A collection $\mathcal{B} = \{{\bf b}_1, \ldots, {\bf b}_k\}$ of vectors is said to be orthogonal if the inner product of any two (different) elements is zero: $$\langle {\bf b}_i | {\bf b}_j \rangle = 0 \hbox{ if } i \ne j.$$ Orthogonal sets are useful. Orthogonal bases are even more useful.

Suppose that $\mathcal{B} = \{{\bf b}_1, \ldots, {\bf b}_n\}$ is an orthogonal basis for an $n$-dimensional vector space $V$. The following video show how to use inner products, rather than solving systems of linear equations, to decompose an arbitrary vector in the $\mathcal{B}$ basis.

If ${\bf x} = c_1 {\bf b}_1 + \cdots + c_n {\bf b}_n$, then $\displaystyle{c_j = \frac{\langle {\bf b}_j | {\bf x}\rangle} {\langle {\bf b}_j | {\bf b}_j\rangle}}$ and the projection of ${\bf x}$ onto ${\bf b}_j$ is $$c_j {\bf b}_j = \left(\frac{| {\bf b}_j \rangle \langle {\bf b}_j |} {\langle {\bf b}_j | {\bf b}_j\rangle} \right ) | {\bf x} \rangle = P_{{\bf b}_j} | {\bf x} \rangle.$$ The sum of all the projections is the identity operator: $$ I = \sum_j P_{{\bf b}_j} = \sum_j \frac{| {\bf b}_j \rangle \langle {\bf b}_j |} {\langle {\bf b}_j | {\bf b}_j\rangle}.$$ If we have an orthonormal basis, with each ${\bf b}_j$ being a unit vector, then the formulas get even simpler. The denominators are all 1, and $$c_j = \langle {\bf b}_j | {\bf x} \rangle, \qquad P_{{\bf b}_j} = | {\bf b}_j \rangle \langle {\bf b}_j |, \qquad I = \sum_j P_{{\bf b}_j} = \sum_j | {\bf b}_j \rangle \langle {\bf b}_j |. $$

Orthonormal bases can also be used to decompose operators, as shown in the second video:

Any $n \times n$ matrix $A = \begin{pmatrix} A_{11} & \cdots & A_{1n} \cr \vdots & \ddots & \vdots \cr A_{n1} & \cdots & A_{nn} \end{pmatrix}$ can be decomposed as $$A = \sum_{i,j} A_{ij} | {\bf e}_i \rangle \langle {\bf e}_j |, \qquad A_{ij} = \langle {\bf e}_i | A {\bf e}_j \rangle.$$ The same sort of thing works for any (real or complex) inner product space $V$. If $\mathcal{B}$ is an orthonormal (not just orthogonal) basis and $L: V \to V$ is an operator, then we define the matrix elements $L_{ij} = \langle {\bf b}_i| L({\bf b}_j) \rangle$ and get the expansions: $$ [L]_{\mathcal{B}} = \begin{pmatrix} L_{11} & \cdots & L_{1n} \cr \vdots & \ddots & \vdots \cr L_{n1} & \cdots & L_{nn} \end{pmatrix}, \qquad L = \sum_{i,j} L_{ij} | {\bf b}_i \rangle \langle {\bf b}_j |.$$