Theorem: Let $\alpha$ be an eigenvalue of $T:V→V$, then the algebraic multiplicity of $\alpha$ is greater than or equal to the geometric multiplicity of $\alpha$.
Comments would be appreciated on whether the proof below is incorrect in any way or could be made more efficient. Alternate proofs would also be appreciated. Many thanks.
Proof
Let the geometric multiplicity of $\alpha$ be $k$, then by definition $\alpha$ corresponds to $k$ linearly independent eigenvectors; $v_1$,...,$v_k$ $∈V$.
Using the theorem that a linearly independent set can be completed to form a basis, extend $v_1$,...,$v_k$ to a basis of $V$. $B$={$v_1$,...,$v_k$,$w_1$,...,$w_r$}. The {$w_i$} are not necessarily eigenvectors.
Apply $T$ to the basis elements in $B$:
$T(v_1)$=$\alpha$$v_1$
$\space$$\space$$\space$$\vdots$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\ddots$
$T(v_k)$=$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\alpha$$v_k$
$T(w_1)$=$a_{11}v_1$+$\cdots$+$a_{k1}v_k$+$c_{11}w_1$+$\cdots$+$c_{r1}w_r$
$\space$$\space$$\space$$\vdots$
$T(w_r)$=$a_{1r}v_1$+$\cdots$+$a_{kr}v_k$+$c_{1r}w_1$+$\cdots$+$c_{rr}w_r$
From the above take the representative matrix of $T$ with respect to the basis $B$ by transposing coefficients:
$[T]_B$ =\begin{pmatrix}\alpha&\cdots&0&a_{11}&\cdots&a_{1r}\\\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\0&\cdots&\alpha&a_{k1}&\cdots&a_{kr}\\0&\cdots&0&c_{11}&\cdots&c_{1r}\\\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\0&\cdots&0&c_{r1}&\cdots&c_{rr}\\\end{pmatrix}
To determine the algebraic multiplicity of $\alpha$ we must take the characteristic polynomial of $T$, which is the characteristic polynomial of $[T]_B$, and determine how many times $\alpha$ appears as a root. If we take $\lambda$ as a general eigenvalue of $T$, then:
$$p_{[T]_B}(\lambda) = \text{det}(\lambda I_{k+r} - [T]_B) = \begin{vmatrix}\lambda-\alpha&\cdots&0&a_{11}&\cdots&a_{1r}\\\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\0&\cdots&\lambda-\alpha&a_{k1}&\cdots&a_{kr}\\0&\cdots&0&\lambda-c_{11}&\cdots&c_{1r}\\\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\0&\cdots&0&c_{r1}&\cdots&\lambda-c_{rr}\\\end{vmatrix}$$
By the theorem that det$(A)$ can be evaluated from any row or column:
$$=\lambda-\alpha\begin{vmatrix}\lambda-\alpha&\cdots&0&a_{21}&\cdots&a_{2r}\\\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\0&\cdots&\lambda-\alpha&a_{k1}&\cdots&a_{kr}\\0&\cdots&0&\lambda-c_{11}&\cdots&c_{1r}\\\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\0&\cdots&0&c_{r1}&\cdots&\lambda-c_{rr}\\\end{vmatrix}$$
Therefore:
$$=(\lambda-\alpha)^k\begin{vmatrix}\lambda-c_{11}&\cdots&c_{1r}\\\vdots&\ddots&\vdots\\c_{r1}&\cdots&\lambda-c_{rr}\\\end{vmatrix}=(\lambda-\alpha)^k \text{ det}(\lambda I_r - C) =(\lambda-\alpha)^k p_C(\lambda)$$
$(\lambda-\alpha)^k$ $ p_C(\lambda)$ is the characteristic polynomial of $T$, therefore $\alpha$ appears $k$ times or more as a root of the characteristic polynomial of $T$. It appears $k$ times as when $\lambda=\alpha$, $p_T (\lambda)=0$. It may appear more than $k$ times as $(\lambda-\alpha)$ may also be a factor of, therefore $\alpha$ may also be a root of, $p_C (\lambda)$.
As the geometric multiplicity of $\alpha$ is $k$, the algebraic multiplicity of $\alpha$ is greater than or equal to the geometric multiplicity of $\alpha$.
Q.E.D.
Algebraic multiplicity =$\dim G (T, \lambda)$ and geometric multiplicity =$\dim E(T, \lambda)$ the first one is the generalized eigenspace and the second one is the normal eigenspace. Now it's well-known that $E(T, \lambda)\subset G(T, \lambda)$ and both of them are the subspaces of $V$. Hence the conclusion follows.