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Artusartos
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Homework Statement
Let [itex]A \in M_n(F)[/itex] and [itex]v \in F^n[/itex]. Let k be the smallest positive integer such that [itex]v, Av, A^2v, ..., A^kv[/itex] are linearly dependent.
a) Show that we can find [itex]a_0, ... , a_{k-1} \in F[/itex] wiht
[itex]a_0v + a_1Av + ... + a_{k-1}A^{k-1}v + A^kv = 0[/itex]
(note that teh coefficient of A^kv is 1). Write
f(x) = x^k + a_{k-1}x^{k-1} + ... + a_0.
Note that f(A)(v)=0.
b) Show that f is a monic polynomial of lowest degree with f(A)(v)=0.
I tried to answer these questions, but I'm not sure if my answers are correct...can anybody please check them for me?
Homework Equations
The Attempt at a Solution
a) Since k be the smallest positive integer such that [itex]v, Av, A^2v, ..., A^kv[/itex] are linearly dependent, we know that [itex]v, Av, A^2v, ..., A^{k-1}v[/itex] is linearly independent. So, for [itex]v, Av, A^2v, ..., A^kv[/itex], we know that for any linear combination of these elements, there exists a coefficient that is nonzero (when the whole equation is zero)...In other words:
[itex]b_0v + b_1Av + ... + b_{k-1}A^{k-1}v + b_kA^kv = 0[/itex] for b_0, ... , b_k not all zero.
Now if we divide the whole equation by b_k, we get
[itex] \frac{b_0}{b_k} v + \frac{b_1}{b_k} Av + ... + \frac{b_{k-1}}{b_k} A^{k-1}v + A^kv = 0[/itex].
If we denote [itex] \frac{b_m}{b_k}[/itex] by [itex]a_m[/itex] (where m is between zero and k), we get...
[itex]a_0v + a_1Av + ... + a_{k-1}A^{k-1}v + A^kv = 0[/itex]
Now to show that f(A)(v)=0 ...
[itex]f(A)(v) = (A^k + a_{k-1}A^{k-1} + ... + a_0v)(v) = a_0v + a_1Av + ... + a_{k-1}A^{k-1}v + A^kv = 0[/itex]
b) [itex]f(A) = f(A)(v) = (A^k + a_{k-1}A^{k-1} + ... + a_0v)(v) = a_0v + a_1Av + ... + a_{k-1}A^{k-1}v + A^kv = 0 [/itex]
Can anybody check if I'm right? I'm also not sure if I understood the questions correctly?