Suppose that $A$ is $3\times 3$ and $\text{rank}(I-A) = 2$ and $\text{rank}(-I-A) = 1$. Then A is diagonalizable. I'm asked to determine the truth or falsity of the following statement:
Suppose that $A$ is $3\times 3$ and $\text{rank}(I-A) = 2$ and $\text{rank}(-I-A) = 1$. Then A is diagonalizable.
I think the answer is yes but I need help confirming this answer.
The most intuitive first step I could think of was given that that $\text{rank}(I-A) = 2 <3$ and $\text{rank}(-I-A) = 1 < 3$, this means that $\det(I-A) = \det(-I-A) = 0$ and thus we have eigenvalues $\pm 1$, with two distinct eigenvectors for $\lambda = -1$. I would think that because the matrix is $3\times 3$ and that we have three distinct eigenvectors, thus $A$ is diagonalizable.
I wasn't satisfied with this answer, because I thought the characteristic polynomial is not given, thus how does one know the algebraic multiplicity of the eigenvalues? Sure, if the eigenvalues were $-1,-1,1$ then it would be true. It would also work if there were a third distinct eigenvalue $\lambda \not= \pm 1$, but what if the eigenvalues were $1,1,-1$?
The fact that $\text{rank}(-I-A) = 1$ means that the columns of $A+I$ are all multiples of one another. Thus I'm guessing the matrix $A$ has to look something like
$$A=\begin{pmatrix}
h\cdot a_{11}-1 & k\cdot a_{11} & l\cdot a_{11} \\
h\cdot a_{21} & k\cdot a_{21}-1 & l\cdot a_{21} \\
h\cdot a_{31} & k\cdot a_{31} & l\cdot a_{31}-1 \\
\end{pmatrix}$$
for $h,k,l \in \mathbb{R}$.
Although taking the determinant might seem complicated at first, many terms end up cancelling and $\det(A) = h\cdot a_{11} + k\cdot a_{21} + l\cdot a_{31} -1$, and since I'm assuming the characteristic polynomial is $-(t-1)^2\cdot(t+1) = -t^3 + t^2 + t -1 = -t^3 + \text{trace}(A)t^2 + t +\det(A)$, $\det(A)=-1$. Thus $h\cdot a_{11} + k\cdot a_{21} + l\cdot a_{31} =0$, but $\text{trace}(A) = h\cdot a_{11} + k\cdot a_{21} + l\cdot a_{31} - 3 = 1$, while also $h\cdot a_{11} + k\cdot a_{21} + l\cdot a_{31} = 4$ and we have a contradiction. Thus we can't have the case where the eigenvalues are $1,1,-1$ and $\text{rank}(-I-A) = 1$.
Am I missing something? I guess if my answer is complete, my question is, is there a quicker way to get the conclusion building on the concept of $\text{rank}(-I-A)=1$? Skipping the calculations above.
Any help is appreciated (:
 A: Yes.
From $\text{rank}(I-A)=2$ and $\text{rank}(-I-A)=1$, we can get $1$ and $-1$ are the eigenvalues of $A$.
Use $\text{Jordan-Form-Theory }$,we can get the number of Jordan block with -1 as the  diagonal elements is $2$ and the number of Jordan block with 1 as the  diagonal elements is $1$.
$A$  is  $3\times 3$,so  there exists an invertible matrix P such that $P^{-1}AP=\left( \begin{matrix}
 1&  0&  0\\
 0&  1&  0\\
 0&  0&  -1\\
\end{matrix} \right) $.
A: Your idea is correct, you mentioned that $\lambda=-1$ has two distinct eigenvectors and $\lambda=1$ has only one eigenvector.
Call $\gamma_\lambda$ the geometric multiplicity of $\lambda$ and $\mu_\lambda$ its algebraic multiplicity, then the following inequality holds for every eigenvalue:
$$1\leq \gamma_\lambda\leq\mu_\lambda\leq3,$$
or in general $n$ when we have an $n\times n$ matrix. If one eigenvalue had algebraic multiplicity $3$, say $\lambda=-1$, then its geometric multiplicity would also be 3 and thus $\lambda=1$ would not have any geometric multiplicity.
You also ask, what would happen if there was a different eigenvalue, but it's impossible for there to be another eigenvalue. As the matrix is $3\times 3$, it can have at most three eigenvalues and the sum of their multiplicities should be less than or equal to $3$. If $\lambda$ was another eigenvalue, then $\gamma_\lambda$ is at least $1$ and summing
$$\gamma_\lambda+\gamma{-1}+\gamma_1\geq1+2+1=4$$
we get an impossibility. I'd rather not work with the determinant because most of this facts can be worked just using the dimensions of certain spaces.
If you'd like me to elaborate more, let me know :)
