Let $A$ be a $(n \times n)$ complex matrix, and for $C$ a $(1 \times n)$ row vector let $W = \{\sum_{j=0}^{n-1}a_jCA^j:j=0,\dots,n-1\}\;.$

Question

(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?

Answer 1

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 $$ A^n = \sum_{j=0}^{n-1}b_jA^j $$

Answer 2

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.