Norm and Rows of Complex Matrix

Click For Summary

Homework Help Overview

The discussion revolves around demonstrating that the rows of an n x n complex matrix A form an orthonormal basis for C^n, given that the norm of Ax equals the norm of x for any vector x in C^n.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various choices for the vector x, including columns of the adjoint of A, to analyze the implications of the norm condition. There are attempts to connect inner products and norms to establish orthogonality and unit length of the rows.

Discussion Status

Participants are actively engaging with the problem, questioning the validity of their approaches and seeking clarification on the connections between their calculations and the properties of the matrix. Some guidance has been offered regarding the implications of inner products, but no consensus has been reached on the next steps.

Contextual Notes

There is an emphasis on avoiding complex notation and ensuring clarity in reasoning, as participants express concerns about becoming bogged down in details. The original poster has not yet established a complete justification for their conclusions.

Yagoda
Messages
45
Reaction score
0

Homework Statement


I want to show that for an n x n matrix A with complex entries, if [itex]\left\|Ax\right\|=\left\|x\right\|[/itex] for any vector x in C^n, then the rows of A are an orthonormal basis of C^n.


Homework Equations





The Attempt at a Solution

All I've managed to do so far is show that the columns of A all have length 1, which you can get by taking x to be e1. Is there a strategy of showing this for the rows and for showing orthogonality without having to write out the terms of the matrix multiplication Ax (I tried this and quickly became bogged down in notation)?
 
Physics news on Phys.org
Can you think of interesting choices for the vector x based on the elements of A?
 
I tried letting x be a column of the adjoint of A. This way when you multiply Ax, if your x is the ith column of the adjoint then the ith entry of the vector Ax is equivalent to the inner product <x,x>. But when you go to take the inner product <Ax,Ax> to find the norm of Ax I end up with a jumble of sums of entries of A being multiplied by each other so I'm not sure where to go from here. Am I on the right track?

Thanks.
 
Okay, suppose I'm setting x to be the first columns of the adjoint of A. Then when I take <Ax,Ax> the first term is <x,x>*<x,x> and the rest of the terms are the inner product of the first row of A with another term of A. Since we know that [itex]\left\|Ax\right\|=\left\|x\right\|[/itex] is it too fast for me to make the jump to saying that <x,x> must equal 1 in order for ||Ax|| to equal ||x|| and subsequently all the other terms must be 0? It seems like there should be more justification in between those steps, but I can't seem to put my finger on it.
 
Yagoda said:
Okay, suppose I'm setting x to be the first columns of the adjoint of A. Then when I take <Ax,Ax> the first term is <x,x>*<x,x> and the rest of the terms are the inner product of the first row of A with another term of A.
Yes. If the row vectors of A are a1..an and x = a1T, you have <a1,a1>2+<a1,a2>2+... You know that <a1,a2>2 etc are non-negative, so...?
 
Ah, yes. I was getting bogged down in notation and missing the bigger picture. Thanks.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
6K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
4
Views
2K