A basis whose vectors are at right angles to one another is the most convenient kind there is. Coordinates in it are found by dot products, one at a time, with no system to solve, and a matrix whose columns form such a basis turns and reflects space without stretching it.

A map drawn on a square grid lets you read east and north independently. On a slanted grid, moving east also changes the northern reading, and every position has to be worked out.

In the orthogonal basis $\mathbf{u}_1=(1,1)$, $\mathbf{u}_2=(-1,1)$, the coordinates of $\mathbf{x}=(3,1)$ are dot products: $\frac{\mathbf{x}\cdot\mathbf{u}_1}{\mathbf{u}_1\cdot\mathbf{u}_1}=\frac42=2$ and $\frac{\mathbf{x}\cdot\mathbf{u}_2}{\mathbf{u}_2\cdot\mathbf{u}_2}=\frac{-2}{2}=-1$. The dashed path is $\mathbf{x}=2\mathbf{u}_1-\mathbf{u}_2$.

A set of vectors is orthogonal when every two different vectors in it are orthogonal, and orthonormal when, in addition, each has length $1$. An orthogonal set of non-zero vectors is independent, the theorem below, so $n$ of them are a basis of $\mathbb{R}^n$. A square matrix $Q$ is orthogonal when its columns are orthonormal, which is the equation $Q^{T}Q=$.

• In an orthogonal basis the coordinates of $\mathbf{x}$ are $c_i=\dfrac{\mathbf{x}\cdot\mathbf{u}_i}{\mathbf{u}_i\cdot\mathbf{u}_i}$: dot with and every other term vanishes. In an orthonormal basis, simply .

Take a relation $c_1\mathbf{u}_1+\cdots+c_k\mathbf{u}_k=\mathbf{0}$ and dot it with . Every term but one vanishes by orthogonality, leaving , and since is not zero, . That holds for every .

Take any relation among the vectors.

Dot it with $\mathbf{u}_i$: the other terms vanish by orthogonality.

The relation says the left side is $\mathbf{0}$.

$\mathbf{u}_i\neq\mathbf{0}$, so $\|\mathbf{u}_i\|^2\neq0$: every weight vanishes.

Check every pair: a set is orthogonal when each dot product between two different members is $0$.

In an orthogonal basis, dot $\mathbf{x}=c_1\mathbf{u}_1+\cdots+c_n\mathbf{u}_n$ with : all the other terms vanish, and is left alone.

A square matrix with orthonormal columns satisfies $Q^{T}Q=I$: entry $(i,j)$ of $Q^{T}Q$ is column dotted with column . So its inverse is its transpose.

• Orthogonal non-zero vectors are independent.
• In an orthogonal basis $c_i=\dfrac{\mathbf{x}\cdot\mathbf{u}_i}{\mathbf{u}_i\cdot\mathbf{u}_i}$; in an orthonormal one .

Great work! You now know how to:

• recognise orthogonal and orthonormal sets;
• find coordinates in an orthogonal basis by dot products;
• use $Q^{-1}=Q^{T}$ for an orthogonal matrix;
• explain why orthogonal matrices keep lengths and angles.