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

  • Thread starter tom08
  • Start date
In summary, the conversation discusses a problem with a 6x3 matrix A, where A is not invertible and A'*A=I, but A*A'<>I. The conversation also explores the concept of orthogonal matrices and the upper bound for |A'*A-A*A'| when A is a rectangular column orthogonal matrix.
  • #1
tom08
19
0

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?
 
Physics news on Phys.org
  • #2
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
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
tom08 said:
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 AT

But read this, orthogonal matrices
 
  • #5
thank u so much.

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

when A is a rectangluar column orthogonal matrix?
 
  • #6
If you have

[tex]A^TA = I[/tex]

then

[tex](A^TA)^T=A^T(A^T)^T=A^TA=I[/tex].

You're starting with [itex]AA^T[/itex], which isn't equal to the identity matrix, so its transpose won't be either.
 
  • #7
Thank u, vela and rock. i realize my mistake.
 

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

Why does A'*A equal I?

When A is multiplied by its transpose A', the resulting matrix contains the dot products of the rows of A with the columns of A'. This means that the diagonal elements of the resulting matrix will be the squared magnitudes of the rows of A, which is equivalent to the diagonal elements of the identity matrix I. Therefore, A'*A=I.

Why is A*A' equal to I?

When A is multiplied by its transpose A', the resulting matrix contains the dot products of the rows of A with the columns of A'. Since A is multiplied by its transpose, the resulting matrix will be square and symmetric. The diagonal elements of this matrix will be the squared magnitudes of the columns of A, which is equivalent to the diagonal elements of the identity matrix I. Therefore, A*A'=I.

Is A'*A always equal to I?

No, A'*A is only equal to I if A is a square matrix with orthogonal columns. If A is not a square matrix or if its columns are not orthogonal, then A'*A will not equal I.

Can A*A' ever equal a matrix other than I?

Yes, A*A' can equal a matrix other than I if A is not a square matrix or if its columns are not orthogonal. In this case, the resulting matrix will not have all diagonal elements equal to 1, making it not equal to I.

What is the significance of A'*A=I and A*A'=I?

A'*A=I and A*A'=I are both examples of orthogonal matrices. This means that the columns of A are perpendicular to each other, and when multiplied by their transpose, they result in an identity matrix. Orthogonal matrices have many important applications in mathematics, physics, and engineering, including in the fields of linear algebra, geometry, and signal processing.

Back
Top