Is the eigenspace of a matrix $A$ equal to its column space? Let $A \in \Bbb R^{n\times n}$. Let $C(A)$ denote the column space of $A$, and $E(A)$ denote the eigenspace, which is the span of all the eigenvectors of $A$. Now I am trying to see if $C(A)=E(A)$.
Denote $\{z_1,\ldots,z_m\}$ as the eigenvectors of $A$. Then $Az_i =\lambda_iz_i$. Since $Az_i\in C(A)$ we have $z_i \in C(A)$. Thus $E(A)\subset C(A)$. Now if $A$ is not singular, $E(A) = \Bbb R^n$, and since $C(A) \subset \Bbb R^{n}$ we have that $E(A)=C(A)$.
But in the case that $A$ is singular, what can we conclude then, if anything? Thanks!
 A: That is false in general: take any invertible matrix $A$. Then $C(A)=\mathbf R^n$. But if the matrix is not diagonalisable, there are less than $n$ linearly independent eigenvectors, so that the span of all eigenvectors is a strict subspace of $\mathbf R^n$.
A: The matrix $A=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ has no real eigenvector, so the span of the eigenvectors is the $\{0\}$ subspace, while the column space is $\mathbb{R}^2$.
The question makes sense only if we work in $\mathbb{C}$, rather than in $\mathbb{R}$. In this case, if $\lambda_1,\dots,\lambda_k$ are the pairwise distinct eigenvalues of $A$, we can write
$$
E(A)=E(A;\lambda_1)\oplus\dots\oplus E(A;\lambda_k)
$$
where $E(A;\lambda)=\{v\in\mathbb{C}^n: Av=\lambda v\}$.
First of all, let's assume $0$ is not an eigenvalue of $A$; then $A$ is invertible and its column space is $C(A)=\mathbb{C}^n$, so $E(A)=C(A)$ amounts to saying that $A$ is diagonalizable.
In case $0$ is an eigenvalue of $A$, the situation is more complicated. 
