Is matrix multiplication transitive?

Question

Let $A, B, C \in M_n(\mathbb R)$ be such that $A$ commutes with $B$, $B$ commutes with $C$ and $B$ is not a scalar matrix. Then $A$ commutes with $C$.

I think it is false, but how can I solve it within 3 minutes?

Answer 1

In $\mathbb R^5$, let $A$ be the matrix for which switches the first two standard basis vectors and preserves the others. That is, $Ae_1=e_2,Ae_2=e_1,$ and $Ae_i=e_i$ for $i=3,4,5$.

Let $C$ be the matrix which switches $e_2$ and $e_3$ (preserving the others), and let $B$ be the matrix which switches $e_4$ and $e_5$ (preserving the other). We have $A$ commutes with $B$, and $B$ with $C$, but not $A$ with $C$.

Answer 2

Take any two non-commuting $2\times2$ matrices$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\text{ and }\begin{pmatrix}a'&b'\\c'&d'\end{pmatrix}.$$Now, take$$A=\begin{pmatrix}a&b&0\\c&d&0\\0&0&0\end{pmatrix},\ B=\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}\text{ and }C=\begin{pmatrix}a'&b'&0\\c'&d'&0\\0&0&0\end{pmatrix}.$$

Answer 3

Take $A,B,C$ block matrices of the form $$ A = \begin{pmatrix} A' & 0 \\ 0 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 0 \\ 0 & B' \end{pmatrix}, \quad C = \begin{pmatrix} C' & 0 \\ 0 & 0 \end{pmatrix}, $$ where $A',C'$ are $k \times k$, $B'$ is $(n-k)\times (n-k)$, and $A'C'-C'A' \neq 0$. Then $AB=BA=CA=AC=0$, but $$AC-CA = \begin{pmatrix} A'C'-C'A' & 0 \\ 0 & 0 \end{pmatrix} \neq 0. $$