Square root of matrix

  • Thread starter aaaa202
  • Start date
  • #1
1,170
3

Main Question or Discussion Point

I'm doing an online course in quantum information theory, but it seems to require some knowledge of linear algebra that I don't have.
A definition that popped up today was the definition of the absolute value of a matrix as:

lAl = √(A*A) , where * denotes conjugate transpose.

Now for a given matrix I can calculate the product A*A, but how is the square root of this defined? I have no idea, though I think it should be basis independent, so maybe square root of the trace.
A*A is clearly positive semidefinite but I don't know if I can use that for anything.
 

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
11,324
4,487
I'm doing an online course in quantum information theory, but it seems to require some knowledge of linear algebra that I don't have.
A definition that popped up today was the definition of the absolute value of a matrix as:

lAl = √(A*A) , where * denotes conjugate transpose.

Now for a given matrix I can calculate the product A*A, but how is the square root of this defined? I have no idea, though I think it should be basis independent, so maybe square root of the trace.
A*A is clearly positive semidefinite but I don't know if I can use that for anything.
What's the definition of the square root of a number?
 
  • #3
1,170
3
Well I guess that the square root of a matrix A is then a matrix B such that B^2 = A. Well wouldn't it be better with a matrix B such that B*B = A. And is this definition basis independent.
 
  • #4
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
11,324
4,487
Well I guess that the square root of a matrix A is then a matrix B such that B^2 = A. Well wouldn't it be better with a matrix B such that B*B = A. And is this definition basis independent.
That would be like requiring the square root of a complex number ##a## to satisfy ##b^*b = a##
 
  • #5
1,170
3
Ok well is the definition basis independent?
 
  • #6
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
11,324
4,487
Ok well is the definition basis independent?
Let me rephrase that question. If a linear transformation T is represented by the matrix A in one basis and A' in another. Then is |A'| = |A|'?

What do you think?
 
  • #7
1,170
3
I am not sure. I need to show that:

(√(A))' = √(A')
B' = √(A')

Now if A' = UAU maybe I can use that somehow...
 
  • #8
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
11,324
4,487
I am not sure. I need to show that:

(√(A))' = √(A')
B' = √(A')

Now if A' = UAU maybe I can use that somehow...
Or, think about diagonalizing A*A. It must be true if you want to accept that and get back to your quantum theory!
 
  • #9
1,170
3
Maybe something like this:

A' = UAU
B' = UBU

Now B2 = A

We want to show that:
√A' = (√A)' = B'

So is B'2 = A'?

Well B'2 = UBUUBU = UB2U = UAU = A'

But all this requires that A and A' and B and B' are related by a basis change with a unitary as above. When is this true?
 
  • #10
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
11,324
4,487
That's essentially the proof. The only technicality is that B' must be the right square root. That probably depends on showing that a positive semi-definite matrix has a unique positive semi-definite square root.
 
  • #11
177
61
The square root of a positive semidefinite Hermitian matrix ##A## is the unique positive semidefinite matrix ##B## such that ##B^2=A##. You can look at Ch. VI s.3 of "Linear algebra done wrong" for an explanation of why such matrix exists and why it is unique.
 

Related Threads for: Square root of matrix

  • Last Post
Replies
4
Views
12K
  • Last Post
Replies
7
Views
5K
Replies
2
Views
2K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
8
Views
36K
  • Last Post
2
Replies
27
Views
7K
Top