Today we learnt: Every linear operator on inner product space has a “mate” such that and are related by

.

How is defined? In today’s lecture we studied the definition of , which is based on the important result: If is a linear transformation, then there exists unique such that .

- where is an orthonormal basis for .
This result tells the relation between the matrix representations of and w.r.t. orthonormal bases.

- If , then is symmetric.
Below we shall see that self-adjoint operator (i.e. satisfying ) are particularly nice.

Recall that for linear operators on vector spaces, we study the concept of similarity and diagonalization: let us review a few important points below. By definition, “ is similar to ” means for some invertible matrix .

- If is a linear operator on vector space (not necessarily inner product space), then is similar to for any ordered bases and .
- If is similar to , then and are matrix representations of the same linear operator.

Next suppose is similar to a diagonal matrix (i.e. ). Write and , then from , we get , i.e.

That means is the eigenvalue and is the corresponding eigenvector. Also, is invertible if and only if form a basis for . The converse is also true: when we diagonalize a matrix (assuming is diagonalizable), how do we find and ? We calculate the eigenvalues to get and then calculate the eigenvectors to get .

Furthermore, we can present the above result in the setting of linear transformation: Suppose where and . Then the linear operator , has the standard matrix representation . If we set (the ordered basis consisting of eigenvectors), then is diagonal. Using this viewpoint, we have another description (or criterion) for diagonalizable matrices:

There exists a basis for such that is diagonal

There exists a basis for such that ,

We can find a basis consisting of eigenvectors of for

Now we turn back to inner product spaces. If is an inner product space, then we can consider more special basis — orthonormal basis (which are much more convenient, at least from the angle of computation). Hence we may consider the following question in the above diagoanlization problem:

Can we find an orthonormal basis such that is diagonal?

In terms of matrices, this is equivalent to finding a set of orthonormal eigenvectors such that such that .

Here we make a nice observation: if where for and for , then

Such a matrix is called an orthogonal matrix (i.e. is orthogonal if . Note that .)

Hence for our problem the condition can be rephrased as . That’s why we invoke the concept of orthogonally diagonalizable.

Now we can state the following key result for self-adjoint linear operators (or in matrix setting, symmetric matrices):

Every symmetric matrix has a set of orthonormal eigenvectors which form a basis for .

In the setting of linear operators, *every self-adjoint operator (on inner product space) has a set of orthonormal eigenvectors which form a basis for . *

See also Lect28-29.pdf in the folder slides.

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