Let $T : R^{3} \to R^{3}$ be the linear transformation defined by $T(x,y,z) = (x+3y+2z,3x+4y+z,2x+y-z)$. Find the rank of $T^{2}$ and $T^{3}$. Let $T : R^{3} \to R^{3}$ be the linear transformation defined by $T(x,y,z) = (x+3y+2z,3x+4y+z,2x+y-z)$. Find the rank of $T^{2}$ and $T^{3}$.
I formed the matrix of linear transformation for $T$ and squared it, and found the rank which is $2$. For $T^{3}$ I found the matrix by multiplying matrix $T$ by $T^{2}$. 
I need to calculate the rank of this matrix.
Is this only way to calculate rank$?$ 
This question is asked as multiple choice problem, which should not take much time. 
But finding rank in this way is really time consuming. Is there any other way$?$
 A: $$T=\begin{bmatrix} 1 & 3 & 2 \\ 3 & 4 & 1 \\ 2 & 1 & -1 \end{bmatrix}$$
$\det{T}=0$ so its rank is < 3. It is actually 2 as $$\begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}=-5 \ne 0.$$
Then $$T^2=\begin{bmatrix} 14 & 17 & 3 \\ 17 & 26 & 9 \\ 3 & 9 & 6 \end{bmatrix}$$
$\det{T^2}=0$, so its rank is < 3. It is actually 2 as $$\begin{bmatrix} 14 & 17 \\ 17 & 26 \\ \end{bmatrix}\ne 0.$$
A: This problem can be solved without ever explicitly computing a power of the matrix of $T$, $$A=\begin{bmatrix}1&3&2\\3&4&1\\2&1&-1\end{bmatrix}.$$ First, it’s easy to spot that the second column of $A$ is the sum of the other two, which themselves are obviously linearly independent, therefore the null space of $A$ is spanned by $(1,-1,1)^T$.  
Now, $$\begin{vmatrix}1&2&1\\3&1&-1\\2&-1&1\end{vmatrix} = \begin{vmatrix}1&2&1 \\ 4&3&0 \\ 1&-3&0\end{vmatrix} \ne 0,$$ so the null space of $A$ and its column space have only the trivial intersection. This means that no vectors are mapped by $T$ to nonzero elements of its null space, which in turn means that the null spaces of every positive power of $A$ are identical.
