I'm working on this problem:

Show that if $(e_1...e_n)$ is an orthonormal basis constructed using the Gram Schmidt process from $(v_1...v_n)$, then for any $j,$ $\langle e_j, v_j \rangle > 0$.

Intuitively this makes sense to me, as the Gram Schmidt procedure takes a vector $v_j$ and removes all of its components that are part of $\langle e_1,..., e_ {j-1} \rangle$, but how would I prove this mathematically?

  • $\begingroup$ do you mean constructed using Gram Schmidt. $\endgroup$ Mar 25, 2016 at 1:26
  • $\begingroup$ I assume you mean that the vectors are the "Q" in a QR decomposition (so they might not be from Gram Schmidt but they are equivalent to the result of Gram Schmidt). For that, just write $v_j$ as a sum of $e_1,\dots,e_j$ and expand out the inner product. $\endgroup$
    – Ian
    Mar 25, 2016 at 1:29
  • $\begingroup$ Sorry, I meant using the Gram Schmidt process. I edited the question. $\endgroup$
    – 100001
    Mar 25, 2016 at 1:40
  • $\begingroup$ What you want to do is write $v_i$ in terms of the basis $e_j$... and this can be done because you know how to write $e_j$ in terms of the basis $v_i$. Now, writing one basis in terms of another has the feeling of a linear transformation, right? ... can you write this out using matrices? Maybe then it will be clear how to proceed. $\endgroup$
    – Elle Najt
    Mar 25, 2016 at 2:18

2 Answers 2


Let $\{w_1,w_2,\ldots,w_n\}$ be obtained from $\{v_1,v_2,\ldots,v_n\}$ by the Gram-Schmidt process, then $w_1=v_1$ and $$w_j=v_j-\sum_{k=1}^{j-1}\frac{\langle v_j,w_k\rangle}{\Vert w_k\Vert^2}w_k \quad\text{ for }2\leq j\leq n.$$ Thus $\langle e_1,v_1\rangle=\langle e_1,w_1\rangle=\Vert w_1\Vert\langle e_1,e_1\rangle>0$ and for $2\leq j\leq n,$ \begin{align*} \langle e_j,v_j\rangle &=\left\langle e_j,w_j+\sum_{k=1}^{j-1}\frac{\langle v_j,w_k\rangle}{\Vert w_k\Vert^2}w_k\right\rangle =\langle e_j,w_j\rangle+\sum_{k=1}^{j-1}\frac{\overline{\langle v_j,w_k\rangle}}{\Vert w_k\Vert^2}\langle e_j,w_k\rangle\\ &=\langle e_j,w_j\rangle+\sum_{k=1}^{j-1}\frac{\overline{\langle v_j,w_k\rangle}\Vert w_k\Vert}{\Vert w_k\Vert^2}\langle e_j,e_k\rangle =\langle e_j,w_j\rangle+\sum_{k=1}^{j-1}\frac{\overline{\langle v_j,w_k\rangle}}{\Vert w_k\Vert}\delta_{jk}\\ &= \langle e_j,w_j\rangle=\Vert w_j\Vert\langle e_j,e_j\rangle>0, \end{align*} where $\delta_{jk}$ is the Kronecker delta function.


By orthonormality of the basis,

$$v_j=\sum_k(e_k\cdot v_j)e^k,$$ and $$v_j^2=\sum_k(e_k\cdot v_j)^2,$$ (the orthogonal transform preserves the $L_2$ norm) so that $$v_j^2\ge\sum_{k<j}(e_k\cdot v_j)^2$$ (a vector is longer than its projection).


$$u_j:=\|u_j\|e_j:=v_j-\sum_{k<j}(e_k\cdot v_j)e_k$$

(the vectors $u_j$, then $e_j$ are obtained by projection) and

$$\|u_j\|e_j\cdot v_j=u_j\cdot v_j=v_j^2-\sum_{k<j}(e_k\cdot v_j)^2.$$


You must log in to answer this question.

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