0

Find all possible $2 \times 2$ matrices A that for any $2 \times 2$ matrix B, AB = BA.

Hint: AB = BA must hold for all B. Try matrices B that have lots of zero entries.

I'm clueless as to how to solve this problem. How should I start it? I tried plugging in values for B that "have lots of zero entries" but didn't seem to see anything that could help.

Did
  • 284,245
Grace
  • 209

4 Answers4

6

Consider the following four matrices:

$$\left(\begin{array}{cc} 1 & 0 \\ 0 & 0\end{array}\right), \quad \left(\begin{array}{cc} 0 & 1 \\ 0 & 0\end{array}\right), \quad \left(\begin{array}{cc} 0 & 0 \\ 1 & 0\end{array}\right), \quad \left(\begin{array}{cc} 0 & 0 \\ 0 & 1\end{array}\right).$$

See what happens when you solve the equation $AB = BA$ for each of those four (let $B$ be each one of those four). To facilitate it, write $A = \left(\begin{array}{cc} a & b \\ c & d\end{array}\right)$. You will get a set of equations for the entries of $a$ which are easily solved. This trick is quite general.

Cameron L. Williams
  • 30,485
  • I endorse your choice for $B$ but I question your variable names for $A$ :) – James S. Cook Jul 22 '15 at 02:08
  • That's fair. Your way is better. I'll edit them accordingly. – Cameron L. Williams Jul 22 '15 at 02:09
  • Oh no need for that, in the $n \times n$ case your notation is a necessity. man you're fast. – James S. Cook Jul 22 '15 at 02:10
  • @CameronWilliams: For each for the four equations, I had these--First equation: a = a, b = 0, c = 0, 0 = 0. Second equation: c = 0, a = d, 0 = 0, c = 0. Third equation: b = 0, 0 = 0, a = d, b = 0. Fourth equation: 0 = 0, b = 0, c = 0, d = d. /// Does this mean a = b = c = d = 0? – Grace Jul 22 '15 at 02:23
  • @Grace: Obviously the 4 equations imply that $a = d$ and $b = c = 0$. – corindo Jul 22 '15 at 02:31
  • @corindo: If that is the case, then $A = \left(\begin{array}{cc} a & 0 \ a & 0\end{array}\right)$, and this is the form for all possible solutions for A? – Grace Jul 22 '15 at 02:42
  • @Grace: $d$ is the one at the lower right corner. – corindo Jul 22 '15 at 02:44
  • @corindo: EDIT: $A = \left(\begin{array}{cc} a & 0 \ 0 & a \end{array}\right)$ – Grace Jul 22 '15 at 02:47
3

Let $A=\left[\begin{array}{cc} a & b \\ c & d\end{array} \right]$ then choose particular matrices $B$ as simple as you like and force $AB=BA$ this gives you all sorts of equations on $a,b,c,d$ which will eventually narrow $A$ to the answer.

James S. Cook
  • 17,257
  • incidentally, the answer for the $2 \times 2$ case is the same as the $n \times n$ case and if you embrace the notation in Cameron's answer fully it provides the path to see it. – James S. Cook Jul 22 '15 at 02:09
  • 1
    I say "answer" but, perhaps I should say "answers" as it is clear that if $A$ works then so does $cA$. We should expect either no answers or infinitely many. – James S. Cook Jul 22 '15 at 02:15
  • The observation in your second comment is a very good one. I hadn't explicitly made that connection myself. (Although you could have one solution - the zero solution.) – Cameron L. Williams Jul 22 '15 at 02:20
0

Hint: if you can't do this abstractly, try writing out the matrices, writing out the products of each entry, and going from there

Race Bannon
  • 1,831
  • 1
  • 12
  • 21
-1

Given a matrix B, consider any matrix that is a polynomial on $B$ with coefficients {$c_k$} in the field you are working with, i.e., take all $A$ with $$A=c_0I+c_1B+c_2B^2+...+c_nB^n$$. I actually think this gives you all such matrices $A$ , but I am not sure.

Gary.
  • 2,492