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I have the following question to answer, but what I would like is a sane and easy to grasp idea of what Normalization is all about(the why, when etc) and then how to do it without it getting over one's head please.

  1. Find an orthonormal basis of $\mathbb{R}^2$ by applying the Gram-Schmidt orthonormalization method to the vectors: $$(1; 2)^T, (2; 1)^T$$

  2. Find a unit vector which is orthogonal to all vectors in the subspace of ${\bf R}^3$ given by: f(x; y; z)T : 3x - 2y + z = 0g

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Do you mean orthogonalization instead of normalization? –  Calle Jul 20 '11 at 4:49
    
I think the description and the attendant picture at Wikipedia are pretty good. Can you use that to do the first problem? I think that if you understand this visually for two vectors, then you are in very good shape. –  Dylan Moreland Jul 20 '11 at 4:53
    
No, I find the wiki explanation a bit too advanced as well. I am in a class to learn this but having a hard time making sense of it. –  user10695 Jul 20 '11 at 4:59
    
I did a little light editing, but I was stumped by f(x;y;z)T:3x-2y+z=0g. I have no idea what this is supposed to be. –  Gerry Myerson Jul 20 '11 at 5:53

4 Answers 4

Normalization of a vector is simply making a vector with the same direction, but of size 1. Thus if you you have a vector $\mathbf v$ and want to find the normalized vector $\hat{\mathbf v}$, you can calculate it by $$\hat{\mathbf v} = \frac{\mathbf v}{|\mathbf v|}$$ where $|\mathbf v|$ is the norm of $\mathbf v$.

I suspect your question is more about orthogonalization or maybe the concept of orthogonality than normalization. Though a bit of confusion maybe has come from that a vector is called normal to a subspace (the vector is called a normal vector to the subspace) if it is orthogonal to all vectors in the subspace.

The Gram-Schmidt procedure for only two vectors $\mathbf v_1$ and $\mathbf v_2$ can be applied as follows. First, set $\mathbf u_1 = \mathbf v_1$ and

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

where $\langle \mathbf v_2, \mathbf u_1 \rangle$ is the dot product of $\mathbf v_2$ and $\mathbf u_1$.

Now, this will give you two vectors $\mathbf u_1$ and $\mathbf u_2$ which are orthogonal, but not normalized. Use the normalization formula above to get an orthonormal system.

If you want to, you can apply the normalization to $\mathbf u_1$ before calculating $\mathbf u_2$. This way, you can skip the denominator in the formula, since if $\hat{\mathbf u}$ is normalized, $\langle \hat{\mathbf u}, \hat{\mathbf u} \rangle = 1$. You will however need to normalize $\mathbf u_2$ when you have calculated it.

For the second part, you want to find a three-dimensional vector $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$ such that the dot product with a vector $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ satisfying $3x - 2y + z = 0$ is zero. Calculate the dot product. Can you identify $a, b, c$? When you have found $a,b,c$, normalize the vector.

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Here is how I teach Gram-Schmidt. You are trying to find an orthogonal basis from a given set. The first vector you can always keep the same because there aren't conditions.

So $u_1=\left(\begin{matrix}1 \\ 2 \end{matrix}\right)$

Next, you know that there is some combination $u_2=\left(\begin{matrix} 2 \\ 1 \end{matrix}\right)+a\left(\begin{matrix} 1 \\ 2 \end{matrix}\right)$ that makes this orthogonal to the first one.

Let's check by taking the dot product $0=u_1^Tu_2=2+2+a(1+4)=4+5a$. This tells us that $a=-4/5$, or that $u_2=\left(\begin{matrix} 2 \\ 1 \end{matrix}\right)-\left(\begin{matrix} 8/5 \\ 4/5\end{matrix}\right)=\left(\begin{matrix} 2/5 \\ -1/5 \end{matrix}\right)$.

At this point we only care about orthogonal, so we can multiply by a constant and not change whether or not they are orthogonal, so let's multiply by $5$ to get $u_2=\left(\begin{matrix} 2 \\ -1 \end{matrix}\right)$. We now have an orthogonal basis using Gram-Schidt. To get an orthonormal basis we just divide by the length to get the set:

$1/\sqrt{5}\left(\begin{matrix} 1 \\ 2 \end{matrix}\right)$ and $1/\sqrt{5}\left(\begin{matrix} 2 \\ -1 \end{matrix}\right)$

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For (2), if the question is to find a normalized vector orthogonal to all vectors in $\mathbf{R}^3$ that satisfy $3x - 2y + z= 0$, then the solution is already there: just normalize $(3,-2,1)$ and the result is $(3/\sqrt{14}, -2/\sqrt{14}, 1/\sqrt{14})$.

Because the given equation is a plane in $\mathbf{R^3}$, and the fundamental theorem of the subspaces of a matrix tells us that the "row space" of the given $1\times 3$ matrix $(3, -2, 1)$ is orthogonal to its nullspace. All the vectors satisfy the equation lie in that nullspace.

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For 1, normalize $(1,2)^T$: what is its length? Then your book should have an equation for the second, if not Wikipedia should help. Can you calculate the length of $(1,2)^T$?

For 2, what is g? Ignoring it, can you find a vector such that $f(\vec{t})=0?$ Then normalize it. Just pick an $x,y$, find $z$...

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