Consider $\mathbb{R}^3$ together with inner product $\langle (x_1, x_2, x_3), (y_1, y_2, y_3) \rangle = 2x_1 y_1+x_2 y_2+3 x_3 y_3$. Use the Gram-Schmidt procedure to find an orthonormal basis for $W=\text{span} \left\{(-1, 1, 0), (-1, 1, 2) \right\}$.

I don't get how the inner product $\langle (x_1, x_2, x_3), (y_1, y_2, y_3) \rangle = 2 x_1 y_1+x_2 y_2+3 x_3 y_3$ would affect the approach to solve this question.. When I did the gram-schmidt, I got $v_1=(-1, 1, 0)$ and $v_2=(0, 0, 2)$ but then realized that you have to do something with the inner product before finding the orthonormal basis. Can someone please help me?

Update: So far I got $\{(\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, 0), (0, 0, \frac{2}{\sqrt{12}})\}$ as my orthonormal basis but I'm not sure if I am doing it right with the given inner product.

  • 1
    $\begingroup$ please try to learn TeX syntax. I did it for you in this case. $\endgroup$
    – Siminore
    Jun 15, 2012 at 13:16
  • $\begingroup$ @Alice You got the same solution I got :) To convince yourself it's right, just compute the pairwise inner products between your solution vectors, and you will find they are orthnormal! $\endgroup$
    – rschwieb
    Jun 15, 2012 at 13:29

2 Answers 2


The choice of inner product defines the notion of orthogonality.

The usual notion of being "perpendicular" depends on the notion of "angle" which turns out to depend on the notion of "dot product".

If you change the way we measure the "dot product" to give a more general inner product then we change what we mean by "angle", and so have a new notion of being "perpendicular", which in general we call orthogonality.

So when you apply the Gram-Schmidt procedure to these vectors you will NOT necessarily get vectors that are perpendicular in the usual sense (their dot product might not be $0$).

Let's apply the procedure.

It says that to get an orthogonal basis we start with one of the vectors, say $u_1 = (-1,1,0)$ as the first element of our new basis.

Then we do the following calculation to get the second vector in our new basis:

$u_2 = v_2 - \frac{\langle v_2, u_1\rangle}{\langle u_1, u_1\rangle} u_1$

where $v_2 = (-1,1,2)$.

Now $\langle v_2, u_1\rangle = 3$ and $\langle u_1, u_1\rangle = 3$ so that we are given:

$u_2 = v_2 - u_1 = (0,0,2)$.

So your basis is correct. Let's check that these vectors are indeed orthogonal. Remember, this is with respect to our new inner product. We find that:

$\langle u_1, u_2\rangle = 3(-1)(0) + (1)(0) + 2(0)(2) = 0$

(here we also happened to get a basis that is perpendicular in the traditional sense, this was lucky).

Now is the basis orthonormal? (in other words, are these unit vectors?). No they arent, so to get an orthonormal basis we must divide each by its length. Now this is not the length in the usual sense of the word, because yet again this is something that depends on the inner product you use. The usual Pythagorean way of finding the length of a vector is:

$||x||=\sqrt{x_1^2 + ... + x_n^2} = \sqrt{x . x}$

It is just the square root of the dot product with itself. So with more general inner products we can define a "length" via:

$||x|| = \sqrt{\langle x,x\rangle}$.

With this length we see that:

$||u_1|| = \sqrt{2(-1)(-1) + (1)(1) + 3(0)(0)} = \sqrt{3}$

$||u_2|| = \sqrt{2(0)(0) + (0)(0) + 3(2)(2)} = 2\sqrt{3}$

(notice how these are different to what you would usually get using the Pythagorean way).

Thus an orthonormal basis is given by:

$\{\frac{u_1}{||u_1||}, \frac{u_2}{||u_2||}\} = \{(\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, 0), (0,0,\frac{1}{\sqrt{3}})\}$

  • $\begingroup$ Thanks so much for the thorough explanation! :) $\endgroup$
    – Alice
    Jun 15, 2012 at 13:36

Perhaps you carried out the Gram-Schmidt algorithm using the ordinary inner product? I think that is the only way you could have gotten through without using the given inner product :)

Anyhow, you need to use the given inner product at each step of the orthonormalization procedure. Changing the inner product will change the output of the algorithm, because different inner products yield different lengths of vectors and report different "angles" between vectors.

For example, when you begin with the first step (normalizing $(-1,1,0)$, you should compute that $\langle(-1,1,0),(-1,1,0)\rangle=3$, and so the first vector would be $\frac{1}{\sqrt{3}}(-1,1,0)$.

  • $\begingroup$ Thanks a lot! I understand it now. Oh, I was also wondering if it asks to find the orthogonal projection of (1, 1, 1) onto the same W, I guess we still apply the given inner product? I got (1/3, 1/3, 1) for the projection and I want to make sure I'm doing it right.. $\endgroup$
    – Alice
    Jun 15, 2012 at 13:37
  • $\begingroup$ @alice I got a negative sign on the middle entry when I tried a moment ago... can you check your work to see if that's missing in your computation? $\endgroup$
    – rschwieb
    Jun 15, 2012 at 13:44
  • $\begingroup$ Oh, oops, I did forget the negative sign in my calculation. Thanks again :) $\endgroup$
    – Alice
    Jun 15, 2012 at 13:46

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