# Homework Help: Linear Algebra (Projection)

1. Jul 18, 2009

I apologize for the excessive use of Latex, but for this particular problem I think the notation would be extremely difficult to read otherwise. I usually try to keep my use of Latex to a minimum.

1. The problem statement, all variables and given/known data

$$\text{Let } \mathbb{C}^3 \text{ be equipped with the standard inner product and let } \text{\textit{W}} \text{ be the subspace of } \mathbb{C}^3 \text{ that is spanned by }$$
$$\pmb{u}_1=(1,0,1) \text{ and } \pmb{u}_2=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right) \text{. Project the vector } \pmb{v}=(1,i,i) \text{ onto } \text{\textit{W}}.$$

2. Relevant equations

None

3. The attempt at a solution

Here is what I have so far. I think I'm missing something, though.

$$\pmb{\hat{\pmb{u}}}_1=\frac{\pmb{u}_1}{\left\|\pmb{u}_1\right\|}=\frac{(1,0,1)}{\sqrt{2}}=\left(\sqrt{2},0,\sqrt{2}\right)$$
$$\pmb{\hat{\pmb{u}}}_2=\frac{\pmb{u}_2}{\left\|\pmb{u}_2\right\|}=\frac{\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)}{1}=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$$
$$\pmb{v}=(1,i,i)=a\pmb{\hat{\pmb{u}}}_1+b\pmb{\hat{\pmb{u}}}_2=a\left(\sqrt{2},0,\sqrt{2}\right)+b\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)=\left(a\sqrt{2}+b\frac{1}{\sqrt{3}},b\frac{1}{\sqrt{3}},a\sqrt{2}-b\frac{1}{\sqrt{3}}\right)$$
$$b\frac{1}{\sqrt{3}}=i\Rightarrow b=i\sqrt{3}$$
$$1=a\sqrt{2}+b\frac{1}{\sqrt{3}}=a\sqrt{2}+i\Rightarrow a=\frac{1-i}{\sqrt{2}}$$
$$v=\frac{1-i}{\sqrt{2}}\pmb{\hat{\pmb{u}}}_1+i\sqrt{3}\pmb{\hat{\pmb{u}}}_2$$

Last edited: Jul 18, 2009
2. Jul 18, 2009

### Office_Shredder

Staff Emeritus
You started off by assuming that v is contained in W, which is a mistake. You need to find a vector orthogonal to W, write v in terms of u1, u2 and that new third vector (which will necessarily be spanning C3 at this point) and then eliminate the component that's not in W.

There's a slick way to do this by noting that since u1, u2 and the third vector are all orthogonal, you can use inner products to determine what the coefficients will be by taking the inner product of v with u1 and u2, this is either described in your text already or you can figure it out in a fairly straightforward manner

3. Jul 18, 2009

### Pengwuino

To add to that, you did not normalize u1 correctly. Also, why are you normalizing them? The question seems to be asking for what the vector is in terms of these vectors that aren't normalized. You can do one of two things - 1) don't normalize and continue with what Office_Shredder suggested and answer with the basis given or 2) normalize and keep in mind that you're dealing with a new basis when answering the question.

4. Jul 18, 2009

Thank you both for your input. I have started reworking the problem using Pengwuino's suggestion of not normalizing the vectors. I also took the advice of Office_Shredder by finding an orthogonal vector.

Things are looking good, but I don't quite understand the trick with inner products. Could you offer a bit more explanation?

Thanks again!

5. Jul 18, 2009

Wait, I think I might understand what you mean now. So, I can take the inner product of v with u1 and u2 and those two values will be the two projection values, right? So do I even need to find that third orthogonal vector?

6. Jul 18, 2009

### Office_Shredder

Staff Emeritus
If v=au1 + bu2+cu3 and the ui's are orthogonal, to find a you can look at <v,u1> = a<u1,u1> (see this by using the linearity of the inner product with the right hand side of the first equation

7. Jul 18, 2009

### Pengwuino

The inner product allows you to find, in a sense, how much of a vector is on another vector. For example, if you have 2 vectors A = (0,0,1) and B = (0,3,0) and you want to find the projection of the vector B onto a new vector C = (0,6,3) , you dot B into C with the usual dot product and divide by the norm of B, that is $$\frac{{B \cdot C}}{{|B|^2 }}$$, you'll get "how much" of B is in C. You'll obviously get 2 which means that you can express C = 2B + 3C if you do the same for the projection of A onto C (getting 3).

8. Jul 18, 2009

Okay, how is this? Is there anything special I should do since I am dealing with complex vectors?

$$\pmb{u}_1=(1,0,1)$$

$$\pmb{u}_2=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)\$$

$$\left\langle \pmb{v},\pmb{u}_1\right\rangle =1+0+i=1+i$$

$$\left\langle \pmb{v},\pmb{u}_2\right\rangle =\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}i-\frac{1}{\sqrt{3}}i=\frac{1}{\sqrt{3}}$$

$$v=(1+i)\pmb{u}_1+\frac{1}{\sqrt{3}}\pmb{u}_2$$

Last edited: Jul 18, 2009
9. Jul 18, 2009