Diagonalization

1. Find eigenvalues, eigenvectors, bases for the eigenspaces, and multiplicities for the eigenspaces
2. Check if for the basis of each eigenspace the algebraic and geometric multiplicity are equal. If they are not, the matrix is not diagonalizable.
3. If the matrix is diagonalizable, is a matrix comprised of each basis vector, and is a diagonal matrix where each entry is an eigenvalue corresponding to to the eigenspace basis vectors.

Diagonalize the matrix given that the eigenvalues are with algebraic multiplicity and with algebraic multiplicity , also

Solution Since and , we see that is diagonalizable, so we take

which is just a matrix comprised of each basis vector.

It then follows that

The matrix has eigenvalues and . However, the basis for is

so , so is not diagonalizable.

An matrix such that for all is called a diagonal matrix and is denoted by

That is, a diagonal matrix is any square matrix that is both upper and lower triangular.

The following matrices are diagonal matrices

Why do we care about diagonal matrices?

As we just observed, a diagonal matrix is simply a dilation, but by a different factor in each direction. Well, what else is only stretched after a linear transformation?

If and then it follows

1. (trivial)

For any positive integer :

In fact, this holds for any integer provided none of are zero, that is, if is invertible

An matrix is diagonalizable if there exists an invertible matrix and an diagonal matrix so that . In this case, we say that diagonalizes to .

This should look familiar. It's change of basis.

A matrix with every eigenvalue being real is diagonalizable if and only if there exists a basis for consisting of eigenvectors of .

We first prove the forward direction, and assume that is diagonalizable. Then there exists an invertible matrix and a diagonal matrix such that , that is, .
Thus

where is a standard basis vector.

We see that for , and since is invertible, it follows from the invertible matrix theorem that set is a basis for so that for , Thus is a basis for consisting of eigenvectors of .

We now prove the backward direction, and assume that there exist a basis of consisting of eigenvectors of . Then for each , for some eigenvalue of . It follows from the invertible matrix theorem that is invertible and thus

which shows that is diagonalizable.

The proof for this theorem is a constructive proof, and provides a method for finding invertible matrix and the diagonal matrix .

Given that is diagonalizable, the th column of will contain the th vector from the basis of eigenvectors, and the the column of the diagonal matrix will contain the corresponding eigenvalue of the -entry.

An matrix is diagonalizable if and only if for every eigenvalue of .

(see algebraic multiplicity, geometric multiplicity)

If an matrix has distinct eigenvalues, then is diagonalizable.

Let say we have a matrix . The actual transformation is quite complicated:

First, we find the eigenvectors and eigenvalues, which are , and , . Now if we transform these eigenvectors, we get:

So how can we interpret a somewhat "complicated" transformation into a simpler one? We can use change of basis!

If we have a matrix , which converts from the eigenbasis to the standard basis, then makes sense. On the left, we convert from the eigenbasis to the standard basis, then apply the transformation. On the right, we apply the diagonalized transformation (which is a "lives in" the eigenbasis), then convert from the eigenbasis to the standard basis.

What about ? is the reverse of , so it converts from the standard basis to the eigenbasis.

1. Convert from the std. basis to the eigenbasis
2. Apply the diagonal transformation which "lives in" the eigenbasis)
3. Convert it back from the eigenbasis to the std. basis.

What about ?

1. Convert from the eigenbasis to the standard basis
2. Apply the transformation (which "lives in" the std. basis)
3. Convert back from the std. basis to the eigenbasis.

So we can interpret as just a stretch of some vectors in the eigenbasis, which can be much simpler.

If we diagonalize , we obtain , , and . Now it makes sense why is composed of eigenvalues, and why is composed of eigenvectors.

So the transformation is equivalent to .

And the result is the same as directly applying .