Is my reasoning/understanding of definitions right? Does the following reasoning make sense?
I have an $n\times n$ matrix $M$ acting on $\mathbb C^n$, 
Then,
The eigenspace of matrix $M$ is $\ker(M-\lambda I)$.
The eigenvalues of  $M^2$ are $\lambda^2$, where $\lambda$ are eigenvalues of $M$.
$Mv=\lambda v \implies M^2 v=\lambda^2v \implies$ eigenvectors of $M$ are eigenvectors of $M^2$
Now $M^2v=\lambda^2 v\implies Mv=\pm\lambda v$ so eigenvectors of $M^2$ are also eigenvectors of $M$
Hence the eigenspaces of $M$ and $M^2$ are the same? (Check: To have the same eigenspace does that mean just having the same set of eigenvectors?)
Thanks.
 A: There is no such thing as "the eigenspace of a matrix". Rather, eigenspaces are associated to eigenvalues, so you should say "The eigenspace of $M$ corresponding to $\lambda$." With that proviso, you are correct that the eigenspace of $M$ corresponding to a given eigenvalue $\lambda$ is $\mathrm{ker}(M-\lambda I)$.
It is also true that if $\lambda$ is an eigenvalue of $M$, then $\lambda^2$ is an eigenvalue of $M^2$. However, it is not true that if $\lambda^2$ is an eigenvalue of $M$, then $\lambda$ is an eigenvalue of $M$ (e.g., $(-1)^2$ is an eigenvalue of $I^2$, but $-1$ is not an eigenvalue of $I$, where $I$ is the identity matrix). 
It is not true in general that eigenvectors of $M^2$ are eigenvectors of $M$. For example, consider the matrix
$$M=\left(\begin{array}{rr}
0 & -1\\
1 & 0
\end{array}\right).$$
Then
$$M^2 = \left(\begin{array}{rr}
-1 & 0\\
0 & -1
\end{array}\right),$$
so $(1,0)$ and $(0,1)$ are eigenvectors of $M^2$. However, neither one is an eigenvector of $M$.
This example also shows that the eigenspaces may be different. $M^2$ has a unique eigenspace, equal to all of $\mathbb{C}^2$, whereas $M$ has two one-dimensional eigenspaces, one associate to $\lambda =i$ and one associated to $\lambda =-i$. 

You can fix the backward implication ("if $\lambda^2$ is an eigenvalue of $M^2$, then $\lambda$ is an eigenvalue of $M$") with some care. For example, because you are working over the complex numbers, the minimal polynomial of $M^2$ splits. Let $\mu(t)$ be the minimal polynomial; if $\lambda_1,\ldots,\lambda_k$ are the distinct eigenvalues of $M^2$, then we know that
$$\mu(t) = (t-\lambda_1)^{a_1}\cdots(t-\lambda_k)^{a_k}.$$
Since $\mu(M^2) = \mathbf{0}$, then we have that
$$\mathbf{0} = \mu(M^2) = (M^2-\lambda_1)^{a_1}\cdots (M^2-\lambda_k)^{a_k} = (M-\rho_1)^{a_1}(M+\rho_1)^{a_1}\cdots (M-\rho_k)^{a_k}(M+\rho_k)^{a_k},$$
where $\rho_i$ is a complex number such that $\rho_i^2 = \lambda_i$ (i.e., a square root of $\lambda_i$).
Let $q(t)$ be the polynomial
$$q(t) = (t-\rho_1)^{a_1}(r+\rho_1)^{a_1}\cdots (t-\rho_k)^{a_k}(t+\rho_k)^{a_k}.$$
Since $q(M)=\mathbf{0}$, the minimal polynomial of $M$ divides $q(t)$; therefore, the eigenvalues of $M$ are among $\pm\rho_1,\pm\rho_2,\ldots,\pm\rho_k$, and those are the only eigenvalues of $M$.
So if $\lambda_1^2,\ldots,\lambda_k^2$ are the eigenvalues of $M^2$, then you can conclude that the eigenvalues of $M$ are among $\pm\lambda_1$, $\pm\lambda_2,\ldots,\pm\lambda_k$.
Note, however, that although you can go from $Mv = \lambda v$ to $M^2v = \lambda^2v$, you cannot go from $M^2v = \lambda^2v$ to $Mv=\pm\lambda v$, even if you know that $\lambda$ is an eigenvalue of $M$, as the example above shows (we have $M^2(1,0) = i^2(1,0)$, and we know that $i$ is an eigenvalue of $M$, but $M(1,0)\neq \pm i(1,0)$); so the line that begins with "Now $M^2v=\lambda^2 v$..." is incorrect.
