Problem: Show that if $X$, a normed vector space over $\mathbb C$, has a finite dimension of at least $2$, then $T$ has a nontrivial inveriant subspace.
Attempt: Here is a solution I have available to the problem:
An operator on a finite dimensional space can be represented by an $n\times n$ matrix, where $\dim(X)=n$. The eigenvalues to such a matrix are given as the roots of the characteristic polynomial. By the fundamental theorem of algebra, we know that every non-constant, single variable polynomial with complex coefficients has at least one complex root. This means that such a matrix will always have at least one non-trivial eigenvalue.
Now, for such a nontrivial eigenvalue $\lambda\in\mathbb C$, there is at least one eigenvector $x\in X:x\neq0$ such that $Tx=\lambda x$, where $T$ is our matrix of interest. Thus, $\mathbb Cx\subseteq X$ is an invariant subspace.
To see that $\mathbb Cx$ is nontrivial, consider: since $\dim(X)\ge2$, we see that $\{0\}\neq\mathbb C x\neq X$ so that $\mathbb Cx$ is not trivial.
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I think I get the jist of the proof, but there are two points in the above I am unsure on, both highlighted in bold.
What is the notation $\mathbb Cx$? Now I think the claim being demonstrated here is that the set of all eigenvectors of $T$ are an invariant subspace for $T$, because clearly, applying $T$ to the set of eigenvectors gives you scalar multiples of those same vectors, which is a subspace of $X$. On the other hand, I think $\mathbb Cx$ is strictly the application of each complex number to every $x$ in $X$. I don't think this is the same as the set of eigenvectors of $T$, although is this what is being hinted at with this notation? Or am I mistaken?
I also don't immediately see the claim that $\{0\}\neq\mathbb C x\neq X$ when the dimension of $X$ is greater than one; could somebody provide a clarification of why this is the case?
Answer to first question:
Here $\mathbb Cx$ means subspace of $X$ generated by $x$.
i.e. $\mathbb Cx = span \{x\} = \{cx:c \in\mathbb C$}
Answer to second question:
As $x \neq 0$ implies dim{$\mathbb Cx$} = $1$ and since dim $X$ is at least 2.
Hence $\{0\}\neq\mathbb C x\neq X$