Anyway, as a practitioner in 2016, not to mention someone about to take the Q, these results are pretty much the most important takeaway of the whole course.
First, distinct eigenvalues yield independent eigenvectors:
If λ1, ..., λk are distinct eigenvalues of A with corresponding eigenvectors x1, ..., xk, then x1, ..., xk are linearly independent.
From this immediately follows the result that makes Principal Component Analysis possible:
An nxn matrix is A is diagonalizable if and only if A has n linearly independent eigenvectors. Furthermore, if A = S-1DS where D is diagonal, then S is composed of the column eigenvectors and the diagonal entries of D are the corresponding eigenvalues.
As mentioned above, the proof is fairly easy exercise in construction:
Suppose A has n linearly independent eigenvectors. Let S be the matrix formed by taking them as column vectors. For each associated eigenvalue, we know that Axj = λjxj so:
Since the eigenvectors are independent, S is non-singular. So, multiplying both sides by its inverse gives A = S-1DS. The converse is basically walking the same thread back the other way.
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