# Diagonal Matrix question

• Jun 11th 2007, 06:29 AM
mister_bluesman
Diagonal Matrix question
The question is that D1 and D2 are any two diagonal 3x3 matrices and

Show that PQ = QP.

Now, I know that D1 and D2 commute, but im not sure how I can show this algebraically. Any ideas?

BTW: A is a 3 x 3 matrix and is the inverse of itself. So, really, P = AD1A and Q = AD2A.

Many thanks

MB
• Jun 11th 2007, 07:32 AM
topsquark
Quote:

Originally Posted by mister_bluesman
The question is that D1 and D2 are any two diagonal 3x3 matrices and

Show that PQ = QP.

Now, I know that D1 and D2 commute, but im not sure how I can show this algebraically. Any ideas?

BTW: A is a 3 x 3 matrix and is the inverse of itself. So, really, P = AD1A and Q = AD2A.

Many thanks

MB

Well:

Now, D1 and D2 commute since they are both diagonal, so D2D1 = D1D2.

Thus
QP = A(D2D1)A = A(D1D2)A = PQ

(Note: The condition that $A^{-1} = A$ is not actually needed here.)

-Dan
• Jun 11th 2007, 08:59 AM
mister_bluesman
Hi there. Thans for that. But could you explain how you got:

I'm not sure how you got PQ = A(D1D2)A and QP = A(D2D1)A. What did you do to do that? Rearrange, multiply out?

Thanks
• Jun 11th 2007, 09:59 AM
topsquark
Quote:

Originally Posted by mister_bluesman
Hi there. Thans for that. But could you explain how you got:

I'm not sure how you got PQ = A(D1D2)A and QP = A(D2D1)A. What did you do to do that? Rearrange, multiply out?

Thanks

Let me be more specific, and let me use $A^{-1}$ since this is a bit more clear.

$PQ = (AD_1A^{-1})(AD_2A^{-1})$

$PQ = AD_1(A^{-1}A)D_2A^{-1}$

$PQ = AD_1 \cdot 1 \cdot D_2A^{-1}$ <-- $A^{-1}A = 1$

$PQ = A(D_1D_2)A^{-1}$

You can do QP the same way and get
$QP = A(D_2D_1)A^{-1}$

And the rest of the argument goes through as before.

(In case it's an issue, matrix multiplication is associative, that is (AB)C = A(BC), so as long as we don't alter the order of the factors we may insert parenthesis wherever we wish.)

-Dan
• Jun 11th 2007, 10:22 AM
mister_bluesman
Ah ok... but, please i dont mean to be stupid, but doesnt Ainv A = I, not 1? Surely multiplying two matrices results in another matrix and not a scalar?

Many thanks (again!)

Andy
• Jun 11th 2007, 03:09 PM
topsquark
Quote:

Originally Posted by mister_bluesman
Ah ok... but, please i dont mean to be stupid, but doesnt Ainv A = I, not 1? Surely multiplying two matrices results in another matrix and not a scalar?

Many thanks (again!)

Andy

You aren't being stupid. My own personal habit is to write a "1" for any (multiplicative) identity element. (In rare cases I use an "e.") I won't say this is standard for Physicists, but I've seen it in a number of Physics texts and adopted the habit. My apologies for the confusion!

-Dan