Orthonormal Diagonalization

We have seen in that under the right conditions a square matrix is similar to a diagonal matrix. We recognize now, via , that a similarity transformation is a change of basis on a matrix representation. So we can now discuss the choice of a basis used to build a matrix representation, and decide if some bases are better than others for this purpose. This will be the tone of this section. We will also see that every matrix has a reasonably useful matrix representation, and we will discover a new class of diagonalizable linear transformations. First we need some basic facts about triangular matrices.

1. Name three broad classes of normal matrices that we have studied previously. No set that you give should be a subset of another on your list.
2. Compare and contrast with .
3. Given an $n\times n$ matrix $A$, why would you desire an orthonormal basis of $\complex{n}$ composed entirely of eigenvectors of $A$?

An upper, or lower, triangular matrix is exactly what it sounds like it should be, but here are the two relevant definitions.

Obviously, properties of a lower triangular matrices will have analogues for upper triangular matrices. Rather than stating two very similar theorems, we will say that matrices are triangular of the same type as a convenient shorthand to cover both possibilities and then give a proof for just one type.

The inverse of a triangular matrix is triangular, of the same type.

Condensed — the full section is in Beezer, A First Course in Linear Algebra.

Not every matrix is diagonalizable, but every linear transformation has a matrix representation that is an upper triangular matrix, and the basis that achieves this representation is especially pleasing. Here is the theorem.

A key step in this proof was the construction of the subspace $W$ with dimension strictly less than that of $V$. This required an eigenvalue/eigenvector pair, which was guaranteed to us by . Digging deeper, the proof of requires that we can factor polynomials completely, into linear factors. This will not always happen if our set of scalars is the reals, $\real{\null}$. So this is our final explanation of our choice of the complex numbers, $\complexes$, as our set of scalars. In $\complexes$ polynomials factor completely, so every matrix has at least one eigenvalue, and an inductive argument will get us to upper triangular matrix representations.

In the case of linear transformations defined on $\complex{m}$, we can use the inner product () profitably to fine-tune the basis that yields an upper triangular matrix representation. Recall that the adjoint of matrix $A$ () is written as $\adjoint{A}$.

Condensed — the full section is in Beezer, A First Course in Linear Algebra.

Normal matrices comprise a broad class of interesting matrices, many of which we have met already. But they are most interesting since they define exactly which matrices we can diagonalize via a unitary matrix. This is the upcoming . Here is the definition.

So a normal matrix commutes with its adjoint. Part of the beauty of this definition is that it includes many other types of matrices. A diagonal matrix will commute with its adjoint, since the adjoint is again diagonal and the entries are just conjugates of the entries of the original diagonal matrix. A Hermitian (self-adjoint) matrix () will trivially commute with its adjoint, since the two matrices are the same. A real, symmetric matrix is Hermitian, so these matrices are also normal. A unitary matrix () has its adjoint as its inverse, and inverses commute (), so unitary matrices are normal. Another class of normal matrices is the skew-symmetric matrices. However, these broad descriptions still do not capture all of the normal matrices, as the next example shows.

A diagonal matrix is very easy to work with in matrix multiplication () and an orthonormal basis also has many advantages (). How about converting a matrix to a diagonal matrix through a similarity transformation using a unitary matrix ( build a diagonal matrix representation with an orthonormal matrix)? That'd be fantastic! When can we do this? We can always accomplish this feat when the matrix is normal, and normal matrices are the only ones that behave this way. Here is the theorem.

We can rearrange the conclusion of this theorem to read $A=UD\adjoint{U}$. Recall that a unitary matrix can be viewed as a geometry-preserving transformation (isometry), or more loosely as a rotation of sorts. Then a matrix-vector product, $A\vect{x}$, can be viewed instead as a sequence of three transformations. $\adjoint{U}$ is unitary, and so is a rotation. Since $D$ is diagonal, it just multiplies each entry of a vector by a scalar. Diagonal entries that are positive or negative, with absolute values bigger or smaller than 1 evoke descriptions like reflection, expansion and contraction. Generally we can say that $D$ stretches a vector in each component. Final multiplication by $U$ undoes (inverts) the rotation performed by $\adjoint{U}$. So a normal matrix is a rotation-stretch-rotation transformation.

The orthonormal basis formed from the columns of $U$ can be viewed as a system of mutually perpendicular axes. The rotation by $\adjoint{U}$ allows the transformation by $A$ to be replaced by the simple transformation $D$ along these axes, and then $D$ brings the result back to the original coordinate system. For this reason is known as the Principal Axis Theorem.

The columns of the unitary matrix in create an especially nice basis for use with the normal matrix. We record this observation as a theorem.

Condensed — the full section is in Beezer, A First Course in Linear Algebra.