Confusion about eigenvalues of an operator

[maNoFchangE]

Suppose ##V## is a complex vector space of dimension ##n## and ##T## an operator in it. Furthermore, suppose ##v\in V##. Then I form a list of vectors in ##V##, ##(v,Tv,T^2v,\ldots,T^mv)## where ##m>n##. Due to the last inequality, the vectors in that list must be linearly dependent. This implies that the equation
$$
0=a_0v+a_1Tv+a_2T^2v+\ldots+a_mT^mv
$$
are satisfied by some nonzero coefficients. For a particular case, assume that the above equation is satisfied by some choice of coefficients where all of them are nonzero.
Now since the equation above forms a polynomial, I can write the factorized form
$$
0=A(T-\mu_1)\ldots(T-\mu_m)v
$$
The last equation suggests that ##T## has ##m## eigenvalues. But this contradicts the fact that ##T## is an operator in a vector space of dimension ##n<m##. Where is my mistake?

 

[suremarc]

##T## is an operator whereas ##\mu_i## is a scalar. Instead you should write ##T-\mu_iI_n##.

Suppose ##m=n+1##, and take ##\mu_1, \ldots, \mu_n## to be the eigenvalues of ##T##. Then clearly the final equation holds for an arbitrary choice of ##\mu_{n+1}##. So your mistake is in concluding that ##T## must have more than ##n## eigenvalues, which is a non sequitur.

 

[maNoFchangE]

Suppose ##m=n+1##, and take ##\mu_1, \ldots, \mu_n## to be the eigenvalues of ##T##. Then clearly the final equation holds for an arbitrary choice of ##\mu_{n+1}##. So your mistake is in concluding that ##T## must have more than ##n## eigenvalues, which is a non sequitur.

Why does it have to be ##m=n+1##? I can take ##m## any value I want such that it is bigger than ##n##.

 

[suremarc]

Why does it have to be ##m=n+1##? I can take ##m## any value I want such that it is bigger than ##n##.

Indeed, you can--I was simply providing a counterexample to get you started.
The eigenvalues of ##T## will always be contained in some proper subset of ##\{\mu_1,\ldots,\mu_m\}##, so at least 1 of the choices for ##\mu_i## will be arbitrary.

 

[Demystifier]

$$
0=A(T-\mu_1)\ldots(T-\mu_m)v
$$
The last equation suggests that ##T## has ##m## eigenvalues.

No it doesn't. From the equation above it does not follow that
$$(T-\mu_i)v=0$$