commutant of an operator when $dim E=2$ Let $B$ an operator on a two-dimensional Hilbert space $E$ such that $B$ is not a scalar operator and let $\mathfrak{C}$ be the commutant of $B$ (i.e. $\mathfrak{C}$ of $B$ is the set of all Operators that commute with $B$.)  If $T$ is in $\mathfrak{C}$, why there exist some complex numbers $a$ and $b$ such that $T = aB + b$ ??
 A: $AB-BA=0$ is a set of $4$ equations with four unknowns. 
$$\begin{bmatrix}a_{12} b_{21} - a_{21}b_{12} & -a_{12} b_{11} + a_{11} b_{12} - a_{22} b_{12} + a{12} b_{22}\\
a_{21} b_{11} - a_{11} b_{21} + a_{22} b_{21} - a_{21} b_{22} & a_{21} b_{12} - a_{12} b_{21}\end{bmatrix} = \begin{bmatrix}0 & 0\\
0 & 0\end{bmatrix}$$
The two equations coming from the diagonal are the same. This is just $tr(AB-BA)=0$.
So, three equations really. Moreover, that equation on the diagonal only has
two unknowns. Therefore it already gives you one unknown in terms of the other. Unless it is identically zero.
Case 1: The diagonal equation is identically zero. Therefore $b_{12}=b_{21}=0$. Since $B$ is non-scalar, it follows that $a_{12}=a_{21}=0$. And the space of solutions is $2$-dimensional, at most.
Case 2: The diagonal equation is not identically zero. In this case $a_{12}$ determines $a_{21}$, or the other way around. This is because only those two unknowns are in present in that equation. Having $a_{12}$ and $a_{21}$, the remaining equations express $a_{11}$ in terms of $a_{21}$ or the other way around.
Therefore the commutant has dimension $2$ at most. Since $aB+b$ has dimension $2$ that is all of it.
