Why doesn't the matrix $\begin{pmatrix} 10 & -20 & -20 \\ -20 & 40 & 40 \\ -20 & 40 & 40 \end{pmatrix}$ have an orthogonal set of eigenvectors?

Question

Why doesn't the matrix $$\begin{pmatrix} 10 & -20 & -20 \\ -20 & 40 & 40 \\ -20 & 40 & 40 \end{pmatrix}$$ have an orthogonal set of eigenvectors?

I have used a matrix calculator to diagonalize the matrix but I see that the eigenvectors:

$$\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} -1/2 \\ 1 \\ 1 \end{pmatrix}$$

are not orthogonal. According the spectral theorem, this matrix should have an orthogonal basis of eigenvectors. What is wrong?

Answer 1

This matrix is a real symmetric matrix. Therefore it does have an orthogonal basis of eigenvectors. How about replacing your first eigenvector by $\pmatrix{0\\1\\-1}$ and your second by $\pmatrix{4\\1\\1}$?

Answer 2

You need to do Gram-Schmidt on the first two eigenvectors to get orthogonal vectors. $$ \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} -1/2 \\ 1 \\ 1 \end{pmatrix}$$

They are eigenvectors associated with the same eigenvalue $\lambda =0$

Answer 3

To complement the other answers, we have a rank-$1$ matrix

$$\begin{bmatrix} 10 & -20 & -20\\ -20 & 40 & 40\\ -20 & 40 & 40\end{bmatrix} = 10 \begin{bmatrix} 1\\-2\\-2\end{bmatrix} \begin{bmatrix} 1\\-2\\-2\end{bmatrix}^\top = 90 \underbrace{\left( \frac 13 \begin{bmatrix} 1\\-2\\-2\end{bmatrix} \right)}_{=: \mathrm v_1} \left( \frac 13 \begin{bmatrix} 1\\-2\\-2\end{bmatrix} \right)^\top = 90 \, \mathrm v_1 \mathrm v_1^\top$$

Since the 2nd and 3rd entries of $\mathrm v_1$ are equal, let

$$\mathrm v_2 := \frac{1}{\sqrt 2} \begin{bmatrix} 0\\ 1\\-1\end{bmatrix}$$

Lastly, we would like to find a unit vector $\mathrm v_3$ that is orthogonal to both $\mathrm v_1$ and $\mathrm v_2$. Hence, we are interested in the null space of

$$\begin{bmatrix} 1 & -2 & -2\\ 0 & 1 & -1\end{bmatrix}$$

which is spanned by $\begin{bmatrix} 4 & 1 & 1\end{bmatrix}^\top$. Hence,

$$\mathrm v_3 := \frac{1}{\sqrt{18}} \begin{bmatrix} 4\\ 1\\ 1\end{bmatrix}$$

Using SymPy, we can verify that vectors $\mathrm v_1$, $\mathrm v_2$ and $\mathrm v_3$ do indeed form an orthonormal basis:

>>> from sympy import *
>>> V = Matrix([[1,0,4],[-2,1,1],[-2,-1,1]]) * diag(3,sqrt(2),sqrt(18))**-1
>>> V
Matrix([
[ 1/3,          0, 2*sqrt(2)/3],
[-2/3,  sqrt(2)/2,   sqrt(2)/6],
[-2/3, -sqrt(2)/2,   sqrt(2)/6]])
>>> V.T * V
Matrix([
[1, 0, 0],
[0, 1, 0],
[0, 0, 1]])