# Let A=((0,1),(0,1)). Let T linier operator toR^(2x2), and T(X)=AX-XA, AAXinR^(2x2). Determine rank(T) ?

##### 1 Answer
Sep 28, 2017

We have:

$\boldsymbol{A} = \left(\begin{matrix}0 & 1 \\ 0 & 1\end{matrix}\right) \setminus \setminus \setminus$; and; $\setminus \setminus \underline{\boldsymbol{T}} \left(\boldsymbol{X}\right) = \boldsymbol{A X} - \boldsymbol{X A} \setminus \setminus \forall \boldsymbol{X} \in {\mathbb{R}}^{2 \times 2}$

Consider a generic element $\boldsymbol{X} \in {\mathbb{R}}^{2 \times 2}$, having terms:

$\boldsymbol{X} = \left(\begin{matrix}{a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22}\end{matrix}\right)$

Then consider the effect of the linear operator $\underline{\boldsymbol{T}}$ on the matrix $\boldsymbol{X}$;

$\underline{\boldsymbol{T}} \left(\boldsymbol{X}\right) = \boldsymbol{A X} - \boldsymbol{X A}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}0 & 1 \\ 0 & 1\end{matrix}\right) \left(\begin{matrix}{a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22}\end{matrix}\right) - \left(\begin{matrix}{a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22}\end{matrix}\right) \left(\begin{matrix}0 & 1 \\ 0 & 1\end{matrix}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}{a}_{21} & {a}_{22} \\ {a}_{21} & {a}_{22}\end{matrix}\right) - \left(\begin{matrix}0 & {a}_{11} + {a}_{12} \\ 0 & {a}_{21} + {a}_{22}\end{matrix}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}{a}_{21} & {a}_{22} - {a}_{11} - {a}_{12} \\ {a}_{21} & {a}_{22} - {a}_{21} - {a}_{22}\end{matrix}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}{a}_{21} & {a}_{22} - {a}_{11} - {a}_{12} \\ {a}_{21} & - {a}_{21}\end{matrix}\right)$

The vectors:

$\left(\begin{matrix}{a}_{21} \\ {a}_{21}\end{matrix}\right) \setminus \setminus \setminus$; and; $\left(\begin{matrix}{a}_{22} - {a}_{11} - {a}_{12} \\ - {a}_{21}\end{matrix}\right)$

Are linearly independent, and therefore:

$r a n k \left(\underline{\boldsymbol{T}}\right) = 2$

even though $r a n k \left(\boldsymbol{A}\right) = 1$