Trouble understanding orthonormal vectors.

Question

I am pretty sure the definition of orthonormal matrix is a matrix whos columns contain vectors that are orthogonal to each other and all of length 1.

In that case I cant understand 17:04 of this lecture:https://www.youtube.com/watch?v=uNsCkP9mgRk&t=974s.

Professor Gilbert strang writes that $Q =\frac{1}{3}\begin{pmatrix}1&-2& 2\\2&-1&-2\\2&2&1\end{pmatrix}$ and says the columns are orthonormal to each other. I dont see how since the dot product of the vectors isn't 0. Is this a mistake in the lecture or am I missing something.

Answer 1

Your matrix is orthogonal:

• every column has norm $1$, since the norm of each column is $\sqrt{\frac49+\frac49+\frac19}=1$;
• the dot product of every two distinct columns is $0$; for instance, the dot product of the first two columns is$$\frac19\bigl(1\times(-2)+2\times(-1)+2\times2\bigr)=0.$$

Answer 2

Firstly, the columns each have length 1. For example, the first column which is $(1/3,2/3,2/3)^T$ has length

$$\sqrt{(1/3)^2 + (2/3)^2 + (2/3)^2} = \sqrt{1/9 + 4/9 + 4/9} = 1.$$

The second and third columns are basically the same. As for being normal, the dot products of any two of the columns is equal to zero. For example,

$$(1,2,2)^T \cdot (-2,-1,2)^T = 1(-2) + 2(-1) + 2(2) = -2 - 2 + 4 = 0. $$

It should be easy to check that the second and third and first and third columns are orthogonal to each other.

Answer 3

The definition of an orthonormal matrix is for

$$Q^T\cdot Q = Id = Q \cdot Q^T$$

to hold.

Calculating this, we get

$$\frac{1}{9}\begin{bmatrix}1 & 2 & 2 \\-2 & -1 & 2 \\ 2 & -2 & 1\end{bmatrix}\cdot\begin{bmatrix}1 & -2 & 2 \\ 2 & -1 & -2 \\ 2 & 2 & 1\end{bmatrix} = \frac{1}{9}\cdot 9 \cdot Id = Id$$

so the matrix is by definition orthonormal.