# A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?

• November 3rd 2012, 04:40 PM
A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?
Dear all, I have checked by computer that for any complex $2\times 2$ matrix $A,B,C$, there we have $A(BC-CB)^2-(BC-CB)^2A=0.$ However, I could not prove it mathematically [without tedious calculation].
• November 3rd 2012, 05:40 PM
chiro
Re: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?

Try expanding all the matrices out and use the fact that (A+B)(C+D) = AC + AD + BC + BD for matrices if they are well defined (dimension wise that is) and if you get zero no matter what, then you're done.
• November 4th 2012, 12:41 PM
Deveno
Re: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?
it's not true.

for a counter-example, let:

$A = \begin{bmatrix}1&1\\0&1 \end{bmatrix};\ B = \begin{bmatrix}1&0\\1&1 \end{bmatrix};\ C = \begin{bmatrix}1&1\\1&0 \end{bmatrix}$.

then:

$BC - CB = \begin{bmatrix}1&1\\2&1 \end{bmatrix} - \begin{bmatrix}2&1\\1&0 \end{bmatrix} = \begin{bmatrix}-1&0\\1&0 \end{bmatrix}$.

so:

$(BC - CB)^2 = \begin{bmatrix}1&0\\-1&0 \end{bmatrix}$.

now:

$A(BC - CB)^2 = \begin{bmatrix}1&1\\0&1 \end{bmatrix}\begin{bmatrix}1&0\\-1&0 \end{bmatrix} = \begin{bmatrix}0&0\\-1&0 \end{bmatrix}$,

but:

$(BC - CB)^2A = \begin{bmatrix}1&0\\-1&0 \end{bmatrix}\begin{bmatrix}1&1\\0&1 \end{bmatrix} = \begin{bmatrix}1&1\\-1&-1 \end{bmatrix}$.

your "computer check" apparently didn't check some "simple matrices".
• November 4th 2012, 03:12 PM
Re: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?
your calculation of BC-CB is wrong....
• November 4th 2012, 05:47 PM
Deveno
Re: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?
*sigh*, this happens to me, more often than i'd like to admit. yes, it should be:

$BC - CB = \begin{bmatrix}-1&0\\1&1 \end{bmatrix}$

this means that:

$A(BC - CB) = \begin{bmatrix}1&1\\0&1 \end{bmatrix} \begin{bmatrix}-1&0\\1&1 \end{bmatrix} = \begin{bmatrix}0&1\\1&1 \end{bmatrix}$

and:

$(BC - CB)A = \begin{bmatrix}-1&0\\1&1 \end{bmatrix} \begin{bmatrix}1&1\\0&1 \end{bmatrix} = \begin{bmatrix}-1&-1\\1&2 \end{bmatrix}$

but these 2 still aren't equal.

*******

in general, unless B and C commute, BC - CB will be non-zero, and unless BC - CB is of nilpotency degree 2, (BC - CB)2 will also be non-zero.

in fact, unless (BC - CB)2 is a scalar multiple of the identity, there will always be SOME matrix A that doesn't commute with it.

so your "theorem" would only be true if (BC - CB)2 = I or 0, which implies BC - CB is either 0, or invertible.

so all we have to do is find matrices B and C so that BC - CB is singular. this isn't hard to do (and is probably what i SHOULD have done).
• November 4th 2012, 06:01 PM
topsquark
Re: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?
Quote:

Originally Posted by Deveno
*sigh*, this happens to me, more often than i'd like to admit. yes, it should be:

$BC - CB = \begin{bmatrix}-1&0\\1&1 \end{bmatrix}$

this means that:

$A(BC - CB) = \begin{bmatrix}1&1\\0&1 \end{bmatrix} \begin{bmatrix}-1&0\\1&1 \end{bmatrix} = \begin{bmatrix}0&1\\1&1 \end{bmatrix}$

and:

$(BC - CB)A = \begin{bmatrix}-1&0\\1&1 \end{bmatrix} \begin{bmatrix}1&1\\0&1 \end{bmatrix} = \begin{bmatrix}-1&-1\\1&2 \end{bmatrix}$

but these 2 still aren't equal.

*******

in general, unless B and C commute, BC - CB will be non-zero, and unless BC - CB is of nilpotency degree 2, (BC - CB)2 will also be non-zero.

in fact, unless (BC - CB)2 is a scalar multiple of the identity, there will always be SOME matrix A that doesn't commute with it.

so your "theorem" would only be true if (BC - CB)2 = I or 0, which implies BC - CB is either 0, or invertible.

so all we have to do is find matrices B and C so that BC - CB is singular. this isn't hard to do (and is probably what i SHOULD have done).

You didn't square the (BC - CB). I checked it on my TI. The sum is indeed zero.

-Dan
• November 5th 2012, 08:39 AM
Deveno
Re: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?
omg. i am...wrong. just....wrong. how embarrassing.

it turns out the theorem IS true, and i can prove it.

it suffices (from a previous post) to prove that (BC - CB)2 is a scalar multiple of the identity (such matrices lie in the center of the ring of 2x2 matrices, and thus will commute with any A).

let:

$B = \begin{bmatrix}b_1&b_2\\b_3&b_4 \end{bmatrix};\ C = \begin{bmatrix}c_1&c_2\\c_3&c_4 \end{bmatrix}$.

then:

$BC - CB = \begin{bmatrix}b_1c_1+b_2c_3&b_1c_2+b_2c_4\\b_3c_1 +b_4c_3&b_3c_2+b_4c_4 \end{bmatrix} - \begin{bmatrix}b_1c_1+b_3c_2&b_2c_1+b_4c_2\\b_1c_3 +b_3c_4&b_2c_3+b_4c_4 \end{bmatrix}$

$=\begin{bmatrix}b_2c_3-b_3c_2&b_1c_2-b_2c_1+b_2c_4-b_4c_2\\b_3c_1-b_1c_3+b_4c_3-b_3c_4&b_3c_2-b_2c_3 \end{bmatrix}$

that is, BC - CB is of the form:

$\begin{bmatrix}u&v\\w&-u \end{bmatrix}$

hence (BC - CB)2 is of the form:

$\begin{bmatrix}u&v\\w&-u \end{bmatrix}\begin{bmatrix}u&v\\w&-u \end{bmatrix} = \begin{bmatrix}u^2+vw&0\\0&u^2+vw \end{bmatrix} = (u^2+vw)I$

so sorry :(
• November 5th 2012, 12:26 PM