Is it possible for an eigenspace to have dimension $0$?

Question

I don't think it is possible because that entails that only the $\mathbf 0$-vector is in the eigenspace, but $\mathbf 0$ is not an eigenvector by definition.

However, my textbook says:

For an $n\times n$ matrix, if there are $n$ distinct eigenvalues, then all eigenspaces have dimension at most $1$.

which seems to imply that eigenspaces of dimension $0$ are possible.

Answer 1

By definition to any eigenvalues correspond at least one eigenvector thus for a n-by-n matrix for each eigenvalue $\lambda_i$ we have $1\le$ dim(eigenspace)$\le n$.

Answer 2

Following the definition, $\lambda$ is an eigenvalue of the matrix $A$ if there exists a non-zero vector $v$ such that:

$$Av = \lambda v.$$

The definition itself assures that, if $\lambda$ is an eigenvalue, then there must be also an eigenvector $v$. The presence of at least one eigenvector implies that the eigenspace relative of $\lambda$ has at least dimension equal to $1$.

You cannot define an eigenvalue without an eigenvector, and viceversa.

Answer 3

It doesn't imply that dimension 0 is possible. You know by definition that the dimension of an eigenspace is at least 1. So if the dimension is also at most 1 it means the dimension is exactly 1. It's a classic way to show that something is equal to exactly some number. First you show that it is at least that number then that it is at most that number.

Answer 4

It is a matter of convention. What everybody should agree on is that $\lambda$ being an eigenvalue of$~A$ means that $\dim(\ker(A-\lambda I))>0$, so the dimension of the eigenspace associated to an eigenvalue is never$~0$. However if the dimension in that formula is$~0$, so if $\lambda$ is not an eigenvalue, then one could still agree that $\ker(A-\lambda I)$ may be called the eigenspace $E_\lambda$ of$~A$ for (the non-eigenvalue)$~\lambda$. This may seem a bit weird, but there are many occasions where it is a convenience to be able to talk about $E_\lambda$ for any scalar$~\lambda$, without first having to ensure that $\lambda$ is an eigenvalue. For instance, if $A$ is a projector (so $A^2=A$) then it is always true that the whole space decomposes as $E_0\oplus E_1$, though it might happen that one of $E_0$ and $E_1$ has dimension$~0$ (namely if $A=I$ respectively $A=0$); this statement would be more complicated to make if it were forbidden to mention $E_\lambda$ unless $\lambda$ was actually an eigenvalue.