Matrix proof: Let A be square w/ A^2 = A; let B be same size. Show (AB - ABA)^2 = 0

[escobarro]

\(\displaystyle 17. \mbox{ Let }\, A\, \mbox{ be a square matrix satisfying }\, A^2\, =\, A\, \mbox{ and let }\, B\, \mbox{ be}\)

\(\displaystyle \mbox{any matrix of the same size. Show that }\, (AB\, -\, ABA)^2\, =\, 0\)

whatever method i try, i end up with lots of variables and it is too complex
anyone got a proof?
thanks

 

[Deleted member 4993]

\(\displaystyle 17. \mbox{ Let }\, A\, \mbox{ be a square matrix satisfying }\, A^2\, =\, A\, \mbox{ and let }\, B\, \mbox{ be}\)

\(\displaystyle \mbox{any matrix of the same size. Show that }\, (AB\, -\, ABA)^2\, =\, 0\)

whatever method i try, i end up with lots of variables and it is too complex
anyone got a proof?
thanks

What are the methods that you tried?

Please share your work with us ...even if you know it is wrong

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[pka]

\(\displaystyle 17. \mbox{ Let }\, A\, \mbox{ be a square matrix satisfying }\, A^2\, =\, A\, \mbox{ and let }\, B\, \mbox{ be}\)

\(\displaystyle \mbox{any matrix of the same size. Show that }\, (AB\, -\, ABA)^2\, =\, 0\)

whatever method i try, i end up with lots of variables and it is too complex

anyone got a proof?

YES a lot do, you need to find a proof for yourself.
If each of \(\displaystyle X~\&~Y\) is a \(\displaystyle n\times n\) matrix then \(\displaystyle (X-Y)^2=X^2-XY-YX+Y^2\)

 

[stapel]

\(\displaystyle 17. \mbox{ Let }\, A\, \mbox{ be a square matrix satisfying }\, A^2\, =\, A\, \mbox{ and let }\, B\, \mbox{ be}\)

\(\displaystyle \mbox{any matrix of the same size. Show that }\, (AB\, -\, ABA)^2\, =\, 0\)

whatever method i try, i end up with lots of variables and it is too complex

What "variables" have you "ended up with"? In what manner has this been "too complex"?

You started by expanding the left-hand side in the usual manner:

. . . . .\(\displaystyle \begin{align} (AB\, -\, ABA)^2\, &=\, (AB\, -\, ABA)\, (AB\, -\, ABA)

\\ \\ &=\, (AB)(AB)\, +\, (-ABA)(AB)\, +\, (AB)(-ABA)\, +\, (-ABA)(-ABA)

\\ \\ &=\, ABAB\, +\, (-1)(ABAAB)\, +\, (-1)(ABABA)\, +\, (ABAABA) \end{align}\)

You took care, as in the above, to keep the matrices in their proper orders (since matrix multiplication is not commutative), and used the rules on scalar multiplication to pull the "minus" signs out front. Then you noted the two clear instances where the "given" could be applied:

. . . . .\(\displaystyle \begin{align} ABAB\, +\, (-1)(ABAAB)\, +\, (-1)(ABABA)\, +\, (ABAABA)\, &=\,ABAB\, +\, (-1)(AB)(A^2)(B)\, +\, (-1)(ABABA)\, +\, (AB)(A^2)(BA)

\\ \\ &=\, ABAB\, +\, (-1)(AB)(A)(B)\, +\, (-1)(ABABA)\, +\, (AB)(A)(BA)\end{align}\)

What did you do next? Where are you bogging down?

Please be complete. Thank you!

 

[escobarro]

What "variables" have you "ended up with"? In what manner has this been "too complex"?

You started by expanding the left-hand side in the usual manner:

. . . . .\(\displaystyle \begin{align} (AB\, -\, ABA)^2\, &=\, (AB\, -\, ABA)\, (AB\, -\, ABA)

\\ \\ &=\, (AB)(AB)\, +\, (-ABA)(AB)\, +\, (AB)(-ABA)\, +\, (-ABA)(-ABA)

\\ \\ &=\, ABAB\, +\, (-1)(ABAAB)\, +\, (-1)(ABABA)\, +\, (ABAABA) \end{align}\)

You took care, as in the above, to keep the matrices in their proper orders (since matrix multiplication is not commutative), and used the rules on scalar multiplication to pull the "minus" signs out front. Then you noted the two clear instances where the "given" could be applied:

. . . . .\(\displaystyle \begin{align} ABAB\, +\, (-1)(ABAAB)\, +\, (-1)(ABABA)\, +\, (ABAABA)\, &=\,ABAB\, +\, (-1)(AB)(A^2)(B)\, +\, (-1)(ABABA)\, +\, (AB)(A^2)(BA)

\\ \\ &=\, ABAB\, +\, (-1)(AB)(A)(B)\, +\, (-1)(ABABA)\, +\, (AB)(A)(BA)\end{align}\)

What did you do next? Where are you bogging down?

Please be complete. Thank you!

Thanks! I ended up getting a proof very similar to yours. I was going wrong initially because I was assigning variables to entries in A and B and attempting to expand and simplify