# matrice 2

• Aug 19th 2009, 01:38 AM
thereddevils
matrice 2
(1)IF A and B are square matrices such that $\displaystyle AB=\omega A$ and $\displaystyle BA=\phi B$ where $\displaystyle \phi$ and $\displaystyle \omega$ are non - zero scalars , prove that $\displaystyle A^2=\phi A$and $\displaystyle B^2=\omega B .$

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

Thanks for helping me out .
• Aug 19th 2009, 03:10 AM
Opalg
Quote:

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

If $\displaystyle AB=\omega A$ then $\displaystyle A = \omega^{-1}AB$, so $\displaystyle A^2 = (\omega^{-1}AB)A =\omega^{-1}A(BA) =\omega^{-1}A(\phi B) = \ldots$.

Quote:

Originally Posted by thereddevils
(2) If A is a 2x2 matrix such that $\displaystyle A^2=\omega A$ where $\displaystyle \omega$ is a non-zero scalar , prove that A is either non-singular or $\displaystyle A=\phi I$ 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 $\displaystyle A^2=\omega A$ by $\displaystyle A^{-1}$ and conclude that $\displaystyle A = \omega I$.
• Aug 19th 2009, 05:09 AM
thereddevils
Quote:

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

I think you mean "singular", not "non-singular". If A is non-singular then you can multiply both sides of the equation $\displaystyle A^2=\omega A$ by $\displaystyle A^{-1}$ and conclude that $\displaystyle A = \omega I$.

Thanks Opalg, but how can i continue from here $\displaystyle \omega^{-1}A(\phi B) = \ldots$ , how to get rid of some variables here to get the required result .
• Aug 19th 2009, 05:19 AM
Opalg
Quote:

Originally Posted by thereddevils
Thanks Opalg, but how can i continue from here $\displaystyle \omega^{-1}A(\phi B) = \ldots$ , 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. (Sadsmile)

$\displaystyle \omega^{-1}A(\phi B) = \phi(\omega^{-1}AB) = \phi A$.
• Aug 19th 2009, 05:38 AM
thereddevils
Quote:

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

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

gosh , sorry to disappoint u ..(Crying)