Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $v_1, \dots, v_m$ be a linearly dependent list of vectors.
If $v_1 \ne 0$, then there is some $v_j$ in the span of $v_1, \dots, v_{j-1}$
If we let j be the smallest integer with this property, and apply the gram-schmidt procedure to produce an orthonormal list $(e_1, \dots, e_{j-1})$ then $v_j$ is in the span of $(e_1, \dots, e_{j-1})$ and
$$v_j = \langle v_j, e_1\rangle e_1+ \dots + \langle v_j, e_{j-1}\rangle e_{j-1}$$
Why does this guarantee that length of $v_j$=0? I'm missing something about linear dependence that should probably be obvious sorry :\

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

The Gram-Schmidt process goes as follows: given $v_1,\ldots,v_n$, you define $$\begin{align*} u_1 &= v_1\\ e_1 &= \frac{1}{||u_1||}u_1;\\ u_2 &= v_2 - \langle v_2,e_1\rangle e_1;\\ e_2 &= \frac{1}{||u_2||}u_2;\\ &\vdots\\ u_{k+1} &= v_{k+1} - \left(\langle v_{k+1},e_1\rangle e_1 + \langle v_{k+1},e_2\rangle e_2 + \cdots + \langle v_{k+1},e_k\rangle e_k\right)\\ e_{k+1}&= \frac{1}{||u_{k+1}||}u_{k+1}\\ &\vdots \end{align*}$$

So when you construct $u_{j}$, you get $\mathbf{0}$, and when you try to construct $e_{j}$ you attempt to divide by $0$. It's not $v_j$ which has length $0$, it's $u_j$.

share|improve this answer
    
I see, thank you. –  fdart17 Mar 29 '11 at 5:04
add comment

It does not guarantee that. What it guarantees is that $v_j - \langle v_j, e_1\rangle e_1- \dots - \langle v_j, e_{j-1}\rangle e_{j-1} = 0$, which means that you cannot define $e_j$ by dividing this vector by its norm. Instead, you throw it away and move on to $v_{j+1}$.

E.g., consider the case $v_1\neq 0$ and $v_2=v_1$.

share|improve this answer
    
Thank you, this also helped with the simple example. –  fdart17 Mar 29 '11 at 5:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.