Help with Norms: x, y, and Dot Product

[Dustinsfl]

If $$\lVert x \rVert=2$$ and $$\lVert y \rVert=3$$, what if anything, can we conclude about the possible values of $$\left\vert \mathbf{x}^T\mathbf{y} \right\vert$$?

I don't think anything can be concluded since the dot product can still end being positive or negative.

 

[VeeEight]

Do you know the formula for the dot product involving cosine?

 

[Mark44]

If $$\lVert x \rVert=2$$ and $$\lVert y \rVert=3$$, what if anything, can we conclude about the possible values of $$\left\vert \mathbf{x}^T\mathbf{y} \right\vert$$?

I don't think anything can be concluded since the dot product can still end being positive or negative.

x $\cdot$ y = ||x|| ||y|| cos($\theta$).

Can you conclude something about |x $\cdot$ y| now?

 

[Dustinsfl]

$$u \cdot v=\lVert v \rVert\lVert u \rVert cos(\theta)$$

 

[Dustinsfl]

It is between 0 and 1, then?

 

[Dustinsfl]

Theta is between -pi/2 and pi/2?

 

[VeeEight]

Plug in all the values you know. Then consider the range of cosine. What values can it take? Knowing this, what values can ||x|| ||y|| cos theta take?

 

[Dustinsfl]

For cosine to be positive, theta is between, and including, -pi/2 to pi/2. Therefore, the right side of equation will be between 0 to 1 times the magnitude of x times the magnitude of y?

 

[VeeEight]

Yes, it is simpler to just write it out, however.

xy = ||x|| ||y|| cos a = 6 cos a.
cos a is between -1 and 1, so xy is in [-6, 6], and so |xy| is just the positive terms in that interval.