Let $A$ be a $(n \times n)$ complex matrix, and for $C$ a $(1 \times n)$ row vector let $W = \{\sum_{j=0}^{n-1}a_jCA^j:j=0,\dots,n-1\}\;.$ (i) Use the Cayley-Hamilton theorem to show that right multiplication $X \mapsto XA$ for $X \in W$ defines a linear operator from $W$ to $W$. 
I'm really not sure to answer this question.  
I know that by the Cayley-Hamilton theorem, $\chi_A(A)=0$ so that
$$A^n = trace(A)A^{n-1}-\dots+(-1)^n(detA)I$$
How do I continue? 
 A: We observe that we may define $W$ as
$W = \displaystyle  \left \{ \sum_0^{n - 1}a_j C A^j \right \} =  \left \{ C\sum_0^{n - 1}a_j A^j \right \}; \tag 1$
that is, $W$ is the set of all row vectors of size $n$ which lie in the image of $C$ under right multiplication by the set of polynomials $\{ \sum_0^{n - 1} a_j A^j \}$ in $A$ of degree at most $n - 1$ with coefficients in $\Bbb C$.  It is clear that $W$ is indeed a subspace of $\Bbb C^n$, considered as the set of $n$-dimensional row vectors $(z_1, z_2, \ldots, z_n)$ where $z_i \in \Bbb C$, $1 \le n \le n$; for example, we trivially have
$\displaystyle C\sum_0^{n - 1}a_j A^j, C\sum_0^{n - 1}b_j A^j \in W \Longrightarrow  C\sum_0^{n - 1}a_jA^j +  C\sum_0^{n - 1}b_jA^j =  C\sum_0^{n - 1}(a_j  + b_j)A^j \in W, \tag{2}$
$\displaystyle  C\sum_0^{n - 1} a_j A^j \in W \Longrightarrow C\sum_0^{n - 1} \alpha a_j A^j \in W \Longrightarrow \alpha C\sum_0^{n - 1}a_j A^j \in W, \tag{3}$
where $\alpha \in \Bbb C$.
Next, let
$V = \displaystyle  \left \{ \sum_0^n a_j C A^j  =  C\sum_0^n a_j A^j \right \}; \tag 4$
that is, $V$ is defined in a similar manner to $W$, but the polynomials in $A$ are allowed to be of degree at most $n$.  It is easy to see that $V$ is also a subspace of $\Bbb C^n$, and by taking $a_n = 0$ in (4) we have that
$W \subset V; \tag 5$
furthermore, it is easy to see from inpsection of (1)-(4) that
$WA \subset V, \tag 6$
and that the map sending $X \in W$ to $XA \in V$ is manifestly linear; if we can show that
$V \subset W, \tag 7$
our proof of the desired result will be complete; this is where Cayley-Hamilton enters into the picture, asserting as it does that $A$ satisfies a monic polynomial of degree $n$:
$A^n = \displaystyle \sum_0^{n - 1} c_i A^i = -\text{Tr}(A) A^{n - 1} + \sum_1^{n - 2} c_i A^i + (-1)^n \det(A); \tag 8$ 
by virtue of (8), we may re-write any element of $V$ in terms of the powers $A^i$, with $i \le n -1$; we have
$\displaystyle C\sum_0^n a_j A^j = C\sum_0^{n - 1} a_j A^j + a_nCA^n =  C\sum_0^{n - 1} a_j A^j + a_nC \sum_0^{n - 1} c_i A^i \in W, \tag 9$
since we have seen in (2), (3) that $W$ is a vector space over $\Bbb C$.  Thus (7) binds and we have
$V = W; \tag{10}$
since we have seen that the right multiplication by $A$, sending $X \to XA$, maps $W \to V$, we conclude from (10) that 
$A:X \to XA, \; W \to W \tag{11}$
as required.
A: Hint: You want to show that if $X \in W$, then it's also true that $XA \in W$.  To that end: by the C-H theorem, there exist coefficients $b_j$ such that
$$
A^n = \sum_{j=0}^{n-1}b_jA^j
$$
