TITLE: Eigenspace and $\ker(T)$
QUESTION [0 upvotes]: It seems like eigenspace and $\ker(T)$ are strongly connected, I have thought about some properties and I would like to make sure I got it right.

for all matrix/transformation there is an Eigenspace $\iff 0\in \ker(T) \iff$ a homogeneous system of linear equations has always a trivial solution.
there are eigenvectors $\iff$ for some $x$, $x I-A$ is not invertible/full rank/etc.
There are eigenvalues $\iff$ there are eigenvectors

Am I right? Am I missing something important?

REPLY [1 votes]: Let me try to answer; please tell me if something is not clear.
0) You're right; the eigenspace and the kernel are related: the eigenspace is the solution to $Ax=\lambda x= x:(A-\lambda I)x =0 $, for $\lambda$ an eigenvalue  of $A$ , so the eigenspace is the kernel of the map $(A-\lambda I)x=0$.
1) $0$ ( the zero vector) is always in the kernel of any linear map, since $T(0)=T(c.0)=0T(c)=0$. And a matrix does not always have an eigenspace , even if $0$ is in Ker(T). Take, e.g., a matrix with Real entries with characteristic polynomial, e.g., $ \lambda^2 +1$, e.g., the antidiagonal $2\times 2$ matrix with entries $-1,1$ . This matrix will contain no (Real ) eigenvalues. 
2) This is false: consider the identity matrix; it is invertible. The whole space is the eigenspace and it has the non-zero eigenvalue $1$. Notice that there is an equivalence between having non-zero determinant and being invertible. An interesting consequence of this is that matrices $M$ with entries over the integers are invertible iff  $Det(M)=1$ or $DetM = -1$.
3)If there I an eigenvalue $\lambda$, then the vector $v$ satisfying $(A -\lambda I)v =0$ is an eigenvector. By construction, $(A- \lambda I)v$ is not invertible, so there are non-trivial solutions , i.e., there are solutions to $(A- \lambda I)v=0$ other than $v=0$.
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