Transformations with All Eigenvalues Distinct
When will a linear transformation have a diagonal matrix? We know that this happens exactly when we can find a basis
of our vector space
so that each basis vector is an eigenvector
. But when does this happen? The full answer can be complicated, but one sufficient condition we can describe right away.
First, remember that we can always calculate the characteristic polynomial of . This will be a polynomial of degree
— the dimension of
. Further, we know that a field element
is an eigenvalue of
if and only if
is a root of the characteristic polynomial. Since the degree of the polynomial is
, we can expect to find no more than
distinct roots of the polynomial — no more than
distinct eigenvalues of
. I want to consider what happens in this generic case.
Now we have field elements
, and for each one we can pick a vector
so that
. I say that they form a basis of
. If we can show that they are linearly independent, then they span some subspace of
. But since there are
distinct vectors here, they span a subspace of dimension
, which must be all of
. And we know, by an earlier lemma that a collection of eigenvectors corresponding to distinct eigenvalues must be linearly independent! Thus, the
form a basis for
consisting of eigenvectors of
. With respect to this basis, the matrix of
is diagonal.
This is good, but notice that there are still plenty of things that can go wrong. It’s entirely possible that two (or more!) of the eigenvalues are not distinct. Worse, we could be working over a field that isn’t algebraically closed, so there may not be roots at all, even counting duplicates. But still, in the generic case we’ve got a diagonal matrix with respect to a well-chosen basis.
[…] space of finite dimension . We take its characteristic polynomial to find the eigenvalues. If all of its roots are distinct (and there are of them, as there must be if we’re working over an algebraically closed […]
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[…] we’ve established that distinct eigenvalues allow us to diagonalize a matrix, but repeated eigenvalues cause us problems. We need to generalize […]
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