# Triangle Inequality in 'Linear Algebra Done Right'

1. Nov 20, 2013

### TheOldHag

I'm stuck on one aspect of the proof on page 105 of the 2nd edition. Equation 6.13 is necessary for the inequality to be an equality as it says but they never seem to account for inequality 6.11. Specifically, I don't see how this satisfies 2 Re<u,v> = 2 |<u,v>|

Thanks for any guidance.

2. Nov 20, 2013

### Office_Shredder

Staff Emeritus
Can you explain your question in a way that doesn't require having the book?

3. Nov 20, 2013

### Jorriss

It's not an equality; it is ≤.

2Re< u, v > ≤ 2|<u, v>|

Last edited: Nov 20, 2013
4. Nov 21, 2013

### TheOldHag

They gave as a condition for this to be an equality that u would be a scalar multiple of v or else v a scalar multiple of u. But they never showed how 2 Re <u,v> would be equal to 2 | <u,v> | in that case.

5. Nov 21, 2013

### R136a1

Just plug in $\mathbf{v}=\alpha \mathbf{u}$ and see what happens.

6. Nov 22, 2013

### TheOldHag

I've given that a try and still have not been able to get it. I'm sure there is some simple connection I'm missing. I'm starting with

|<u, au>| =
| a*u1*comp(u1) + a*u2*comp(u2) + ... + a*uN*comp(uN) | =
| a |u1|^2 + a |u2|^2 + ... + a |uN|^2 |

And from here I have no idea where that is taking me. Similar result form the other side.

7. Nov 22, 2013

### TheOldHag

I think I got it but still a bit confused. This assumes a and hence complex conjugate of a are real numbers and hence the imaginary part in the equations below ends up being zero and dropping off. So is the case that this is an equality only if one vector is a real scalar multiple of the other?

$|<u, au>|$ =
$\sqrt{(Re <u, au>)^{2} + (Im <u, au>)^{2}}$ =
$\sqrt{(Re \overline{a} <u, u>)^{2} + (Im \overline{a} <u, u>)^{2}}$ =
$\sqrt{(Re \overline{a} \left\|u\right\|^{2})^{2} + (Im \overline{a} \left\|u\right\|^{2})^{2}}$ =
$\sqrt{(Re \overline{a} \left\|u\right\|^{2})^{2}}$ =
$Re \overline{a} \left\|u\right\|^{2}$ =
$Re <u, au>$