I am stuck in finding the generalised condition for two matrices of any order to commute over each other i.e.
$$AB=BA$$
Asked
Active
Viewed 339 times
0
-
See here. – cqfd Jun 12 '19 at 05:10
-
Also here, which talks about the non-diagonalizable case. – eyeballfrog Jun 12 '19 at 05:12
1 Answers
1
Matrix multiplication is always commutative when (i) one of the matrices is the Identity matrix (ii) One of the matrix is the Zero matrix. (iii) both matrices are Diagonal matrices.
Moreover, Two matrices that are simultaneously diagonalizable are always commutative.
Becase $A$ and $B$ are simultaneously diagonalizable, then a matrix $T$ exists, such that $D_A = S^{-1} \cdot A \cdot S$ and $D_B = S^{-1} \cdot B \cdot S$ \begin{align} A \cdot B &= S \cdot D_A S^{-1} \cdot S \cdot D_B \cdot S^{-1} \rm {~~as ~}(S^{-1}S = I)\\ &= S \cdot D_A \cdot D_B \cdot S^{-1} \\ &= S \cdot D_B \cdot D_A \cdot S^{-1} \\ &= S \cdot D_B \cdot S^{-1} \cdot S \cdot D_A \cdot S^{-1} \\ &= B \cdot A \end{align}
Vineet
- 846
-
1
-
1I knew 2D rotation matrix are commutative. Thanks for point out the mistake. Appreciate it :-) – Vineet Jun 13 '19 at 05:26