Prove ||w + z|| <= ||w|| + ||z|| for complex vectors

[s_j_sawyer]

Homework Statement

Let z, w be complex vectors of C^n.

Prove ||w + z|| <= ||w|| + ||z||

(using the standard inner product for C^n)
(i.e. <w,z> = w*z', where * is the dot product and ' denotes the complex conjugate)

The Attempt at a Solution

Well, I found that

||w + z||
= sqrt( w*w' + z*z' + z*w' + w*z')
= sqrt( w*w' + z*z' + <w,z> + <w,z>' )

and

||w|| + ||z||
= sqrt(w*w' + z*z')

So by showing that <w,z> + <w,z>' <= 0 then I guess that will finish the proof but I am unsure of how to do this.

Any help?

 

[Mark44]

Homework Statement

Let z, w be complex vectors of C^n.

Prove ||w + z|| <= ||w|| + ||z||

(using the standard inner product for C^n)
(i.e. <w,z> = w*z', where * is the dot product and ' denotes the complex conjugate)

The Attempt at a Solution

Well, I found that

||w + z||
= sqrt( w*w' + z*z' + z*w' + w*z')
= sqrt( w*w' + z*z' + <w,z> + <w,z>' )

and

||w|| + ||z||
= sqrt(w*w' + z*z')

The equation above isn't true. ||w|| = sqrt(w*w') and ||z|| = sqrt(z*z'), but the two square roots don't add to sqrt(w*w' + z*z'). This is like saying that the lengths of two sides of a right triangle add up to the length of the hypotenuse.

So by showing that <w,z> + <w,z>' <= 0 then I guess that will finish the proof but I am unsure of how to do this.

Any help?

 

[s_j_sawyer]

The equation above isn't true. ||w|| = sqrt(w*w') and ||z|| = sqrt(z*z'), but the two square roots don't add to sqrt(w*w' + z*z'). This is like saying that the lengths of two sides of a right triangle add up to the length of the hypotenuse.

Oooohhh deeaarr how embarrassing. Thank you for that insightful information. I should be able to get it now.

 

[s_j_sawyer]

Sorry to double post but I didn't think simply editing would bump this up.

Alright so I played around with this some more but still could not get it to work.

I tried to prove this using the method of squaring both sides for the regular 'scalar triangle inequality' proof like so:

||w + z||^2 = (sqrt<w + z,w + z>)^2
= <w + z, w + z>
= <w,w> + <w,z> + <z,w> + <z,z>
= ||w||^2 + <w,z> + <w,z>' + ||z||^2

and so if I can show that

<w,z> + <w,z>' <= 2*||w||*||z||

then I would have

||w + z||^2 <= (||w|| + ||z||)^2

but I have been unable to do so. Any assistance would be greatly appreciated!

 

[lanedance]

have you tried assuming a complex form, say w = aj + b and expanding the inner products?

 

[xeno_gear]

$$\|w + z\|^2$$ is not equal to $$\langle w + z, w + z \rangle$$ unless you're in a real vector space. But since these are complex, then

$$\|w + z\|^2 = \langle w + z, \overline{w + z} \rangle$$

Some hints:
If $$w$$ is a complex number (or vector, or whatever), then $$w = a + bi$$, where $$a$$ and $$b$$ are real numbers (or vectors), and

$$w + \overline{w} = a + bi + a - bi = 2a = 2\operatorname{Re}(w).$$

Also, $$\operatorname{Re}(w) \le \|w\|$$

These should hopefully get you on the right track.