Let $(V, \langle \cdot,\cdot \rangle)$ be a real or complex inner product space. Let $\{x_1, ..., x_n\}$ be a linearly independent set. Define:

$$e_1 = \frac{x_1}{||x_1||}, z_k = x_k - \sum_{j=1}^{k-1} \langle x_k,e_j \rangle e_j, e_k = \frac{z_k}{||z_k||}$$

for all $2 \le k \le n$.

How to see that: $\langle e_i, e_j \rangle = \delta_{ij} = \begin{cases} 1, j=k \\ 0, j \neq k \end{cases}$?

I did direct calculations but I reached nothing. I remember that there's some lemma involved with this. Could someone point me to the right direction?


1 Answer 1


One can show by induction (on $j$) that $\langle e_j , e_i\rangle$ satisfy your condition for all $i\le j$.

Precisely, let $S(k)$ be the statement that

$$\langle e_i, e_j\rangle = \begin{cases} 1 &\text{if }i=j \\0 &\text{if }i\neq j\end{cases}$$

for all $i,j\le k$.

Then $S(1)$ is true by definition of $e_1$. Assume that $S(k)$ is true. By definition, $\langle e_{k+1}, e_{k+1}\rangle = 1$. When $i< k+1$,

$$\begin{split} \langle e_{k+1}, e_i\rangle &= \frac{1}{\|z_k\|} \langle x_{k+1} - \sum_{m=1}^k \langle x_{k+1}, e_m\rangle e_m, e_i\rangle \\ &= \frac{1}{\|z_k\|}\left( \langle x_{k+1}, e_i\rangle - \langle x_{k+1}, e_i\rangle\right)\\ &= 0\end{split} $$

(Note that we used the induction hypothesis). Thus $S(k+1)$ is also true. By (finite) induction we are done.

Note that the fact that $\{x_1, x_2, \cdots, x_n\}$ are linear independent is used to guaranteed that $z_k \neq 0$ for all $k$ (So that you can define $e_k$)

This process is called the Gram-Schmidt process.

  • $\begingroup$ The kind of induction we used here was: "assuming that for all $t\le m$, it holds that for all $j<t$: $<e_j,e_t> = 0$, we prove that for all $t \le m+1$, for all $j<t$ it holds that $<e_j,e_t> = 0$?" $\endgroup$
    – user230734
    Commented Sep 20, 2015 at 10:15
  • $\begingroup$ Is this strong induction or weak induction? $\endgroup$
    – user230734
    Commented Sep 20, 2015 at 10:18
  • 1
    $\begingroup$ @BolzWeir Please see the edit. $\endgroup$
    – user99914
    Commented Sep 20, 2015 at 10:22
  • $\begingroup$ Makes perfect sense now. Thanks a lot. $\endgroup$
    – user230734
    Commented Sep 20, 2015 at 10:40

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