eigen vectors and linear independence I know that it is the case that if we have eigenvectors $v_1....v_r$ that correspond to distinct eigen values $λ_1....λ_r$ of an $n$ x $n$ matrix, then {$v_1....v_r$} is linearly independent.
But is is it possible to have $a_1$ linearly independent eigen vectors corresponding to $λ_1$, $a_2$ linearly independent eigen vectors corresponding to $λ_2$... $a_p$ linearly independent eigen vectors corresponding to $λ_n$, and the set of all $r$ eigenvectors where $r=a_1+a_2+...+a_p$ be linearly dependent?
For example: consider a $3x3$ matrix. Imagine we have one vector(call it $v_1$) forming a basis for the eigenspace of a one eigen value, but two vectors(call them $v_2$ and $v_3$) forming a basis for different eigen value. are we guaranteed that {$v_1,v_2,v_3$} are linearly independent?
Suppose  is any field. For example, it could be ℝ or ℂ.
Suppose $M$ is an n×n -matrix.
Suppose $\lambda_1, \dots, \lambda_r$ is a list of distinct eigenvalues of $M$.
Suppose:

$v_{1, 1}, \dots, v_{1, k_1}$ is a linearly independent list of eigenvectors of $M$ corresponding to $\lambda_1$.
$v_{2, 1}, \dots, v_{2, k_2}$ is a linearly independent list of eigenvectors of $M$ corresponding to $\lambda_2$.
$\vdots$
$v_{r, 1}, \dots, v_{r, k_r}$ is a linearly independent list of eigenvectors of $M$ corresponding to $\lambda_r$.

Suppose $\alpha_{1, 1}, \dots, \alpha_{1, k_1}, \alpha_{2, 1}, \dots, \alpha_{2, k_2}, \dots, \alpha_{r,1}, \dots, \alpha_{r, k_r} \in \mathbb{F}$ are scalars such that
$$(\alpha_{1, 1} v_{1,1} + \dots + \alpha_{1, k_1} v_{1, k_1}) +
  (\alpha_{2, 1} v_{2,1} + \dots + \alpha_{2, k_2} v_{2, k_2}) + \dots +
(\alpha_{r,1} v_{r,1} + \dots + \alpha_{r, k_r} v_{r, k_r}) = 0.$$
Define $u_1 = (\alpha_{1, 1} v_{1,1} + \dots + \alpha_{1, k_1} v_{1, k_1}), \dots,
u_r=(\alpha_{r,1} v_{r,1} + \dots + \alpha_{r, k_r} v_{r, k_r})$.
Clearly, $u_1$ is either the zero vector or a $\lambda_1$-eigenvector of $M$. Analagously, $u_2$ is either the zero vector or a $\lambda_2$-eigenvector of $M$. And so on for $u_3, \dots, u_r$.
By the theorem you say you know, it must be the case that for each $i \in \{1, \dots, r\}$, $u_i$ is actually the zero vector.
But then for each $i \in \{1, \dots, r\}$, all the coefficients $\alpha_{i,1}, \dots, \alpha_{i, k_i}$ must be zero, because the vectors $v_{i,1}, \dots, v_{i, k_i}$ are linearly independent.
We have shown that any linear combination of $v_{1, 1}, \dots, v_{1, k_1}, \dots, v_{r, 1}, \dots, v_{r, k_r}$ equal to zero must actually be the trivial combination (i.e., with all coefficients zero). ∎
We claim that the set of all $r$ eigenvectors is linearly independent. The main proof ideas are already provided in the comments. For each $i=1,\dots, p$ let $v_{i1},\dots, v_{ia_i}$ be linearly independent system of eigenvectors corresponding to the eigenvalue $\lambda_i$ and  $\mu_{i1},\dots, \mu_{ia_i}$ be scalars such that $$\sum_{i=1}^p \sum_{j=1}^{a_i} \mu_{ij} v_{ij}=0.$$
For each $i$ put $v_i=\sum_{j=1}^{a_i} \mu_{ij} v_{ij}$. Then $v_i$ is an eigenvectors corresponding to $\lambda_i$ and
$$\sum_{i=1}^p v_i=\sum_{i=1}^p \sum_{j=1}^{a_i} \mu_{ij} v_{ij}=0.$$
The linear independence property formulated in the first paragraph of the question implies that $0=v_i=\sum_{j=1}^{a_i} \mu_{ij} v_{ij} $ for each $i$. Then the linear independence property formulated in the second  paragraph of the question implies that $\mu_{ij}=0$ for each $j$.