The following is from an old exam.
Let $V$ be a vector space, and $T:V\rightarrow V$ a linear transformation that is not diagonalizible.
Show that if $1$ and $2$ are eigenvalues of $T$, then $dimV>2$.
I don't understand the given answer. There is a theorem in our book which states that:
If $\lambda_1,\dots ,\lambda_k$ are distinct eigenvalues of a linear transformation $T$, and if $v_1,\dots ,v_k$ are their corresponding eigenvectors, then $v_1,\dots ,v_k$ are linearly independent.
The answer then refers to the above theorem and concludes that there are at least 2 linearly independent eigenvectors and therefore $dimV\geq 2$.
From where do they conclude that there are 2 linearly independent eigenvectors? As far as I know there is a theorem which states that the geometric multiplicity is always less than or equal to the algebraic multiplicity, but I'm not aware of anything which states that the geometric multiplicity is always at least 1 ( since that's not true ).
The only thing I can think of is that the existence of two distinct non-zero eigenvalues might imply $dimN(A-\lambda I)\geq1$ for each $\lambda_i$.
But of course the point is they don't say anything about that subject and I just don't get how they conclude that there are at least two linearly independent eigenvectors.