Given $$A = \begin{bmatrix}
3 & 1 &0 \\
0 &3 & 1\\
0 &0 & 3
\end{bmatrix}$$ find matrices $B$ such that $AB=BA$.
Trivially $B=A^{-1}$ and $B=kI$ are the solutions
Also we have Characteristic Polynomial as
$$A^3-9A^2+27A-27I=0$$ $\implies$
$$(A-3I)^3=0$$
Is it possible to find other $B's$ using above Nilpotency of $A-3I$?
Best Answer
By just writing out the matrix multiplication and simplifying you get: \begin{align*} AB &= BA\\ \begin{bmatrix} 3 & 1 &0 \\ 0 &3 & 1\\ 0 &0 & 3 \end{bmatrix} \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \\ \end{bmatrix} &= \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \\ \end{bmatrix} \begin{bmatrix} 3 & 1 &0 \\ 0 &3 & 1\\ 0 &0 & 3 \end{bmatrix}\\ \begin{bmatrix} 3b_{11} + b_{21} & 3b_{12} + b_{22} & 3b_{13}+b_{23} \\ 3b_{21} + b_{31} & 3b_{22} + b_{32} & 3b_{23}+b_{33} \\ 3b_{31} & 3b_{32} & 3b_{33} \end{bmatrix} &= \begin{bmatrix} 3b_{11} & b_{11}+3b_{12} & b_{12} + 3b_{13} \\ 3b_{21} & b_{21}+3b_{22} & b_{22} + 3b_{23} \\ 3b_{31} & b_{31}+3b_{32} & b_{32} + 3b_{33} \end{bmatrix} \\ \begin{bmatrix} b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \\ 0 & 0 & 0 \end{bmatrix} &= \begin{bmatrix} 0 & b_{11} & b_{12} \\ 0 & b_{21} & b_{22} \\ 0 & b_{31} & b_{32} \end{bmatrix} \end{align*} Hence, $b_{21},b_{31},b_{32}=0$, $b_{11}=b_{22}=b_{33}$ and $b_{12}=b_{23}$, confirming the solutions are exactly those given by Robert.