# Can anybody check my answers (linear algebra)?

1. Oct 13, 2012

### Artusartos

1. The problem statement, all variables and given/known data

Let $A \in M_n(F)$ and $v \in F^n$.

Also...$[g \in F[x] : g(A)(v)=0] = Ann_A (v)$ is an ideal in F[x], called the annihilator of v with respect to A. We know that $g \in Ann_A(v)$ if and only if f divides g in F[x]. f is the monic polynomial of lowest degree in the set...so it is the minimal polynomial and divides everything in the set. Let $V = Span(v, Av, A^2v, ... , A^{k-1}v).$. V is the smallest A-invariant subspace containing v. We denote the fact by writing V=F[x]v. This corresponds to the F[x]-module structure on $F^n$ induced by multiplication by A. We also know that $v, Av, A^2v, ... , A^{k-1}v$ is a basis, B, of V.

Now these are the questions...

1) Define $Ann_A(V) =[g \in F[x] : g(A)(w) = 0 for all w \in V].$ Show that $Ann_A(V)=Ann_A(v)$

2) Let T: V -> V be induced by multiplication by A: T(w)=Aw for $w \in V$. Show that $Ann_A(V) = [g \in F[x] : g(T) = 0] = [g \in F[x]: g([T]_B)=0]$.

Here the first one means that g(T): V -> V is the 0-transformation and the second one means that g([T]_B) is the 0-matrix. Since $Ann_A(V) = (f) = [g \in F[x]: f|g]$, we write $f=min_T(x)$, the monic polynomial of lowest degree with f(T)=0.

2. Relevant equations

3. The attempt at a solution

1) In order to show that $Ann_A(V)=Ann_A(v)$, I need to show that $Ann_A(V) \subset Ann_A(v)$ and $Ann_A(v ) \subset Ann_A(V)$. It is clear that $Ann_A(V ) \subset Ann_A(v)$.

In order to show that $Ann_A(V) \subset Ann_A(v)$...

we need to show that g.v=0 implies that g.w=0 for all w in V. Since a field is an integral domain, we know that either g or v must be zero. We know that v cannot be zero, because...in the set $[g \in F[x] : g(A)(v)=0] = Ann_A (v)$, we know that f is the monic polynomial of lowest degree, and that f dividees every element in that set. If v was equal to zero, then all polynomials with coefficients in F[x] would be in that set...since zero times anything is zero. So we would also have constant polynomials, but f cannot divide a constant polynomial...so that would be a contradiction. So g(A) must be zero. Since g(A) is zero, g(A) times anything is zero...so g(A)w=0 for all w in V.

2) We say g(T) = g(A), where A is restricted to v.

$g(x) = c_0 + c_1x +... + c_tx^t$ and $g(T)w= (c_0 + c_1T + ... + c_tT^t)(w) = c_0w + c_1Tw +... + c_t(T)^t)(w) = c_0w + c_1Aw + ... + c_t(A)^tw = (c_0 + c_1A + ... + c_tA^t)(w) = g(A)w$.

So $[g \in F[x]: g(A).w = 0 for all w \in V]$ = $[g \in F[x]: g(T).w=0 for all w \in V]$ = $[g \in F: g(T) = 0]$.

Now for the second one, since we know that $[g \in F: g(T) = 0]$, we know that g sents every T to zero. Since T is in V, we also know that $[T]_B$ is also in V...so g must also sent $[T]_B$ to zero.

Do you think my answers are correct? If not, then can you tell me why?