1. ## matrice 2

(1)IF A and B are square matrices such that $AB=\omega A$ and $BA=\phi B$ where $\phi$ and $\omega$ are non - zero scalars , prove that $A^2=\phi A$and $B^2=\omega B .$

(2) If A is a 2x2 matrix such that $A^2=\omega A$ where $\omega$ is a non-zero scalar , prove that A is either non-singular or $A=\phi I$ where I is an identity .

Thanks for helping me out .

2. Originally Posted by thereddevils
(1)IF A and B are square matrices such that $AB=\omega A$ and $BA=\phi B$ where $\phi$ and $\omega$ are non - zero scalars , prove that $A^2=\phi A$and $B^2=\omega B .$
If $AB=\omega A$ then $A = \omega^{-1}AB$, so $A^2 = (\omega^{-1}AB)A =\omega^{-1}A(BA) =\omega^{-1}A(\phi B) = \ldots$.

Originally Posted by thereddevils
(2) If A is a 2x2 matrix such that $A^2=\omega A$ where $\omega$ is a non-zero scalar , prove that A is either non-singular or $A=\phi I$ where I is an identity.
I think you mean "singular", not "non-singular". If A is non-singular then you can multiply both sides of the equation $A^2=\omega A$ by $A^{-1}$ and conclude that $A = \omega I$.

3. Originally Posted by Opalg
If $AB=\omega A$ then $A = \omega^{-1}AB$, so $A^2 = (\omega^{-1}AB)A =\omega^{-1}A(BA) =\omega^{-1}A(\phi B) = \ldots$.

I think you mean "singular", not "non-singular". If A is non-singular then you can multiply both sides of the equation $A^2=\omega A$ by $A^{-1}$ and conclude that $A = \omega I$.
Thanks Opalg, but how can i continue from here $\omega^{-1}A(\phi B) = \ldots$ , how to get rid of some variables here to get the required result .

4. Originally Posted by thereddevils
Thanks Opalg, but how can i continue from here $\omega^{-1}A(\phi B) = \ldots$ , how to get rid of some variables here to get the required result .
I thought you'd be able to manage the rest of it.

$\omega^{-1}A(\phi B) = \phi(\omega^{-1}AB) = \phi A$.

5. Originally Posted by Opalg
I thought you'd be able to manage the rest of it.

$\omega^{-1}A(\phi B) = \phi(\omega^{-1}AB) = \phi A$.

gosh , sorry to disappoint u ..