# LINEAR ALGEBRA: What is the relationship between ||y||^2 and y . x?

1. Nov 6, 2006

### VinnyCee

QUESTION (5.1, #30 -> Bretscher, O.):

Consider a subspace V of $$\mathbb{R}^n$$ and a vector $$\overrightarrow{x}$$ in $$\mathbb{R}^n$$. Let $$\overrightarrow{y}\,=\,proj_v\,\overrightarrow{x}$$. What is the relationship between the following quantities?

$$||\overrightarrow{y}||^2$$ and $$\overrightarrow{y}\,\cdot\,\overrightarrow{x}$$

My work so far:

I only know this for sure...

$$||\overrightarrow{y}||\,\leq\,||\overrightarrow{x}||\,\iff\,\overrightarrow{x}\,\in\,V$$

2. Nov 6, 2006

### quasar987

I assume $\hat{v}$ is a unit vector? Then $\vec{y}=(\vec{x}\cdot \hat{v})\hat{v}$. I think this "substitution" is a good place to start.

3. Nov 6, 2006

### VinnyCee

Ok, I use that version and write:

$$\,\left[\left(\overrightarrow{v}\,\cdot\,\overrightarrow{x}\right)\,\overrightarrow{v}\right]\,\cdot\,\left[\left(\overrightarrow{v}\,\cdot\,\overrightarrow{x}\right)\,\overrightarrow{v}\right]\,$$

$$\left[\left(\overrightarrow{v}\,\cdot\,\overrightarrow{x}\right)\,\overrightarrow{v}\right]\,\cdot\,\overrightarrow{x}$$

This is two equations in two variables.

Is the second one equal to zero?

Last edited: Nov 6, 2006
4. Nov 6, 2006

### quasar987

write the squared norm of y as $||\vec{y}||^2=\vec{y}\cdot\vec{y}$.

and simplify <y,x> a little also.

the relationship should become clear.

(Why would you think the second one equals to zero?)

5. Nov 6, 2006

### VinnyCee

The squared norm of y is $$||\,\left(\overrightarrow{v}\,\cdot\,\overrightarrow{x}\right)^2\,\overrightarrow{v}^2\,||$$?

$$\overrightarrow{y}\,\cdot\,\overrightarrow{y}\,=\,||\overrightarrow{y}||^2$$

Does that sound right?

Last edited: Nov 6, 2006
6. Nov 6, 2006

### quasar987

As I said, the squared norm (or norm squared if you'd like) of $\vec{y}$ is $||\vec{y}||^2=\vec{y}\cdot\vec{y}$.

And as I said in post #2, $\vec{y}=(\vec{x}\cdot \hat{v})\hat{v}$. So

$$||\vec{y}||^2=(\vec{x}\cdot \hat{v})\hat{v}\cdot(\vec{x}\cdot \hat{v})\hat{v}$$.

Now use the properties of the scalar product to make that look nicer (e.g. $\vec{a}\cdot \alpha\vec{b}= \alpha\vec{a}\cdot \vec{b}= \alpha(\vec{a}\cdot \vec{b})$)

7. Nov 6, 2006

### quasar987

As a general rule, stay away from notational shortcuts like $\vec{y}^2$ in a mathematics course. Always write the ||

8. Nov 6, 2006

### VinnyCee

Like the first equation in post #3?

Scalar product properties?

So...

$$\hat{v}\,\left(\overrightarrow{x}\,\cdot\,\hat{v}\right)^2$$ is the norm squared?

Last edited: Nov 6, 2006
9. Nov 6, 2006

### VinnyCee

I am guessing that this might have something to do with the Cauchy-Schwarz inequality since it is in the section and all, right?

$$|\vec{x}\,\cdot\,\vec{y}|\,\leq\,||\vec{x}||\,||\vec{y}||$$

How do I use the inequality for this problem? Just start substituting?

