Thread: AB = BA when B = nI

1. AB = BA when B = nI

Hi everyone, I've been stuck on this question:

A =
a b
c d

B =
x y
z w

Show that AB = BA for any A only if B is a multiple of the identity matrix.

---

AB =
(ax+bz) (ay+bw)
(cx+dz) (cy+dw)

BA =
(xa+yc) (xb+yd)
(za+wc) (zb+wd)

equating elements gives
ax + bz = xa + yc
ay + bw = xb + yd
cx + dz = za + wc
cy + dw = zb + wd

cancel out common terms
bz = yc
ay + bw = xb + yd
cx + dz = za + wc
cy = zb, same as first equation

and I'm stuck here.
bz = yc
ay + bw = xb + yd
cx + dz = za + wc

For these things to be equal, we would need
a = d, w = x, and bz = yc
but I don't know what to do next.

2. Re: AB = BA when B = nI

Originally Posted by registered0
Hi everyone, I've been stuck on this question:

A =
a b
c c

B =
x y
z w

Show that AB = BA for any A only if B is a multiple of the identity matrix.

---

AB =
(ax+bz) (ay+bw)
(cx+dz) (cy+dw)

BA =
(xa+yc) (xb+yd)
(za+wc) (zb+wd)

equating elements gives
ax + bz = xa + yc
ay + bw = xb + yd
cx + dz = za + wc
cy + dw = zb + wd

cancel out common terms
bz = yc
ay + bw = xb + yd
cx + dz = za + wc
cy = zb, same as first equation

and I'm stuck here.
bz = yc
ay + bw = xb + yd
cx + dz = za + wc

For these things to be equal, we would need
a = d, w = x, and bz = yc
but I don't know what to do next.
Should your \displaystyle \displaystyle \begin{align*} A \end{align*} actually be \displaystyle \displaystyle \begin{align*} \left[\begin{matrix} a & b \\ c & d \end{matrix}\right] \end{align*}?

3. Re: AB = BA when B = nI

Yes, thankyou for pointing that out.

4. Re: AB = BA when B = nI

since your equations have to hold for every choice of a,b,c,d, you can use particular choices of a,b,c and d to narrow down the possibilities.

so, for example, we are free to choose a = 1, b = 0, c = 1, d = 1. this leads to:

0 = y
y = y
x + z = z + w → x = w.

this tells us B has to be of the form:

[x 0]
[z x], or else it won't commute with the particular matrix A =

[1 0]
[1 1].

similarly, we can also pick a = 1, b = 1, c = 0, d = 1, leading to:

z = 0
y + w = x + y → x = w
z = z

so that B must also be of the form:

[x y]
[0 x].

so if B is to commute with all matrices A, it must commute with these 2 matrices in particular, meaning B has to be of the form:

[x 0]
[0 x] = xI.

the only thing left to prove now is that if indeed B = xI, then AB = BA, regardless of A.