Showing algebraic multiplicity of eigenvalue is at least two I'm currently working on the following question:

Let $T$ be a linear operator that has two linearly independent eigenvectors with the same eigenvalue $\lambda$. Prove that $\lambda$ is a multiple root of the characteristic polynomial of $T$. In other words, show that $(t-\lambda)^2$ divides the characteristic polynomial.

I find problems like this incredibly frustrating, because I see it's clearly true - but proving it is another matter.
Say that $T : V \rightarrow V$, then there is a basis $\bf{B}$ 
 of $V$ such that the matrix of $T$ with respect to that basis is upper triangular and since $\lambda$ has two linearly independent eigenvectors $v_1, v_2$ we can find a $\bf{B}$ such that $v_1, v_2 \in \bf{B}$ and so this gives:
$\chi_A(t) = \det(tI - A) = (t-a_{11})(t-a_{22})...(t-a_{nn})$
Since $v_1, v_2$ are column vectors of the matrix $A$, we have that two of $a_{ii} = \lambda$ so
$\det(tI - A) = (t-a_{11})(t-a_{22})...(t-\lambda)(t-\lambda)...(t-a_{nn})$
$\det(tI - A) = (t-a_{11})(t-a_{22})...(t-\lambda)^2...(t-a_{nn})$
and so $(t-\lambda)^2$ divides the characteristic polynomial. This is my attempt at it, but I'm not convinced it even works. Thanks for any help.

There are two issues with your proof. First, it is not necessarily true that you can find a basis $B$ such that the matrix of $T$ w.r.t. $B$ is upper triangular. You can always find one over $\mathbb{C}$, but not over $\mathbb{R}$, not to mention other fields. Second, even if such bases exist, why can you find one that contains $v_1$ and $v_2$?
Actually, you may prove the statement in a legitimate but slightly different way. Extend $\{v_1,v_2\}$ to a basis $B$ of $V$. Then the matrix of $T$ w.r.t. $B$ is block upper triangular, where the upper left block is $\lambda I_2$. Note that for a block matrix $M=\begin{pmatrix}A&B\\0&C\end{pmatrix}$ where $A,C$ are square matrices, $\det M=(\det A)(\det C)$. Now the rest is straightforward.
