matrice 2

**thereddevils** · August 19th 2009, 01:38 AM

(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 .

**Opalg** · August 19th 2009, 03:10 AM

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 $A = \omega I$ .

**thereddevils** · August 19th 2009, 05:09 AM

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 .

**Opalg** · August 19th 2009, 05:19 AM

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.

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

**thereddevils** · August 19th 2009, 05:38 AM

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 ..

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