Question

If $A$ and $B$ are two square matrices of order $n$ and $A$ and $B$ commute then for any real number $k$ prove that $A-kI,B-kI$ also commute

Answer 1

Given the matrices $A,B$ commute then $AB=BA$.....(1).

Now,

$(A-kI)(B-kI)$

$=AB-k(A+B)+k^{2}I$ [ Since $I^2=I$]

$=BA-k(B+A)+k^{2}I$ [ Since $A+B=B+A$] [ Using (1)]

$=(B-kI)(A-kI)$.

So the matrices $(A-kI),(B-kI)$ commute.