# Dependency of vectors in eigenspace corresponding with eigenvalue zero

The eigenspace corresponding with the eigenvalue zero is the same as the null space of the original matrix. All vectors in the null space are linearly independent so the eigenvectors of zero are also independent.

Is this conclusion right?

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The last sentence is not correct, however. Certainly not all vectors in the nullspace are linearly independent. If $Ax=0$ so that $x$ is in the nullspace, then $2x$ is another vector in the nullspace that is linearly dependent with $x$.
@Jef Here's another try at breaking your bridge between bases and subspaces. For finite dimensional vector spaces, the base is going to have only finitely many elements. However, if your field is infinite, then nonzero subspaces are always going to have infinitely many elements. Certainly they can't all be in the basis. If the field is finite, then a subspace with dimension $k$ is going t have $q^k$ elements, (definitely more than $k$) where $q$ is the size of the field. – rschwieb Dec 28 '12 at 19:19
If you solve $Ax=0$ for $x$, you do not have a base, you have a single element $x$ in the kernel. You can work to find several linearly independent $x_i$ such that $Ax_i=0$. When you find sufficiently many, you will have a base for the kernel $\{x_1\dots. x_k\}$. Any linear combination of those $x_i$ will also be in the kernel. – rschwieb Dec 28 '12 at 20:37