# Help with norms

1. Apr 11, 2010

### 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.

2. Apr 11, 2010

### VeeEight

Re: Norms

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

3. Apr 11, 2010

### Staff: Mentor

Re: Norms

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

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

4. Apr 11, 2010

### Dustinsfl

Re: Norms

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

5. Apr 11, 2010

### Dustinsfl

Re: Norms

It is between 0 and 1, then?

Last edited: Apr 11, 2010
6. Apr 11, 2010

### Dustinsfl

Re: Norms

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

7. Apr 11, 2010

### VeeEight

Re: Norms

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?

8. Apr 11, 2010

### Dustinsfl

Re: Norms

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?

9. Apr 11, 2010

### VeeEight

Re: Norms

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.