# Matrix Question: A*B=A (B is non-identity)

1. Nov 30, 2011

### thinktank2

If two ℝeal valued, non-identity matrices A (dimension MxN) and B (dimension NxN)
satisfy the condition A * B = A
Is there a name for the relationship between B and A ?

For example for: $A = \left( \begin{array}{cc} \frac{1}{3} & \frac{2}{3} \end{array} \right), \quad B = \left( \begin{array}{cc} \frac{3}{5} & \frac{2}{5} \\ \frac{1}{5} & \frac{4}{5} \end{array} \right) \quad A * B = \left( \begin{array}{cc} \frac{1}{3} & \frac{2}{3} \end{array} \right)$

My Observations:
• If such A exists for B, then for any real number k, k*A is also a solution for B
since (k*A)*B = k*(A*B)=k*A
implies, there will be infinite solutions for B.
.
• $p\to+\infty{\left( \begin{array}{cc} \frac{3}{5} & \frac{2}{5} \\ \frac{1}{5} & \frac{4}{5} \end{array} \right)}^p = \left( \begin{array}{cc} \frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{2}{3} \end{array} \right)$

My Question: Given a matrix B, how do we find A such that A*B=A ?

2. Nov 30, 2011

### D H

Staff Emeritus
This looks a bit too much like homework for me to give a direct answer (now).

I can give a hint in the form of a question: How is this related to eigenvalues and eigenvectors?

3. Nov 30, 2011

### thinktank2

Thank you very much for the hint. I swear this not my homework. I know I missed something very basic but didn't know which one.

4. Nov 30, 2011

### thinktank2

I am hitting a wall. May I request another hint or two?

5. Dec 2, 2011

### PhDorBust

D_H gave a pretty good hint. I'll give you a little more.

First consider the problem AB=B. Given A you can easily find all such B using eigenthings.

You can transform AB=A into this form by taking the transpose of both sides.

Very good problem if you came up with it yourself.

6. Dec 2, 2011

### D H

Staff Emeritus
And one more hint: Other than the trivial solution A=0, there are no solutions if one is not an eigenvalue of B.