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

• VinnyCee
In summary: The dot product (or inner product) between two vectors is just the product of the vectors themselves.
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...

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

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.

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?

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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?)

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?

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

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

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?

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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?

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!

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

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

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VinnyCee said:
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: $\vec{a}\cdot \alpha\vec{b}= \alpha\vec{a}\cdot \vec{b}= \alpha(\vec{a}\cdot \vec{b})$.

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

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

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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.

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

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

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

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

Yes! phew. lol

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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?

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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?

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

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

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

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)

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

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

Pythagoras.

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

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

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

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

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

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

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

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

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

|y|2
= y.y
= (projvx).(projvx)
= [(v.x/v.v)v].[(v.x/v.v)v]
= COMPUTE THIS

y.x
= (projvx).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 $\beta$.

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

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

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

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

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.

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## What is the relationship between ||y||^2 and yx?

The relationship between ||y||^2 and yx is that ||y||^2 is equal to the dot product of y with itself, or y⋅y. This is also equivalent to yx multiplied by its transpose, yx^T.

## How is ||y||^2 calculated?

||y||^2 is calculated by taking the dot product of y with itself. This involves multiplying each element in y with its corresponding element in y, and then summing all of these products together.

## What is the significance of ||y||^2 in linear algebra?

||y||^2 is significant in linear algebra because it represents the magnitude or length of the vector y. It is also used in various calculations, such as finding the distance between two vectors or determining if two vectors are orthogonal.

## Is ||y||^2 always positive?

Yes, ||y||^2 is always positive. This is because it is the sum of squared values, which cannot be negative. Additionally, the square root of ||y||^2, which is ||y||, is always positive or zero.

## Can ||y||^2 be greater than yx?

No, ||y||^2 cannot be greater than yx. This is because yx is calculated by multiplying the elements in y with their corresponding elements in x, and then summing these products together. Since ||y||^2 is equal to the dot product of y with itself, it would always be equal to or less than yx.

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