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

In summary, if you want to show that two complex numbers are equal, you need to show that their inner products are equal.
  • #1
s_j_sawyer
21
0

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?
 
Physics news on Phys.org
  • #2
s_j_sawyer said:

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.
s_j_sawyer said:
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?
 
  • #3
Mark44 said:
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.
 
  • #4
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!
 
  • #5
have you tried assuming a complex form, say w = aj + b and expanding the inner products?
 
  • #6
[tex] \|w + z\|^2 [/tex] is not equal to [tex] \langle w + z, w + z \rangle [/tex] unless you're in a real vector space. But since these are complex, then

[tex]
\|w + z\|^2 = \langle w + z, \overline{w + z} \rangle
[/tex]

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

[tex]
w + \overline{w} = a + bi + a - bi = 2a = 2\operatorname{Re}(w).
[/tex]

Also, [tex]\operatorname{Re}(w) \le \|w\| [/tex]

These should hopefully get you on the right track.
 

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

What does ||w + z|| represent in the inequality ||w + z|| <= ||w|| + ||z|| for complex vectors?

The notation ||w + z|| represents the magnitude or length of the complex vector w + z. This is calculated using the Pythagorean theorem, where the square root of the sum of the squares of the real and imaginary components of w + z is taken.

How is the inequality ||w + z|| <= ||w|| + ||z|| proven for complex vectors?

The inequality can be proven using the triangle inequality property, which states that the length of any side of a triangle must be less than or equal to the sum of the lengths of the other two sides. In this case, the vector w + z can be thought of as the hypotenuse of a triangle with sides w and z, and the inequality follows from the triangle inequality property.

Can the inequality ||w + z|| <= ||w|| + ||z|| be extended to more than two complex vectors?

Yes, the inequality can be extended to any number of complex vectors using the same logic. The magnitude of the sum of all the vectors will always be less than or equal to the sum of the magnitudes of each individual vector.

Is the inequality ||w + z|| <= ||w|| + ||z|| true for all complex vectors?

Yes, the inequality holds true for all complex vectors. This can be proven using mathematical induction, where the base case for two vectors is established and then the inequality is shown to hold for any additional vectors added to the sum.

How is the inequality ||w + z|| <= ||w|| + ||z|| useful in mathematics and science?

The inequality has various applications in mathematics and science, such as in proving the convergence of series and in the analysis of complex systems. It is also used in vector calculus and linear algebra to establish important properties of vector spaces. In physics, the inequality is used in the analysis of forces and velocities in complex systems.

Similar threads

Replies
7
Views
1K
Replies
13
Views
2K
Replies
10
Views
2K
Replies
17
Views
2K
Replies
2
Views
3K
Replies
10
Views
2K
Replies
6
Views
978
Back
Top