Orthonormality

Orthonormality is a property of a set of vectors that combines two independent conditions: the vectors must be mutually perpendicular (orthogonality) and each must have unit length (normality). Orthonormal vectors form the building block for the QR decomposition introduced in the next section.

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Orthogonality

Two vectors $x$ and $y$ are orthogonal if their dot product is zero:

$$x^T y = 0$$

Geometrically, orthogonal vectors are perpendicular to each other.

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Normality

A vector $x$ is normal (or a unit vector) if its dot product with itself equals 1:

$$x^T x = 1$$

This is equivalent to saying $\|x\| = 1$ — the vector has length 1.

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Orthonormality

A set of vectors is orthonormal if every vector in the set is normal and every pair of distinct vectors is orthogonal. Together:

$$q_i^T q_j = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases}$$

This compact notation is known as the Kronecker delta:

$$\delta_{ij} = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases}$$

So the orthonormality condition is simply $q_i^T q_j = \delta_{ij}$.

Note

For a set of three vectors $\{x, y, z\}$ to be orthonormal, you must verify:

Normality (3 conditions): $x^T x = 1, \quad y^T y = 1, \quad z^T z = 1$

Orthogonality (3 conditions): $x^T y = 0, \quad y^T z = 0, \quad z^T x = 0$

All six conditions must hold.

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Worked Example: Verifying Orthonormality

Problem: Determine whether the following set is orthonormal.

$S = \left\{\, \frac{1}{\sqrt{5}}\begin{pmatrix}2\\1\end{pmatrix},\; \frac{1}{\sqrt{5}}\begin{pmatrix}1\\-2\end{pmatrix} \,\right\}$

Let $u = \dfrac{1}{\sqrt{5}}\begin{bmatrix}2\\1\end{bmatrix}$ and $v = \dfrac{1}{\sqrt{5}}\begin{bmatrix}1\\-2\end{bmatrix}$.

We must verify both normality and orthogonality.

Step 1 — Check normality of $u$

$$u^T u = \frac{1}{\sqrt{5}}\begin{bmatrix}2 & 1\end{bmatrix} \cdot \frac{1}{\sqrt{5}}\begin{bmatrix}2\\1\end{bmatrix} = \frac{1}{5}\begin{bmatrix}2 & 1\end{bmatrix}\begin{bmatrix}2\\1\end{bmatrix} = \frac{1}{5}(4 + 1) = \frac{5}{5} = 1$$

$u^T u = 1$ (True)

Step 2 — Check normality of $v$

$$v^T v = \frac{1}{\sqrt{5}}\begin{bmatrix}1 & -2\end{bmatrix} \cdot \frac{1}{\sqrt{5}}\begin{bmatrix}1\\-2\end{bmatrix} = \frac{1}{5}\begin{bmatrix}1 & -2\end{bmatrix}\begin{bmatrix}1\\-2\end{bmatrix} = \frac{1}{5}(1 + 4) = \frac{5}{5} = 1$$

$v^T v = 1$ (True)

Step 3 — Check orthogonality

$$u^T v = \frac{1}{\sqrt{5}}\begin{bmatrix}2 & 1\end{bmatrix} \cdot \frac{1}{\sqrt{5}}\begin{bmatrix}1\\-2\end{bmatrix} = \frac{1}{5}\begin{bmatrix}2 & 1\end{bmatrix}\begin{bmatrix}1\\-2\end{bmatrix} = \frac{1}{5}(2 - 2) = 0$$

$u^T v = 0$ (True)

Conclusion

All three conditions are satisfied: $u^T u = 1$, $v^T v = 1$, and $u^T v = 0$.

$$\boxed{S \text{ is orthonormal.}}$$