📂Linear Algebra

Orthogonal Basis and Its Coordinates

Definition¹

An inner product space basis $V$ that is an orthogonal set is called an orthogonal basis. If $S$ is an orthonormal set, it is called an orthonormal basis.

Theorem

If $S = \left\{ \mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n} \right\}$ is an orthogonal basis of the inner product space $V$, and let $\mathbf{u} \in V$. Then, the following equation holds.

$$\begin{equation} \begin{aligned} \mathbf{u} &= \dfrac{\langle \mathbf{u}, \mathbf{v}_{1} \rangle}{\| \mathbf{v}_{1} \|^{2}} \mathbf{v}_{1} + \dfrac{\langle \mathbf{u}, \mathbf{v}_{2} \rangle}{\| \mathbf{v}_{2} \|^{2}} \mathbf{v}_{2} + \cdots + \dfrac{\langle \mathbf{u}, \mathbf{v}_{n} \rangle}{\| \mathbf{v}_{n} \|^{2}} \mathbf{v}_{n} \\ &= \sum \limits _{i=1}^{n} \dfrac{\langle \mathbf{u}, \mathbf{v}_{i} \rangle}{\| \mathbf{v}_{i} \|^{2}} \mathbf{v}_{i} \end{aligned} \label{thm1} \end{equation}$$

If $S$ is an orthonormal basis, the following equation holds.

$$\begin{equation} \begin{aligned} \mathbf{u} &= \langle \mathbf{u}, \mathbf{v}_{1} \rangle\mathbf{v}_{1} + \langle \mathbf{u}, \mathbf{v}_{2} \rangle\mathbf{v}_{2} + \cdots + \langle \mathbf{u}, \mathbf{v}_{n} \rangle\mathbf{v}_{n} \\ &= \sum \limits _{i=1}^{n} \langle \mathbf{u}, \mathbf{v}_{i} \rangle \mathbf{v}_{i} \end{aligned} \label{thm2} \end{equation}$$

Explanation

From the above theorem, the following vector is said to be the coordinates of $\mathbf{u} \in V$ with respect to the basis $S$.

$$(\mathbf{u})_{S} = \left( \langle \mathbf{u}, \mathbf{v}_{1} \rangle, \langle \mathbf{u}, \mathbf{v}_{2} \rangle, \dots, \langle \mathbf{u}, \mathbf{v}_{n} \rangle \right)$$

Proof

Since $S$ is a basis of $V$, $\mathbf{u} \in V$ has the following unique linear combination representation.

$$\begin{equation} \mathbf{u} = c_{1} \mathbf{v}_{1} + c_{2} \mathbf{v}_{2} + \cdots + c_{n} \mathbf{v}_{n} \label{eq1} \end{equation}$$

Taking the inner product of $\mathbf{u}$ with each of $\mathbf{v}_{i}$ yields the following.

$$\begin{align*} \\ \langle \mathbf{u}, \mathbf{v}_{i} \rangle &= \langle c_{1} \mathbf{v}_{1} + c_{2} \mathbf{v}_{2} + \cdots + c_{n} \mathbf{v}_{n} , \mathbf{v}_{i} \rangle \\ &= c_{1} \langle \mathbf{v}_{1}, \mathbf{v}_{i} \rangle + c_{2} \langle \mathbf{v}_{2}, \mathbf{v}_{i} \rangle + \cdots c_{i} \langle \mathbf{v}_{i}, \mathbf{v}_{i} \rangle +\cdots + c_{n} \langle \mathbf{v}_{n}, \mathbf{v}_{i} \rangle \\ &= c_{i} \langle \mathbf{v}_{i}, \mathbf{v}_{i} \rangle \\ &= c_{i} \| \mathbf{v}_{i} \|^{2} \end{align*}$$

$$\\ \implies c_{i} = \dfrac{\langle \mathbf{u}, \mathbf{v}_{i} \rangle }{\| \mathbf{v}_{i} \|^{2}}$$

Since the $c_{i}$ that satisfy the above equations are unique, substituting them into $\eqref{eq1}$ gives the following.

$$\begin{align*} \mathbf{u} &= \dfrac{\langle \mathbf{u}, \mathbf{v}_{1} \rangle}{\| \mathbf{v}_{1} \|^{2}} \mathbf{v}_{1} + \dfrac{\langle \mathbf{u}, \mathbf{v}_{2} \rangle}{\| \mathbf{v}_{2} \|^{2}} \mathbf{v}_{2} + \cdots + \dfrac{\langle \mathbf{u}, \mathbf{v}_{n} \rangle}{\| \mathbf{v}_{n} \|^{2}} \mathbf{v}_{n} \\ &= \sum \limits _{i=1}^{n} \dfrac{\langle \mathbf{u}, \mathbf{v}_{i} \rangle}{\| \mathbf{v}_{i} \|^{2}} \mathbf{v}_{i} \end{align*}$$

If $S$ is an orthonormal set, then $\| \mathbf{v}_{i} \|^{2}=1$ holds, so $\eqref{thm2}$ is satisfied.

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