Kernel of $(I-A)^2$ where $A$ has unique real eigenvalue $1$ Assume I have a real square matrix $A$ with a simple eigenvalue $1$ and corresponding eigenvector $v$. Then $v$ should span the kernel of the matrix $I-A$, where $I$ is the identity matrix. Now I am wondering the following: does $v$ also span the kernel of $(I-A)^2$? Writing $(I-A)^2 x = 0 \iff (I-A)x = A(I-A)x$ this question should be equivalent to asking whether $\operatorname {span}(v) = \ker(I-A)$ can lie in the image of $I-A$.
Thanks in advance for any help on this matter.
 A: Yes it does, and it seems you have kind of stumbled on to the fascinating concept of generalized eigenvectors. If we think of $(I-A)$ as a linear transformation then applying it once to the eigenvector $v$ yields the following:
$$(I-A)v=0$$
Applying that linear transformation twice yields the following:
$$(I-A)^2v=(I-A)(I-A)v=(I-A)0=0$$
so if $v$ is eigenvector of $A$ then necessarily we have that $v$ lies in the kernel of both $(I-A)$ and $(I-A)^2$.
About Generalized Eigenvectors, if you look at the following matrix:
$$A=\begin{pmatrix}
2&1&0\\
0&2&1\\
0&0&3\\
\end{pmatrix}$$
We find that we have eigenvalues:$\lambda_1=2,\lambda_2=3$, where $\lambda_2$ has multiplicity $2$. Now this is a $3\times3$ matrix, but we only have two eigenvalues, and if we work out the eigenvectors through the usual process we find that we have only two genuine eigenvectors. However there is a way of "generalizing" the concept of eigenvector so that we obtain a basis of general eigenvectors. First you find the genuine for $\lambda_1$:
$$(A-2I)v_1=0\Rightarrow v_1=v_1=\begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix}t$$
Now, since the genuine eigenspace for an eigenvalue of multiplicity 2 is one dimensional this implies we have a general eigenvector. So we must find the general eigenvector by doing the following:
$$(A-2I)^2v_2=0\Rightarrow v_2=v_2=\begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix}t+\begin{pmatrix}
0\\
1\\
0\\
\end{pmatrix}s$$
As you can see the generalized eigenvector is 2 dimensional, and contains the genuine eigenvector. This generalized eigenvector combined with the other genuine eigenvector forms a complete basis of our vector space. Also, with generalized eigenvectors we have the following relation as well:
$$(A-\lambda_1I)v_2=v_1$$
This was a very rough and dirty crash course on generalized eigenvectors, and actually rigorously  developing the concept takes more time than I have to type out an answer, however many upper level linear algebra textbooks go into this topic and I'm sure there a quite a few online resources that could help as well if you're interested.
Now to Finally answer the question at hand since I misunderstood it the first time:
First let $A$ represent a linear transformation $T:V\rightarrow W$, for some $\mathbb{R}$-linear spaces $V,W$. Since $A$ is diagonalizable we know that it has eigenvalues $\lambda_i$ where $i\in\{1,2,...,m\}$ where $m\leq n$. We also know that $A$ has eigenvalues of the form:
$$Av_j=\lambda_iv_j$$
Where $j\in\{1,2,...,m\}$ and $i=j$. By applying $A$ to both sides of the equation from the left we obtain:
$$A^2v_j=\lambda_i^2v_j$$
Which implies that $A^2$ and $A$ have the same eigenvectors. Since $A$ is diagonlizable this means that we can form a basis of $A$ from it's and eigenvectors, and since $A^2$ has the same eigenvectors as $A$ we know that $A^2$ is diagonalizable and has the same same basis as $A$. We now want to show that $N_{A^2}=N_A$. We already know that $N_A\subset N_{A^2}$, so we just need to show the other direction. Let $u\in N_{A^2}$, we know that $N_{A^2}\subset V$, thus we know that $u\in V$. Since the eigenvectors of $A^2$ form a basis they span the space and we can write $u$ as a linear combination of the eigenvectors which are linearly independent:
$$u=a_1v_1+a_2v_2+...+a_mv_m$$
For some $a_1, a_2,...,a_j\in\mathbb{R}$. And since $u\in N_{A^2}$:
$$A^2u=A^2(a_1v_1+a_2v_2+...+a_mv_m)=0$$
$$\Rightarrow a_1\lambda_1^2v_1+a_2\lambda_2^2v_2+...+a_m\lambda_m^2v_m=0$$
However these vectors are linearly independent, thus they can only combine to equal the zero vector in a trivial way. Thus we have that:
$$a_1\lambda_1^2,a_2\lambda_2^2,...,a_m\lambda_m^2=0$$
$$\Rightarrow a_j\lambda_i^2=0$$
$$\Rightarrow a_j\lambda_i=0$$
This leaves us with two cases, either $a_j=0$ or $\lambda_i=0$. In the case of $a_j=0$ we have the zero vector, in the case $\lambda_i=0$ we have a non zero vector. Either way we can write the following:
$$ a_1\lambda_1v_1+a_2\lambda_2v_2+...+a_m\lambda_mv_m=0$$
$$\Rightarrow A(a_1v_1+a_2v_2+...+a_mv_m)=0$$
$$\Rightarrow Au=0$$
Thus if $u\in N_{A^2}$ we have that $u\in N_{A}$ which implies that $N_{A^2}\subset N_{A}$ and we obtain $N_{A^2}=N_A$. Thus we have proven that $Ax^2=0\Rightarrow Ax=0$.
A: Consider the $2 \times 2$ real matrix $A = \bigl(\begin{smallmatrix} 1 & 0 \\ 0 & 2 \end{smallmatrix} \bigr)$. Observe that $I - A = \bigl(\begin{smallmatrix} 0 & \phantom -0 \\ 0 & -1 \end{smallmatrix} \bigr)$ has the property that $\ker(I - A) = \operatorname{span}_\mathbb R \{(0, 1)^T\}.$ Likewise, we have that $(I - A)^2 = \bigl(\begin{smallmatrix} 0 & 0 \\ 0 & 1 \end{smallmatrix} \bigr)$ has the propery that $\ker(I - A)^2 = \operatorname{span}_\mathbb R \{(0, 1)^T\}.$ Consequently, the matrix $A$ has a simple eigenvalue $1$ with an eigenvector $v$ such that $v$ is also an eigenvector of $(I - A)^2$ corresponding to a simple eigenvalue.
On the other hand, there is nothing that guarantees this. Consider the $3 \times 3$ real matrix $$B = \begin{pmatrix} 2 & 0 & \phantom -0 \\ 0 & 0 & -1 \\ 0 & 1 & \phantom -2 \end{pmatrix}.$$ One can show that $v = (0, -1, 1)^T$ is an eigenvector corresponding to the eigenvalue $1$ of geometric multiplicity $1$ so that $\ker(I - B) = \operatorname{span}_\mathbb R \{v\};$ however, we have that $$(I - B)^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ so that $\dim_\mathbb R \ker(I - B)^2 = 2.$ Consequently, it is not true that $v$ spans the kernel of $(I - B)^2.$
