The set of bases of all the eigenspaces is linearly independent

The bases of the different eigenspaces are still linearly independent is cuz these eigenspaces do not overlap (except $\vec{0}$).

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• We have our distinct eigenvalues, ${\lambda }_1,{\lambda }_2,...{\lambda }_k$.

  • Since some eigenvalues could be duplicates, I have used $k$ instead of $n$.
  • When $k=n$, each eigenvalue is unique and we have $n$ distinct eigenspaces. When this happens, our eigenvectors span $R^n$.

• Each eigenvalue ${\lambda }_i$ has the associated eigenspace $E_{{\lambda }_i}$.

• Each eigenspace will have it’s basis, $B_i$. Each element in the basis set is an eigenvector (obv).

• Creating a set which is union of all the $B_i$s.

• This new set is still linearly independent.

• The reason why the bases of these different eigenspaces are still linearly independent is cuz these eigenspaces do not overlap (except $\vec{0}$).

• Eigenspaces are not random spans on the grid, they are specific spans that remain unchanged during a matrix operation.

• Since two eigenspaces do not overlap, the bases also do not overlap.