# Confusion about eigenvalues of an operator

Tags:
1. Feb 21, 2016

### 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?

Last edited: Feb 21, 2016
2. Feb 21, 2016

### 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.

3. Feb 21, 2016

### maNoFchangE

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

4. Feb 21, 2016

### suremarc

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.

5. Mar 1, 2016

### Demystifier

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