LINEAR ALGEBRA: What is the relationship between ||y||^2 and yx?

[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...

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

But how do I use this to relate the two quantities? Please help, thanks!

 

[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.

 

[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?

 

[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?)

 

[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?

 

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

 

[quasar987]

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

 

[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?

 

[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?

 

[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!

You need to re-read your notes carefully I think!

 

[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 $$||\vec{y}||^2$$)

 

[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
$$\overrightarrow{y}\,\cdot\,\overrightarrow{x}$$

 

[quasar987]

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 $$||\vec{y}||^2$$)

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

 

[VinnyCee]

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

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

 

[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.

 

[VinnyCee]

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

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

 

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

 

[VinnyCee]

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

 

[quasar987]

Yes! phew. lol

 

[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?

 

[quasar987]

But you can simplify even more, because $\vec{a}=\hat{v}$ is unitary. That means that $||\vec{a}||=||\hat{v}||=1$. And what is $\vec{a}\cdot\vec{a}$ is terms of $||\vec{a}||$ again?

 

[VinnyCee]

$$\vec{a}\,\cdot\,\vec{a}\,=\,||\vec{a}||^2$$

$$(1)^2\,=\,1$$

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

 

[quasar987]

Beautiful. Beautiful.

Now $\vec{x}\cdot\vec{y}$.

How can you rewrite $\left[\left(\overrightarrow{v}\,\cdot\,\overrightarrow{x }\right)\,\overrightarrow{v}\right]\,\cdot\,\overrightarrow{x}$ using the associativity property?

(By associativity property I just mean $\vec{a}\cdot \alpha\vec{b}= \alpha\vec{a}\cdot \vec{b}= \alpha(\vec{a}\cdot \vec{b})$ because it does not matter with what you "associate" alpha, the answer is the same)

 

[VinnyCee]

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

But you cannot take the dot product with a scalar, right?

 

[mathwonk]

Pythagoras.

 

[VinnyCee]

$$||\vec{x}\,+\,\vec{y}||^2\,=\,||\vec{x}||^2\,+\,||\vec{y}||^2\,\iff\,\vec{x}\,and\,\vec{y}\,are\,orthogonal.$$

But how do I use that for this relation: $$\left[(\hat{v}\,\cdot\,\vec{x})\,\hat{v}\right]\,\cdot\,\vec{x}$$?

 

[quasar987]

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

But you cannot take the dot product with a scalar, right?

Right. Which proves that you didn't apply the property correctly. Replace $(\vec{x}\cdot \hat{v})$ by $\alpha$ as you did before if it confuses you. You will get an expression that allows you to apply directly $\alpha\vec{a}\cdot \vec{b}= \alpha(\vec{a}\cdot \vec{b})$

 

[VinnyCee]

$$\alpha\,=\,\hat{v}\,\cdot\,\vec{x}$$

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

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

$$\alpha\,(\alpha)\,=\,\alpha^2$$

 

[AKG]

This is really getting over-complicated. What is the definition of proj_vx? You should probably have something like (^v.x/_v.v)v. So simply compute:

|y|²
= y.y
= (proj_vx).(proj_vx)
= [(^v.x/_v.v)v].[(^v.x/_v.v)v]
= COMPUTE THIS

y.x
= (proj_vx).x
= [(^v.x/_v.v)v].x
= COMPUTE THIS

The only facts you need to use are:

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

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

where $\alpha ,\, \beta$ are scalars and $\vec{a} ,\, \vec{b}$ are obviously vectors. The second fact is really just the first fact with $\beta = 1$. Just use the fact that v.v, v.x, (^v.x/_v.v), etc. are just scalars, like $\alpha$ and $\beta$.

 

[quasar987]

$$\alpha\,=\,\hat{v}\,\cdot\,\vec{x}$$

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

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

$$\alpha\,(\alpha)\,=\,\alpha^2$$

Yep. Now that the confusing algebra is over, you can substitute back $\alpha \rightarrow (\hat{v}\cdot \vec{x})$, and compare with what you obtained for $||\vec{y}||^2$ (post #22).

30+ posts for a 2 lines problem, nice, lol.