Orthogonal Diagonalization

Recall that for diagonalization, we want to find matrix , and recall for orthogonal matrices, we have . For orthogonal diagonalization, we want to find matrix

Orthogonal Subspaces

Definition

Let and be two subspaces of . We say and are orthogonal if for every and

Orthogonally Diagonalizable Matrix

Definition

An is orthogonally diagonalizable if there exists an orthogonal matrix and an diagonal matrix so that .

In this case, we say orthogonally diagonalizes to .

Lemma

, is orthogonally diagonalizable is diagonalizable.

However, the converse is not true, in that not every diagonalizable matrix is orthogonally diagonalizable.

Symmetric Matrices have Real Eigenvalues

Theorem

Let be a symmetric matrix. Then every eigenvalue of is real.

Dot Product Commutes with Matrix Vector Product Iff Symmetrical

Lemma

Let . Then is symmetric if and only if for all .

Proof

First, we prove the forward direction, and assume is symmetric, so . Then for any ,

Now we prove the backwards direction, and assume for all . We must show that is symmetric.
Since

holds for every , we see that by the matrices equal theorem. From this, we have that for every :

so that by matrices equal theorem, so is symmetric.

Eigenspaces of a Symmetric Matrix are Orthogonal Subspaces

Theorem

Let be a symmetric matrix. If are distinct eigenvalues of , then and are orthogonal subspaces of .

Proof

Let and . Then and . We must show that . This is trivial if either or , so we assume . Then

Since , we have that . However, since and are distinct, , so . Thus and are orthogonal subspaces of

Symmetric Matrices are Orthogonally Diagonalizable

Theorem

Let . Then is symmetric if and only if is orthogonally diagonalizable.

Proof

First, we prove the backwards direction, that if is orthogonally diagonalizable, then is symmetric. There is an orthogonal matrix and diagonal matrix so that . It follows that and

where we have used the fact that being diagonal implies . Hence so is symmetric.

The forward direction is out of the scope of MATH 115 and thus omitted.

Giant Problem

Example

Orthogonally diagonalize .

Solution
We first find the characteristic polynomial.

So we have

Eigenvalue	algebraic multiplicity

We now find an a basis for each eigenspace.

Hence a basis for is

and . For , we solve .

Hence a basis for is

and .

Now let

We should verify that , , and , and also verify that is orthogonal to both and . Note that is diagonalizable since and .

We now find an orthogonal basis for by applying the Gram-Schmidt Procedure to . Let

and let .
Thus is an orthogonal basis for and is an orthogonal basis for . Moreover, since the eigenspace is orthogonal, is an orthogonal basis for (since it is a linearly independent set of 3 vectors in ).

We then normalize the vectors to obtain an orthonormal basis for :

Finally, we see that where