Show that A and B are Communitative

**sparky** · February 19th 2011, 07:34 AM

I need some guidance on how to start answering this question please:

Question: Suppose C is a square matrix such that $c^2=I$ . Let $CD_1C and B=CD_2C$ where $D_1$ and $D_2$ are diagonal Matrices. Show that A and B are communitative

**FernandoRevilla** · February 19th 2011, 07:47 AM

Originally Posted by sparky

Question: Suppose C is a square matrix such that $c^2=I$ . Let $CD_1C and B=CD_2C$ where $D_1$ and $D_2$ are diagonal Matrices. Show that A and B are communitative

I suppose you meant $A=CD_1C$ . Use $C^2=I$ and $D_1D_2=D_2D_1$

$AB=CD_1CCD_2C=CD_1C^2D_2C=\ldots$

$BA=\ldots$

Fernando Revilla

**sparky** · February 19th 2011, 08:37 AM

Yes you are right, Let A = CD1C, thanks

**sparky** · February 20th 2011, 02:30 PM

Originally Posted by FernandoRevilla

I suppose you meant $A=CD_1C$ . Use $C^2=I$ and $D_1D_2=D_2D_1$

$AB=CD_1CCD_2C=CD_1C^2D_2C=\ldots$

$BA=\ldots$

Fernando Revilla

Thanks for your reply FernandoRevilla.

Ok, is this correct?

$AB = CD_1CCD_2C$

$= CD_1C^2D_2C$

$= D_1C^4D_2$

$BA = CD_2CCD_1C$

$= CD_2C^2D_1C$

$= D_2C^4D_1$

$Therefore, AB = BA$

If this is correct, then what was the purpose of $C^2 = I$ ? To confuse me? What about the $D_1 and D_2$ being diagonal matrices - was that designed to confuse me as well? Or did they have some real purpose in this question?

**topsquark** · February 20th 2011, 02:44 PM

Originally Posted by sparky

Thanks for your reply FernandoRevilla.

Ok, is this correct?

$AB = CD_1CCD_2C$

$= CD_1C^2D_2C$

$= D_1C^4D_2$

$BA = CD_2CCD_1C$

$= CD_2C^2D_1C$

$= D_2C^4D_1$

$Therefore, AB = BA$

If this is correct, then what was the purpose of $C^2 = I$ ? To confuse me? What about the $D_1 and D_2$ being diagonal matrices - was that designed to confuse me as well? Or did they have some real purpose in this question?

You aren't quite done yet. You know that C^2 = I, so therefore C^4 = I. Then we have that AB = D1D2 and BA = D2D1. What do you know about the matrices D1 and D2?

-Dan

**LoblawsLawBlog** · February 20th 2011, 02:55 PM

I'm not sure how you went from line 2 to line 3, sparky. It looks like you're assuming that $C$ commutes with $D_1$ and $D_2$ , which you haven't proved (and which, without thinking too much about it, I doubt is true).

The facts that $C^2=I$ and the $D$ 's are diagonal both play key roles in the proof. Can you show that $D_1D_2=D_2D_1$ as topsquark suggests?

**sparky** · February 20th 2011, 03:01 PM

Thanks for the reply Topsquark

What I know about D1 and D2 is that they are diagonal matrices, which means that they are square matrices with zeros outside of the diagonal entries, like this:

$ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} $

**topsquark** · February 20th 2011, 03:12 PM

Originally Posted by sparky

Thanks for the reply Topsquark

What I know about D1 and D2 is that they are diagonal matrices, which means that they are square matrices with zeros outside of the diagonal entries, like this:

$ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} $

Right. Do two such matrices commute?

-Dan

**sparky** · February 20th 2011, 03:44 PM

Originally Posted by topsquark

Right. Do two such matrices commute?

-Dan

Ok, I was not sure, so to answer your question, I worked out a simple example (see below) and yes they do commute:

$ \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} $ $ \begin{pmatrix} 3 & 0 \\ 0 & 4 \end{pmatrix} $

= $ \begin{pmatrix} 3 & 0 \\ 0 & 8 \end{pmatrix} $

$ \begin{pmatrix} 3 & 0 \\ 0 & 4 \end{pmatrix} $ $ \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} $

= $ \begin{pmatrix} 3 & 0 \\ 0 & 8 \end{pmatrix} $

**sparky** · February 20th 2011, 03:59 PM

Thanks for your reply LoblawsLawBlog.

I think I now understand. Correct me if I'm wrong:

Since $C^2 = I$ and I is an identity matrix, therefore I is a diagonal matrix. C is also equal to I, therefore C is also diagonal. And since $D_1$ and $D_2$ are also diagonal, therefore since all of them are diagonal, AB = BA

**topsquark** · February 20th 2011, 07:51 PM

Originally Posted by sparky

Thanks for your reply LoblawsLawBlog.

I think I now understand. Correct me if I'm wrong:

Since $C^2 = I$ and I is an identity matrix, therefore I is a diagonal matrix. C is also equal to I, therefore C is also diagonal. And since $D_1$ and $D_2$ are also diagonal, therefore since all of them are diagonal, AB = BA

Hmmmm... If C = -I would this still work? You don't need this anyway.
$AB = C D_1 C C D_2 C = C D_1 C^2 D_2 C = C D_1 I D_2 C = C D_1 D_2 C$

Do this for BA. What do you get?

-Dan

**sparky** · February 21st 2011, 03:11 AM

Ok,

$BA=CD_2CCD_1C$

$BA=CD_2C^2D_1C$

$BA=CD_2ID_1C$

$BA=CD_2D_1C$

which is the same thing as $AB=CD_1D_2C$

Thanks!

**HallsofIvy** · February 21st 2011, 04:41 AM

Originally Posted by sparky

Thanks for your reply FernandoRevilla.

Ok, is this correct?

$AB = CD_1CCD_2C$

$= CD_1C^2D_2C$

$= D_1C^4D_2$

Where did you get this? How did the "C"s from the sides move inside? Better to use $C^2= I$ here and write
$= CD_1C_2C$ .

$BA = CD_2CCD_1C$

$= CD_2C^2D_1C$

$= D_2C^4D_1$

same thing: $= CD_2D_1C$ and, since $D_2$ and $D_1$ are diagonal matrices they commute: this is the same as $CD_1D_2C$ that you had before.

$Therefore, AB = BA$

If this is correct, then what was the purpose of $C^2 = I$ ? To confuse me? What about the $D_1 and D_2$ being diagonal matrices - was that designed to confuse me as well? Or did they have some real purpose in this question?

**sparky** · February 21st 2011, 07:13 AM

Thanks a lot HallsofIvy

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