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

by VinnyCee
Tags: algebra, linear, relationship, ||y||2
 P: 492 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... $$||\overrightarrow{y}||\,\leq\,||\overrightarrow{x}||\,\iff\,\overrighta rrow{x}\,\in\,V$$ But how do I use this to relate the two quantities? Please help, thanks!
 Sci Advisor HW Helper PF Gold P: 4,765 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.
 P: 492 Ok, I use that version and write: $$\,\left[\left(\overrightarrow{v}\,\cdot\,\overrightarrow{x}\right)\,\overrighta rrow{v}\right]\,\cdot\,\left[\left(\overrightarrow{v}\,\cdot\,\overrightarrow{x}\right)\,\overrighta rrow{v}\right]\,$$ $$\left[\left(\overrightarrow{v}\,\cdot\,\overrightarrow{x}\right)\,\overrighta rrow{v}\right]\,\cdot\,\overrightarrow{x}$$ This is two equations in two variables. Is the second one equal to zero?
HW Helper
PF Gold
P: 4,765

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

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?)
 P: 492 The squared norm of y is $$||\,\left(\overrightarrow{v}\,\cdot\,\overrightarrow{x}\right)^2\,\over rightarrow{v}^2\,||$$? $$\overrightarrow{y}\,\cdot\,\overrightarrow{y}\,=\,||\overrightarrow{y}| |^2$$ Does that sound right?
 Sci Advisor HW Helper PF Gold P: 4,765 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. $\vec{a}\cdot \alpha\vec{b}= \alpha\vec{a}\cdot \vec{b}= \alpha(\vec{a}\cdot \vec{b})$)
 Sci Advisor HW Helper PF Gold P: 4,765 As a general rule, stay away from notational shortcuts like $\vec{y}^2$ in a mathematics course. Always write the ||
 P: 492 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?
 P: 492 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?
 Sci Advisor HW Helper PF Gold P: 4,765 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!
 P: 492 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 $$||\vec{y}||^2$$)
 P: 197 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 $$\overrightarrow{y}\,\cdot\,\overrightarrow{x}$$
 Quote by 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 $$||\vec{y}||^2$$)
By the properties of the scalar product (or inner product or whatever you call the $\cdot$ operation). More precesily, use this property: $\vec{a}\cdot \alpha\vec{b}= \alpha\vec{a}\cdot \vec{b}= \alpha(\vec{a}\cdot \vec{b})$.
 P: 492 $$||\vec{y}||^2=(\vec{x}\cdot \hat{v})\hat{v}\cdot(\vec{x}\cdot \hat{v})\hat{v}$$ $$||\vec{y}||^2=(\vec{x} \cdot ||\vec{v}||^2)(\vec{x} \cdot ||\vec{v}||^2)=(||\vec{x}||^2 \cdot ||\vec{v}||^4)$$
 Sci Advisor HW Helper PF Gold P: 4,765 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.
 P: 492 $$||\vec{y}||^2=\alpha(\vec{a}) \cdot \alpha(\vec{a})$$ But what is the result of a dot product between two identical vectors?
 Sci Advisor HW Helper PF Gold P: 4,765 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: $$\vec{a}\cdot \alpha\vec{b}= \alpha\vec{a}\cdot \vec{b}= \alpha(\vec{a}\cdot \vec{b})$$
 P: 492 $$||\vec{y}||^2=\alpha^2 \vec{a} \cdot \vec{a}$$?