Proving Linear Algebra: AB=BA for All B(2x2), A=àI for All à € Real Number

[annoymage]

let A € M(2x2)..
if AB=BA for all B(2x2),
show that A=àI for all à € real number

its really hard to send new post using phone..

 

[CompuChip]

Do you understand the question? Basically, it's asking you to prove that the only matrix that commutes with any other matrix is the identity matrix.

Well, if you don't see any clever way to solve it (like me at first reading), you could simply take two matrices,

$$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \qquad B = \begin{pmatrix} x & y \\ u & v \end{pmatrix}$$
and see what equations you get when you multiply AB = BA. Then note that this equations should hold for any values of x, y, u and v - allowing you to solve for a, b, c and d.

 

[annoymage]

before that,its hard to give my answer, so please teach me how to make that latex matrix.

 

[CompuChip]

If you click on it, or quote my post, you can see the code I used.

(If you want to use pmatrix in your own LaTeX documents, you have to \usepackage{amsmath}).