Are the eigenvectors in the bases for the eigenspaces of a matrix linearly independent? Okay, so I know that a basis consists of linearly independent vectors. But when you're talking about multiple bases for multiple eigenspaces of a matrix, I want to know if all vectors in all these bases are linearly independent? For example, if a basis for an eigenspace of a matrix A is B = {v1, v2), and a basis for another eigenspace of A is B' = {w1}, are v1, v2, and w1 necessarily linearly independent?
I'm trying to understand a few concepts regarding eigenspaces in my book, and it seems like if this is true it would make a few things I'm confused on make sense. So I'm just wondering if this is, in fact, the case?
 A: Yes, they are linearly independent.
Be $B_1,\ldots,B_n$, $B_k=\{v_{k1},\ldots,v_{k|B_k|}\}$, the bases of distinct eigenspaces $V_k$ of $A$, and $\lambda_1,\ldots,\lambda_n$ the corresponding eigenvectors. Obviously $\lambda_i=\lambda_k\iff i=k$, or else the eigenspaces would not be distinct.
Now assume the vectors in $B_1\cup\dots\cup B_n$ were linearly dependent. Then you'd have numbers $\alpha_{kl}$ such that
$$\sum_{k=1}^n\sum_{l=1}^{|B_k|}\alpha_{kl}v_{kl}=0\,. \tag{1}$$
To simplify notation, we define
$$w_k = \sum_{l=1}^{|B_k|}\alpha_{kl}v_{kl}\,.\tag{2}$$
With this the equation $(1)$ simplifies to
$$\sum_{k=1}^n w_k = 0\,.$$
Obviously $w_k\in V_k$. Therefore we get
$$A\sum_{k=1}^n w_k = \sum_{k=1}^n \lambda_k w_k=0\,. \tag{3}$$
Now we consider two cases for $\lambda_n$. The first case is that $\lambda_n=0$. Then obviously in equation $(3)$ the term $k=n$ vanishes and therefore can be omitted. That is, in that case, $B_1\cup\dots\cup B_{n-1}$ is already linearly dependent.
Now consider $\lambda_n\ne 0$. Then we can divide $(3)$ by $\lambda_n$ and subtract $(2)$ to obtain
$$\sum_{k=1}^n\left(\frac{\lambda_k}{\lambda_n}-1\right)w_k = 0\,.$$
Again, we see that the coefficient of $w_n$ vanishes, therefore already $B_1\cup\dots\cup B_n$ is linearly dependent.
So we have shown that if the basis vectors of a set of eigenspaces are linearly dependent, then we can remove one of the eigenspaces and still have a linearly dependent set. But we can do the same step with the new set, until we arrive at just one eigenspace, for which we again have to conclude linear dependence. But we chose a basis for the eigenspace, so that cannot be linearly dependent. Therefore our original assumption must have been wrong, and $B_1\cup\dots\cup B_n$ indeed is linearly independent.
A: Hint: If $B$ and $B'$ are bases for subspaces $W$ and $W'$ respectively, then $B \cup B'$ is a linearly independent set if and only if $W \cap W' = \{0\}$.

Edit: to be clear, you should prove the above claim first.
But once we know the claim is true, it remains to check that two eigenspaces [for different eigenvalues] have intersection $\{0\}$.
If this were not true, then there would exist a vector $v \ne 0$ that is both a $\lambda$-eigenvector and a $\lambda'$-eigenvector where $\lambda \ne \lambda'$. Is this possible?

As pointed out in the comments, this only works for two eigenspaces at a time (and induction would require more properties than the 2 eigenspace case). The linked duplicate question provides more detail. Apologies for my mistake.
