# Square root of matrix

1. Jan 18, 2016

### aaaa202

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.

2. Jan 18, 2016

### PeroK

What's the definition of the square root of a number?

3. Jan 18, 2016

### aaaa202

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. Jan 18, 2016

### PeroK

That would be like requiring the square root of a complex number $a$ to satisfy $b^*b = a$

5. Jan 18, 2016

### aaaa202

Ok well is the definition basis independent?

6. Jan 18, 2016

### PeroK

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. Jan 18, 2016

### aaaa202

I am not sure. I need to show that:

(√(A))' = √(A')
B' = √(A')

Now if A' = UAU maybe I can use that somehow...

8. Jan 18, 2016

### PeroK

Or, think about diagonalizing A*A. It must be true if you want to accept that and get back to your quantum theory!

9. Jan 18, 2016

### aaaa202

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. Jan 18, 2016

### PeroK

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. Jan 18, 2016

### Hawkeye18

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.