Let $$A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 3 \end{pmatrix},$$ where $A \in M_{3 \times 3} (\Bbb R)$. Test $A$ for diagonalizability.
$$\det(A-\operatorname{\lambda I})=\det \begin{pmatrix} 1-\lambda & 1 & 0 \\ 0 & 1-\lambda & 2 \\ 0 & 0 & 3-\lambda \end{pmatrix}=(1-\lambda)^2(3-\lambda).$$
Compare algebraic and geometric multiplicity.
For $\lambda=1$, I get the eigenvector $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, and for $\lambda=3$, I get $\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$
Since we can't find enough eigenvectors to fill the eigenspace for $3 \times 3$ matrix, it is not diagonalizable.
Am I on the right track? Also, for an eigenvalue with algebraic multiplicity of, say, $2$ , must it have $2$ independent eigenvectors? But I can only find $1$.
Your calculation of the eigenvalues and eigenvectors is correct, and your conclusion that $A$ is not diagonalizable is also correct.
$A$ is diagonalizable iff the geometric multiplicity of every eigenvalue is equal to its algebraic multiplicity. The eigenvalue $\lambda = 1$ has algebraic multiplicity $2$, but geometric multiplicity $1 \neq 2$, so (as you ask at the end) there should necessarily be two independent eigenvectors, which is not the case.