Results 1 to 4 of 4

Math Help - AB = BA when B = nI

  1. #1
    Newbie
    Joined
    Aug 2012
    From
    Australia
    Posts
    2

    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.
    Last edited by registered0; August 5th 2012 at 02:34 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,409
    Thanks
    1294

    Re: AB = BA when B = nI

    Quote Originally Posted by registered0 View Post
    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 \begin{align*} A \end{align*} actually be \displaystyle \begin{align*} \left[\begin{matrix} a & b \\ c & d \end{matrix}\right] \end{align*}?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2012
    From
    Australia
    Posts
    2

    Re: AB = BA when B = nI

    Yes, thankyou for pointing that out.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,325
    Thanks
    699

    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.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum