# Show that A and B are Communitative

• Feb 19th 2011, 08:34 AM
sparky
Show that A and B are Communitative
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
• Feb 19th 2011, 08:47 AM
FernandoRevilla
Quote:

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
• Feb 19th 2011, 09:37 AM
sparky
Yes you are right, Let A = CD1C, thanks
• Feb 20th 2011, 03:30 PM
sparky
Quote:

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

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?
• Feb 20th 2011, 03:44 PM
topsquark
Quote:

Originally Posted by sparky

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
• Feb 20th 2011, 03:55 PM
LoblawsLawBlog
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?
• Feb 20th 2011, 04:01 PM
sparky

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:

$$\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}$$
• Feb 20th 2011, 04:12 PM
topsquark
Quote:

Originally Posted by sparky

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
• Feb 20th 2011, 04:44 PM
sparky
Quote:

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:

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

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

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

= $$\begin{pmatrix} 3 & 0 \\ 0 & 8 \end{pmatrix}$$
• Feb 20th 2011, 04:59 PM
sparky

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
• Feb 20th 2011, 08:51 PM
topsquark
Quote:

Originally Posted by sparky

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
• Feb 21st 2011, 04:11 AM
sparky
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!
• Feb 21st 2011, 05:41 AM
HallsofIvy
Quote:

Originally Posted by sparky

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
$= CD_1C_2C$.

Quote:

$BA = CD_2CCD_1C$

$= CD_2C^2D_1C$

$= D_2C^4D_1$
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.

Quote:

$Therefore, AB = BA$

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?
• Feb 21st 2011, 08:13 AM
sparky
Thanks a lot HallsofIvy