# Homework Help: Why A'*A=I but A*A' !=I

1. Jan 22, 2010

### tom08

1. The problem statement, all variables and given/known data

I encounter a strange problem.

Let A= [1.0000 0 0
0 0 0
0 0 0.4472
0 0.3162 0
0 0.9487 0
0 0 0.8944]

I am surprise to find that A'*A=I, but A*A'<>I . Can anyone give me an explanation?

2. Jan 22, 2010

### rock.freak667

A is a 6x3 matrix, so A is not square. A is not invertible.

If AB=I, then A=B-1 provided that A and B are invertible to begin with.

3. Jan 23, 2010

### tom08

Thanks for ur kind reply. but must A and B be square such that the following equation holds ?

(A'B)'=B'*A

I think that if A*A'=I (wheter is square or not), we take transpose operator on both sides of the equation, and obtain that

(A*A')' = I'

then

A*A' = I

4. Jan 23, 2010

### rock.freak667

Sorry I was reading A' as A-1 not AT

5. Jan 23, 2010

### tom08

thank u so much.

BTW, is there an upper bound for |A'*A-A*A'|

when A is a rectangluar column orthogonal matrix?

6. Jan 23, 2010

### vela

Staff Emeritus
If you have

$$A^TA = I$$

then

$$(A^TA)^T=A^T(A^T)^T=A^TA=I$$.

You're starting with $AA^T$, which isn't equal to the identity matrix, so its transpose won't be either.

7. Jan 23, 2010

### tom08

Thank u, vela and rock. i realize my mistake.