Let $e_1, e_2, e_3$ be an orthonormal basis. How to prove that for any vector $u$,

$$u = (u, e_1)e_1 + (u, e_2)e_2 + (u, e_3)e_3,$$

i.e., the $i$th coordinate of $u$ in the basis equals $(u,e_i)$?

  • $\begingroup$ By $(u, e_i)$, you mean inner product, right? $\endgroup$
    – apnorton
    Commented Jan 22, 2015 at 22:58
  • $\begingroup$ @anorton Yes I do. $\endgroup$
    – J.D.
    Commented Jan 23, 2015 at 0:05

1 Answer 1


Since $\{e_1,e_2,e_3\}$ is a basis, for every $u$ there exist unique scalars $\lambda_1$, $\lambda_2$, and $\lambda_3$ such that $$ u=\lambda_1\cdot e_1+\lambda_2\cdot e_2+\lambda_3\cdot e_3\tag{1} $$ This implies \begin{align*} \langle u,e_1\rangle &= \langle \lambda_1\cdot e_1+\lambda_2\cdot e_2+\lambda_3\cdot e_3,e_1\rangle \\ &= \lambda_1\cdot\langle e_1,e_1\rangle+\lambda_2\cdot\langle e_2,e_1\rangle+\lambda_3\cdot\langle e_3,e_1\rangle \\ &= \lambda_1\cdot 1+\lambda_2\cdot 0+\lambda_3\cdot 0 \\ &= \lambda_1 \end{align*} Similarly one shows that $\lambda_2=\langle u,e_2\rangle$ and $\lambda_3=\langle u,e_3\rangle$. Substituting these values into (1) gives your result.


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