Is the multiplicity of an eigenvalue equal to the dimension of it's associated eigenspace? This is a silly question, but if someone could provide a short proof as to why if it is true, or a counterexample and short explanation if it's false I would appreciate it.
EDIT:
By multiplicity I mean, when solving the characteristic polynomial, the roots (i.e the eigenvalues) can be repeated, so I was referring to the algebraic multiplicity. 
 A: Maybe. It depends on the definition of multiplicity of eigenvalue you mean. In fact there are two notions of the multiplicity of an eigenvalue. The geometric multiplicity is defined to be the dimension of the associated eigenspace. The algebraic multiplicity is defined to be the highest power of $(t-\lambda)$ that divides the characteristic polynomial. The algebraic multiplicity is not necessarily equal to the geometric multiplicity. In fact the two are equal for all eigenvalues of the linear transformation if and only if the linear transformation is diagonalizable. 
Essentially the algebraic multiplicity counts the maximum number of eigenvectors we could have for that eigenvalue, while the geometric multiplicity counts how many eigenvectors we actually have. I.e. we have that $m_a(\lambda)\ge m_g(\lambda)$ (where for simplicity I've used $m_a$ to denote algebraic multiplicity and $m_g$ to denote geometric multiplicity).
As an example, the matrix
$$ \begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}$$
has a single eigenvalue, 1, which has algebraic multiplicity 2, since the characteristic polynomial is $(1-t)^2$, but only a single eigenvector, the first basis vector, call it $e_1$. You can check that any other eigenvector will be a scalar multiple of $e_1$. 
As the commenter below pointed out, one can view the algebraic multiplicity as the dimension of the generalized eigenspace associated to that eigenvalue. Let $T$ be the linear transformation, let $\lambda$ be a scalar. Then $v$ is an eigenvector for $T$ if $Tv=\lambda v$, or if $(T-\lambda)v =0$. Hence the $\lambda$-eigenspace of $T$ is $\ker (T-\lambda)$. Then a generalized $\lambda$-eigenvector of rank $n$ is a vector $v$ with $(T-\lambda)^nv=0$. The generalized $\lambda$-eigenspace of $T$ is the vector space of all generalized $\lambda$-eigenvectors. 
Let $E_\lambda^n$ denote the space of all generalized $\lambda$-eigenvectors of rank at most $n$. Then $E_\lambda^n = \ker (T-\lambda)^n$. We can see this gives us an ascending chain of subspaces
$$ E_\lambda^1 \subset E_\lambda^2\subset \cdots \subset E_\lambda^n \subset \cdots $$
In a finite dimensional vector space, this chain of subspaces has to stabilize eventually, since the dimension can't increase forever. It stabilizes at the generalized $\lambda$-eigenspace for $T$, and it turns out as mentioned at the beginning of this paragraph that this space has dimension $m_a(\lambda)$. Since the first space in the chain, $E_\lambda^1$, is the usual $\lambda$-eigenspace of $T$, which has dimension $m_g(\lambda)$, this is one way to see that $m_g(\lambda)\le m_a(\lambda)$ (at least intuitively; one would have to justify that the generalized $\lambda$-eigenspace has dimension $m_a(\lambda)$).
Anyway, there's a really interesting story about generalized eigenspaces and the Jordan canonical form that might be useful to read about, but I can't go into it here, so see the links.
A: Since it was not shared how to actually prove that eigenvalue's geometric multiplicity cannot exceed its algebraic multiplicity, I will show it briefly. Let $m_g(\lambda)$ be the geometric multiplicity of $\lambda$ and $m_a(\lambda)$ its algebraic multiplicity. Let $\{v_i\}_{i=1}^{m_g(\lambda)}$ be a basis for the $\text{kern} (A-\lambda I)$ with $m_g(\lambda)$ elements, being $A \in \mathbb{R}^{n \times n}$ a $n$ by $n$ matrix. Completing $\{v_i\}_{i=1}^{m_g(\lambda)}$  to a $\mathbb{R}^{n}$ basis, say $\{v_i\}_{i=1}^{m_g(\lambda)}\cup \{u_j\}_{j=1}^{n-m_g(\lambda)}$, we can construct a matrix $P$ such that $P=\left[v_1\ v_2\ \cdots\ v_{m_g(\lambda)}\ u_1\ u_2\ \cdots\ u_{n-m_g(\lambda)}\right].$ Hence,
$$\begin{split}
P^{-1} A P =&  P^{-1}\left[A v_1\ A v_2\ \cdots\ A v_{m_g(\lambda)}\ A u_1\ A u_2\ \cdots\ A u_{n-m_g(\lambda)}\right]\\
=& P^{-1}\left[\lambda v_1\ \lambda v_2\ \cdots\ \lambda v_{m_g(\lambda)}\ A u_1\ A u_2\ \cdots\ A u_{n-m_g(\lambda)}\right]\\ 
=&  \left[\lambda P^{-1} v_1\ \lambda P^{-1} v_2\ \cdots\ \lambda P^{-1} v_{m_g(\lambda)}\ P^{-1} P^{-1} A u_1\ A u_2\ \cdots\ P^{-1} A u_{n-m_g(\lambda)}\right]\\ 
 =& \left[\lambda e_1\ \lambda e_2\ \cdots\ \lambda e_{m_g(\lambda)}\ P^{-1} A u_1\ P^{-1} A u_2\ \cdots\ P^{-1} A u_{n-m_g(\lambda)}\right]\\
=& \begin{pmatrix} \lambda I  &  U \\ \\ 0 & V\\ \end{pmatrix}
\end{split}, $$ where $I$ is the identidy of order $m_g(\lambda)$ by $m_g(\lambda)$, $V$ is a arbitrary $n-m_g(\lambda)$ by $n-m_g(\lambda)$ $n-m_g(\lambda)$ matrix, $U$ a suitable matrix,  lastly, 0 is a suitable zero matrix. Thus, $P^{-1} A P$ is an upper triangular block matrix, and, denoting by $p_A$ the characteristic polynomial of $A$, we have that $$\begin{split}p_A (\mu)=\text{det}(A-\mu I)=& \text{det}(P^{-1})\text{det}(A-\mu I) \text{det}(P)\\ 
=& \text{det}(P^{-1}\left(A-\mu I\right)P)\\ 
=& \text{det}\left(P^{-1}A P-\mu I\right)\\
=& \text{det}\left( \begin{pmatrix} \lambda I  &  U \\ \\ 0 & V\\ \end{pmatrix} - \begin{pmatrix} \mu I  &  0 \\ \\ 0 & \mu I \\ \end{pmatrix} \right)\\
=& (\lambda-\mu)^{m_{g}(\lambda)}\text{det} \left( V-\mu I \right) \\
\end{split}.$$ From which we conclude that $m_{a}(\lambda) \geq m_{g}(\lambda).$
Observing that $m_{a}(\lambda) \geq m_{g}(\lambda)$ for all eigenvalues $\lambda$, it's clear why the matrix is diagonalizable if and only if for all eigenvalues the geometric multiplicity is equal to its algebraic multiplicity.
