📂Linear Algebra

Relationship Between Orthogonality and Linear Independence

Definition¹

An inner product space $V$’s two vectors $\mathbf{u}, \mathbf{v}$ are said to be orthogonal if they satisfy $\langle \mathbf{u}, \mathbf{v} \rangle = 0$.

A set made up of elements of $V$ where each element is orthogonal to every other element is called an orthogonal set.

If the norm of every element in an orthogonal set is $1$, then it is called an orthonormal set.

Theorem

A subset $S = \left\{ \mathbf{v}_{1}, \mathbf{v}_{2}, \cdots, \mathbf{v}_{n} \right\}$ of an inner product space $V$ that includes no zero vector and is an orthogonal set implies that $S$ is linearly independent.

Proof

To prove that $S$ is linearly independent, it suffices to show that the only solution to the following equation

$$k_{1} \mathbf{v}_{1} + k_{2} \mathbf{v}_{2} + \cdots + k_{n} \mathbf{v}_{n} = \mathbf{0}$$

is $k_{1}=k_{2}=\cdots=k_{n}=0$. Taking the inner product of each vector $\mathbf{v}_{i}$ with the above equation, we have

$$\begin{align*} \\ 0 &= \langle \mathbf{0}, \mathbf{v}_{i} \rangle \\ &= \langle k_{1} \mathbf{v}_{1} + k_{2} \mathbf{v}_{2} + \cdots + k_{n} \mathbf{v}_{n}, \mathbf{v}_{i} \rangle \\ &= k_{1} \langle \mathbf{v}_{1}, \mathbf{v}_{i} \rangle + k_{2} \langle \mathbf{v}_{2}, \mathbf{v}_{i} \rangle + \cdots k_{i} \langle \mathbf{v}_{i}, \mathbf{v}_{i} \rangle +\cdots + k_{n} \langle \mathbf{v}_{n}, \mathbf{v}_{i} \rangle \\ &= k_{i} \langle \mathbf{v}_{i}, \mathbf{v}_{i} \rangle \end{align*}$$

However, since $S$ is an orthogonal set that does not include the zero vector, $\langle \mathbf{v}_{i}, \mathbf{v}_{i} \rangle>0$ is true. Therefore,

$$k_{i} = 0,\quad \forall 1\le i \le n$$

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