Show that in an n-dimensional vector space V over the universal set with orthogonal basis {$a_1, a_2,..., a_n$}, each vector B can be expressed as:

B = $\frac{<B,a_1>a_1}{||a_1||^2}$ + $\frac{<B,a_2>a_2}{||a_2||^2}$ +.... + $\frac{<B,a_n>a_n}{||a_n||^2}$

I tried something along the lines of using B = $\frac{(B.V)V}{||V||^2}$ but I couldn't get to the final answer.

  • $\begingroup$ That a vector is the sum of its components along the basis vector is true of any basis—indeed it’s an essential part of the definition of a basis. Are you really asking about the sum of orthogonal projections onto the basis vectors? That’s a different matter. $\endgroup$
    – amd
    May 27, 2020 at 20:26

1 Answer 1


Let $B$ be a vector of $V$. Since $\{a_1,\ldots, a_n\}$ is a basis for $V$, you can find coefficients $\alpha_\, \ldots, \alpha_n$ such that $$B=\alpha_1 a_1+\ldots +\alpha_n a_n$$ Note that \begin{align*} \langle B, a_i\rangle&=\langle \alpha_1 a_1+\ldots +\alpha_n a_n, a_i\rangle \nonumber\\ &=\alpha_1 \langle a_1, a_i\rangle +\ldots +\alpha_n \langle a_n,a_i\rangle\nonumber \end{align*} Since the basis is orthogonal, the above expression simplifies as \begin{align*} \langle B, a_i\rangle&=\alpha_i\langle a_i, a_i\rangle\nonumber\\ &=\alpha_i ||a_i||^2\nonumber \end{align*} Hence $$\alpha_i=\frac{\langle B,a_i\rangle}{||a_i||^2}$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .