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

[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 = <br /> \left( \begin{array}{cc}<br /> \frac{1}{3} &amp; \frac{2}{3} \end{array} \right), \quad B = \left( \begin{array}{cc}<br /> \frac{3}{5} &amp; \frac{2}{5} \\<br /> \frac{1}{5} &amp; \frac{4}{5} \end{array} \right) \quad A * B = \left( \begin{array}{cc}<br /> \frac{1}{3} &amp; \frac{2}{3} \end{array} \right)<br />

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.
  .
• <br /> p\to+\infty{\left( \begin{array}{cc}<br /> \frac{3}{5} &amp; \frac{2}{5} \\<br /> \frac{1}{5} &amp; \frac{4}{5} \end{array} \right)}^p = \left( \begin{array}{cc}<br /> \frac{1}{3} &amp; \frac{2}{3} \\<br /> \frac{1}{3} &amp; \frac{2}{3} \end{array} \right)<br />

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

 

[D H]

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?

 

[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.

 

[thinktank2]

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

 

[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.

 

[D H]

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