Linear transformation basis problem So the question asks: given $A=\begin{bmatrix}0&2&0\\ 0&0&0\\4&1&0\end{bmatrix}$
(a) Compute $A^2$ and $A^3$
(b) Find a vector $x$ such that $A^2x ≠ 0$.
(c) Show that if $v$ is any vector satisfying $A^2v ≠ 0$, then the set $B_v=\{v, Av, A^2v\}$ is a basis of $\mathbb{R}^3$. 
(d) Let $T_A$ be the linear transformation associated to the matrix A (that is, $T_A(x) = Ax,\ \forall x ∈ \mathbb{R}^3$). Find the matrix $\left[T_A\right]_{B_v}$ of the linear transformation $T_A$ with respect to the base $B_v$ (where $v$ is any vector satisfying $A^2v = 0$ and $B_v$ is as deﬁned in part (c) above).
So so far I have: 
(a) $A^2=\begin{bmatrix}0&0&0\\ 0&0&0\\0&8&0\end{bmatrix},\ A^3=\begin{bmatrix}0&0&0\\ 0&0&0\\0&0&0\end{bmatrix}$
(b)Suppose $x = \begin{bmatrix}a\\ b\\c\end{bmatrix}$, then $A^2x$=$\begin{bmatrix}0\\ 0\\8b\end{bmatrix}$. So
$A^2x\neq 0$ when $b \neq 0$
i.e. $x = \begin{bmatrix}0\\ 1\\0\end{bmatrix}$
(c) Suppose $B_v=\{v, Av, A^2v\}$ spans $\mathbb{R}^3$
Suppose $v = \begin{bmatrix}a\\ b\\c\end{bmatrix}$, then $Av = \begin{bmatrix}2b\\ 0\\4a+b\end{bmatrix}$, $A^2v=\begin{bmatrix}2ab\\ 0\\4a^2+ab\end{bmatrix}$
$$
Bv=\left\{\begin{bmatrix}a\\ b\\c\end{bmatrix}, \begin{bmatrix}2b\\ 0\\4a+b\end{bmatrix}, \begin{bmatrix}2ab\\ 0\\4a^2+ab\end{bmatrix}\right\}
$$
Since the only coefficients $c_i$ that satisfy the equation 
$$
c_1\begin{bmatrix}a\\ b\\c\end{bmatrix}+c_2\begin{bmatrix}2b\\ 0\\4a+b\end{bmatrix}+c_3\begin{bmatrix}2ab\\ 0\\4a^2+ab\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
$$
are $c_1=c_2=c_3=0$
So $v$ and $Av$ and $A^2v$ are linearly independent. So it is a basis of $\mathbb{R}^3$
But I am not sure what the question (d) is asking for. Is it asking for the similar matrix of $A$, like $AS=SB$? And I am not confident about my proof in (c). Can I just suppose they span $\mathbb{R}^3$, and only prove they are linearly independent, or I have to prove they span $\mathbb{R}^3$ as well? 
 A: For item (c), you must prove that $B_v$ spans everything, or that it is linearly independent, so you can't start assuming that is spans everything. Write $$av + bAv + cA^2v = 0,$$and we now must check that $a=b=c= 0$. This is easily done as following: applying $A^2$ on both sides we'll get $a = 0$. Then we go back at the linear combination and apply $A$ to get $b = 0$. Now we're left with $cA^2v = 0$, which gives $c=0$. Notice how we're using everytime that $A^2v \neq 0$.
For item (d), recall that if $B = \{u_i\}_{i=1}^n$ and $C=\{v_i\}_{i=1}^m$ are bases, then $[T]^B_C = [a_{ij}]_{m \times n}$ is characterized by $T(u_j) = \sum_{i=1}^m a_{ij}v_i$.
With this in mind, to find $[T_A]_{B_v}$, we compute: $$\begin{align} T_A(v) &= Av = 0 \cdot v + 1 \cdot Av + 0 \cdot A^2v \\ T_A(Av) &= A^2v = 0 \cdot v + 0 \cdot Av + 1 \cdot A^2v  \\ T_A(A^2v)&= A^3v = 0 =0 \cdot v + 0 \cdot Av + 0\cdot A^2v,\end{align}$$so: $$[T_A]_{B_v} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}.$$
