Showing a $2 \times 2$ matrix that commutes with two non-commuting matrices is a scalar matrix Let $A, \; B,$ and $C$ be $2 \times 2$ matrices such that $AB = BA$ and $AC = CA$, but $BC \neq CB$.  Show that $A$ is a scalar matrix, that is show that for some $k \in \mathbb{R}$, that $A = kI$.
Now, here is what I want to do.  I want to use the face that since $A$ and $B$ commute, that they can be simultaneously upper triangularized.  To show this, we have that,
\begin{align}
A = P
\left(\begin{matrix}
a_{1} & a_{2} \\
0 & a_{4}
\end{matrix}\right)P^{-1}
\end{align}
\begin{align}
B = P
\left(\begin{matrix}
b_{1} & b_{2} \\
0 & b_{4}
\end{matrix}\right)P^{-1}.
\end{align}
I want to use this to show that if the matrices commute, then I can rule out eventually that $a_{2} = 0$, and then further show that $a_{1} = a_{4}$.  Now, my issue is that I can do this if I use the fact that $A$ and $C$ commute upper-triangularly, but not necessarily with the same matrix $P$.  
 A: Define a map $\psi_A \colon M_2(\mathbb{R}) \rightarrow M_2(\mathbb{R})$ by $\psi_A(M) = [A,M] = AM - MA$. You want to show that $\psi_A \equiv 0$. We immediately see that $A, B, C, I \in \ker(\psi_A)$. If $\psi \neq 0$, then necessarily $A \neq 0$. The matrix $B$ is not a multiple of $A$ (for if it were, $BC = CB$ would follow from $AC = CA$), so $\{A,B\}$ is a linearly independent set of vectors in $M_2(\mathbb{R})$. Similarly, check that $C$ is not linear combination of $A$ and $B$ and that $I$ is not a linear combination of $A,B,C$ and so $\dim \ker(\psi_A) = 4$ and hence $\psi_A \equiv 0$, a contradiction.
A: Hint: If $A$ is not diagonalizable, then without loss of generality you can assume that $A$ is upper triangular with $1$ in the upper right corner and a constant $\lambda$ on the diagonal. Use this to show that if $A$ commutes with both $B$ and $C$, and $B,C$ are not the zero matrix then $B$ and $C$ are both scalar multiples of the the same form as $A$ (possibly with different diagonal constants) and then show this implies that $B$ commutes with $C$, a contradiction.  If $A$ is diagonalizable then it's much easier. Show that if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$ then so is $Bv$. Thus if $v$ is not an eigenvector of $B$ then $A$ is a scalar multiple of the identity. Same for $C$. Thus $A,B,C$ all have the same eigenvectors unless $A$ is a scalar multiple of the identity, so if $A \neq kI$ then $B$ and $C$ be diagonalized by a common basis $P$ so they commute, a contradiction.
