matrix for cyclic subspace

Question

For the linear operator $T$ on the vector space $V$, find an ordered basis for the $T$-cyclic subspace generated by the vector $z$.

$V=P_3(\Bbb{R})$, $T(f(x))=f"(x)$, and $z=x^3$

So I have: $$T(z)=6x$$ $$T^2(z)=0$$

Thus, the ordered basis for the cyclic subspace generated by the vector $z$ is $\{z,T(z)\}$

However, I am having trouble finding the matrix for the cyclic subspace. For some reason, I am stuck. Thanks!

Answer 1

Presumably, you mean the matrix of the linear operator with respect to this ordered basis. Denote $v_1 = z, v_2 = T(z)$, so that our basis is $\mathcal B = \{v_1,v_2\}$. We have $$ T(v_1) = 0v_1 + 1v_2\\ T(v_2) = 0v_1 + 0v_2 $$ use each of these rows to construct a column of the matrix. In particular, we have $$ [T]_{\mathcal B} = \pmatrix{0&0\\1&0} $$