '18 at 0:28