Let $A$ be a $3 \times 3$ matrix such that
$\lambda_1=-1$ and $\lambda_2=2$ as two of its eigenvalues.
rank$(A+I)=1$.
I am trying to show that A is diagonalizable.
I know that from the nullity theorem that the multiplicity of eigenvalue -1 is 2.
And I need 3 linearly independent eigenvectors to prove that A is diagonalizable.
So my questions are:
How can I say that 2 independent eigenvectors correspond to the eigenvalue -1? Or can I say something like that?
How can I be sure what will correspond to the eigenvalue 2?
I am a bit confused.
Since $\operatorname{rank}(A + I) = 1$, we have that $\dim(\operatorname{null}(A + I)) = 2$. In other words, the $0$ eigenvalue of $A + I$ has a geometric multiplicity of $2$.
Let $v_1$ and $v_2$ be two linearly independent vectors spanning $\operatorname{null}(A + I))$ (i.e., the $0$ eigenspace of $A + I$).
The claim is that $v_1$ and $v_2$ are eigenvectors of $A$ corresponding to the eigenvalue $\lambda_1 = -1$. This is easily checked through: $$\begin{align} A v_i &= (A + I - I) v_i\\ &= (A + I)v_i - v_i\\ &= (-1) v_i \end{align}$$
Edit: Thus, the eigenspace corresponding to $\lambda_1 = -1$ has a geometric dimension of two. For your second question, there exists at least one eigenvector corresponding $\lambda_2 = 2$ (otherwise it wouldn't be an eigenvalue). That is, every eigenspace has a geometric dimension of at least one.