So guys,
A is a n by n matrix
Lets say $Ax= \lambda x$ where $\lambda$ denotes the eigenvalue. If $\lambda=0$ then $x$ is in the nullspace $N(A)$. If $\lambda \ne 0$ then $x$ is in the column space $C(A)$. Those spaces have dimensions $(n-r)+r=n$
let $r$ denote the rank of the matrix.
So Why doesn't every square matrix have n linearly independent eigenvectors?
Best Answer
Forget about the $\lambda=0$ or $\lambda\ne 0$ distinction, it's not essential here; let's assume $\lambda\ne 0$, and the matrix is full rank.
If I got you right, your reasoning, applied to this particular case, is this: Given that the dimension of the column space $C(A)$ is $n$ (true) and that the eigenvectors are in the colum space (true), then we should be able to find $n$ LI eigenvectors.
But the conclusion does not follow. The eigenvectors are in the column space, true, but not all vectors in the column space are eigenvectors.
Example : $A=\begin{pmatrix} 2 &1 \\ 0 & 2\end{pmatrix}$ Here the column space is the full space, but the eigenvectors lie on a one-dimensional subspace $x= (a, 0)^t$