10. Nov 6, 2006

### quasar987

The first equation in post #3 doesn't make much sense because the ||a|| notation means "norm of vector a", but in your first equation of post 3, you have a scalar in btw the ||.||. You don't even need absolute values either because $\vec{y}\cdot\vec{y}=y_x^2+y_y^2+y_z^2$ can never be negative.

So the equation to work with is $||\vec{y}||^2=(\vec{x}\cdot \hat{v})\hat{v}\cdot(\vec{x}\cdot \hat{v})\hat{v}$

And if that is the norm squared, how did you pass from that to $\hat{v}\,\left(\overrightarrow{x}\,\cdot\,\hat{v}\ \right)$ as the norm squared?! This is not even a scalar!

Last edited: Nov 6, 2006
11. Nov 6, 2006

### VinnyCee

So we have:

$$||\vec{y}||^2=(\vec{x}\cdot \hat{v})\hat{v}\cdot(\vec{x}\cdot \hat{v})\hat{v}$$

But how do you simplify that expression? (without just going back to $$||\vec{y}||^2$$)

Last edited: Nov 6, 2006
12. Nov 6, 2006

### xAxis

You can simplify it in many ways, but you must compare your RHS
$$(\vec{x}\cdot \hat{v})\hat{v}\cdot(\vec{x}\cdot \hat{v})\hat{v}$$
with
$$\overrightarrow{y}\,\cdot\,\overrightarrow{x}$$

Last edited: Nov 6, 2006
13. Nov 6, 2006

### quasar987

By the properties of the scalar product (or inner product or whatever you call the $\cdot$ operation). More precesily, use this property: $\vec{a}\cdot \alpha\vec{b}= \alpha\vec{a}\cdot \vec{b}= \alpha(\vec{a}\cdot \vec{b})$.

Last edited: Nov 6, 2006
14. Nov 6, 2006

### VinnyCee

$$||\vec{y}||^2=(\vec{x}\cdot \hat{v})\hat{v}\cdot(\vec{x}\cdot \hat{v})\hat{v}$$

$$||\vec{y}||^2=(\vec{x} \cdot ||\vec{v}||^2)(\vec{x} \cdot ||\vec{v}||^2)=(||\vec{x}||^2 \cdot ||\vec{v}||^4)$$

Last edited: Nov 6, 2006
15. Nov 6, 2006

### quasar987

Sniff. Where do you pull all these equations from? Your last equation doesn't make sense and you should see it! You're taking the dot product of scalars!

You have the template right here:

$$\vec{a}\cdot \alpha\vec{b}= \alpha\vec{a}\cdot \vec{b}= \alpha(\vec{a}\cdot \vec{b})$$

Except in your case, $\alpha = (\vec{x}\cdot \hat{v})$ and $\vec{a}=\vec{b}=\hat{v}$. And also, both a and b are multiplied by a constant.

16. Nov 6, 2006

### VinnyCee

$$||\vec{y}||^2=\alpha(\vec{a}) \cdot \alpha(\vec{a})$$

But what is the result of a dot product between two identical vectors?

Last edited: Nov 6, 2006
17. Nov 6, 2006

### quasar987

You've only rewritten the equation by changing the notation. You're not more advanced.

Use the property of the dot product that allows you to take the alpha's and pull them in front: $$\vec{a}\cdot \alpha\vec{b}= \alpha\vec{a}\cdot \vec{b}= \alpha(\vec{a}\cdot \vec{b})$$

18. Nov 6, 2006

### VinnyCee

$$||\vec{y}||^2=\alpha^2 \vec{a} \cdot \vec{a}$$?

19. Nov 6, 2006

### quasar987

Yes! phew. lol

Last edited: Nov 6, 2006
20. Nov 6, 2006

### VinnyCee

Now what? Do the same for $$\overrightarrow{y}\,\cdot\,\overrightarrow{x}$$?

$$(\vec{x}\cdot \hat{v})^2\,\hat{v} \cdot \hat{v}$$

What qualities does the $$\vec{a} \cdot \vec{a}$$ part have?

Last edited: Nov 6, 2006