Understanding the Inverse of a Matrix: A'*A=I but A*A'<>I

[tom08]

Homework Statement

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?

 

[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.

 

[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

 

[rock.freak667]

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

Sorry I was reading A' as A^-1 not A^T

But read this, orthogonal matrices

 

[tom08]

thank u so much.

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

when A is a rectangluar column orthogonal matrix?

 

[vela]

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.

 

[tom08]

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