# What are the angle brackets in Linear Algebra?

In my linear algebra book, they have angle brackets around two different vectors, so it looks like this: $\langle\mathbf{u_2},\mathbf{v}_1\rangle$. They don't use angle brackets to define vectors, but use regular parenthesis instead.

For the Gram-Schmidt process, they define

$\mathbf{v}_1 = \mathbf{u}_1 = (1,1,1)$

and

$\mathbf{v}_2 = \mathbf{u}_2 = \mathbf{u}_2 - \dfrac{\langle\mathbf{u}_2, \mathbf{v}_1\rangle}{\|\mathbf{v}_1\|^2} \mathbf{v}_1$

where $\mathbf{u}_2 = (0,1,1)$

They conclude that that formula is equal to

$(0,1,1) - \dfrac{2}{3}(1,1,1)$.

What operation is the angle brackets to have that result?

• It's the inner product, probably the standard inner product $\langle u,v\rangle = \sum u_i\cdot v_i$ (if real, conjugate one factor if complex). – Daniel Fischer Nov 17 '13 at 20:20
• There should be a definition for the symbol in the textbook. – egreg Nov 17 '13 at 20:24
• I'm sure it's somewhere. I must have missed it when it was introduced. – JFA Nov 17 '13 at 20:27
• @DanielFischer if you want to go ahead and make that an answer since you were the first one to propose it, I can make that the correct answer. – JFA Nov 17 '13 at 20:50
• @Jack Nah, you got a pretty good answer, I think. Unless you're not satisfied with that, I see no reason to add another one. – Daniel Fischer Nov 17 '13 at 20:55

The angled brackets represent an inner product. The best known one is the scalar product or the dot product. If ${\bf u} = (u_1,u_2,u_3)$ and ${\bf v} = (v_1,v_2,v_3)$, then the dot product is given by $$\langle {\bf u},{\bf v} \rangle = u_1v_1 + u_2v_2+u_3v_3$$ It has many useful properties. First $\langle {\bf u},{\bf u} \rangle = \|{\bf u}\|^2$, and second if ${\bf u}$ and ${\bf v}$ are both non-zero then $\langle {\bf u},{\bf v}\rangle = 0$ if and only if ${\bf u}$ and ${\bf v}$ are orthogonal. In general:
$$\langle {\bf u}, {\bf v} \rangle = \|{\bf u}\| \|{\bf v}\| \cos\theta$$
where $\theta$ is the angle between ${\bf u}$ and ${\bf v}$. This idea can be generalised. Notice that $\langle {\bf u},{\bf v} \rangle = {\bf u}E{\bf v}^{\top}$, where $E$ is the 3-by-3 identity matrix and ${\bf u}$ and ${\bf v}$ are being thought of as 1-by-3 matrices. For any 3-by-3 matrix, say $M$, we can define $\langle {\bf u},{\bf v}\rangle_M := {\bf u}M{\bf v}^{\top}$. Different matrices gives rise to different $\langle {\bf u},{\bf v}\rangle_M$. We usually assume that $M$ is a positive definite matrix.