Eigenspace problem I am trying to solve the following problem:
Let $(\lambda,v)$ be an eigenpar of $A \in \mathbb{R}^{n \times n}$ such that the algebraic multiplicity of $\lambda$ is $2$, but with geometrical multiplicity equal to $1$. Show that $v \in$ Im$(A-\lambda I) $, where Im$(A-\lambda I)$ denotes the image space of the matrix $A-\lambda I$.
My strategy was to suposse that $v \notin$ Im$(A - \lambda I)$, this way i conclude using the Kernel Image Theorem that $\mathbb{R}^{n}$ can be made as the direct sum of Kernel$(A-\lambda I)$ and Im$(A-\lambda I)$. I thought this way i could get some contradiction, but if it is the case i do not noticed how.
 A: Suppose $v\notin \text{Im}(A-\lambda I)$. As you noticed, we have $\mathbb{R}^n=\ker (A-\lambda I)\oplus\text{Im}(A-\lambda I)=K\oplus J.$ Now, observe that $A$ is invariant under $K$ and under $J$. So the restrictions $A|_K:K\rightarrow K$, $A|_J:J\rightarrow J$ are well defined linear maps.
Now choose a basis $\{v_1,\dots,v_{n-1}\}$ of $J$ and let $[A]_B$  be the matrix of $A$ with respect to the basis $B=\{v,v_1,\dots,v_{n-1}\}$. Then $[A]_B$ has the form
$$ [A]_B = \left( \begin{matrix} \lambda & * & \dots & *\\ 0 & a_{11} & \dots & a_{1(n-1)}\\\vdots & \vdots & \ddots & \vdots\\ 0 & a_{(n-1)1} & \dots & a_{(n-1)(n-1)}\end{matrix} \right). $$
Let $p$ be the characteristic polynomial of $A$. We can see by cofactor expansion of the determinant that
$$
p(t)=det(A-t I)=det([A-t I]_B)=det(A|_K-tI|_K)det(A|_J-tI|_J)=p_K(t)p_J(t),
$$
where $p_L$ is the characteristic polynomial of the restriction $A|_L$, for $L=K,J$.
Note that $p_J(\lambda)\neq 0$. If not, then there exists $0\neq x\in \text{Im}(A-\lambda I)$ such that $Ax=\lambda x$, which contradicts the fact that $\ker(A-\lambda I)=\langle v\rangle$. Moreover, as $K$ is one-dimensional vector space, we have $A|_K=(\lambda)$, so $p_K(t)=\lambda-t$, and so $$p(t)=(\lambda-t)p_J(t),\quad p_J(\lambda)\neq 0$$
and this contradicts the fact that $\lambda$ has multiplicity 2 as a root of $p$.
