Section 1.4 Operators
When a linear function maps \(V\) into itself, special things happen. First, the matrix that represents \(T: \F^n \to \F^n\) is square. There are a large number of equivalences between the structure of square matrices, linear maps, and sets of vectors. Many of these are captured in the invertible matrix theorem, one of the central objects of study in elementary linear algebra.
Theorem 1.4.1. Invertible matrix theorem.
Let \(A\) be an \(n \times n\) matrix. If any of the following conditions hold, all of them do. If any of them are false, they are all.
\(A\) is invertible.
\(A\) row reduces to the identity matrix \(I\text{.}\)
\(A\) has \(n\) pivot positions.
\(rank A = n\text{.}\)
The equation \(A \vec x = \vec 0\) has only the trivial solution.
The columns of \(A\) are linearly independent.
The function \(T(\vec x) = A \vec x\) is one-to-one.
The equation \(A \vec x = \vec v\) is consistent for all \(b \in \F^n\text{.}\)
The columns of \(A\) span \(\F^n\text{.}\)
The function \(T(\vec x) = A \vec x\) is onto.
There is a matrix \(C\) so that \(C A = I\text{.}\)
There is a matrix \(D\) so that \(A D = I\text{.}\)
\(A^T\) is invertible.
\(\displaystyle \det A \neq 0.\)
Operators contain more information than the invertibility of the functions that they represent. For the following discussion, let us fix a basis of a vector space \(V\) and let \(A\) be the matrix that represents a function \(T: V \to V\text{.}\) A scalar \(\la\) and a vector \(v\) are said to be an eigenpair for \(A\) if
\begin{equation*}
A v = \la v.
\end{equation*}
It is straightforward to see that the set of all vectors \(v\) for which the eigenvector equation holds is a subspace of \(V\text{,}\) called the eigenspace associated with \(\la\text{.}\) The eigenspaces of the matrix \(A\) are its invariant subspaces, which is to say that a vector in an eigenspace is mapped by \(A\) to the same eigenspace. It turns out that knowing the invariant subspaces of \(A\) are often enough to completely characterize \(A\text{.}\) If \(A\) is \(n\times n\) and \(A\) has \(n\) linearly independent eigenvectors (that is, one can find a basis of \(\F^n\) consisting of eigenvectors of \(A\)), then
\begin{equation*}
A = S D S\inv,
\end{equation*}
where \(S\) is a matrix of eigenvectors and \(D\) is a diagonal matrix of the associated eigenvalues (including repetition of course). (One should think of \(S\) as a change of basis matrix under which the operator \(A\) becomes diagonal.)
Many operators are not diagonalizable, even very simple ones. For example, \(A = \bbm 1 \amp 1 \\ 0 \amp 1 \ebm\) only has a one-dimensional eigenspace. Diagonalizability is so useful that we give characterizations of those operators a special name, the Spectral Theorem. An operator on a real vector space is called symmetric if \(A^T = A\text{.}\) An operator on a complex vector space is called Hermitian (or conjugate symmetric) if \(A\ad = \cc{A^T} = A\text{.}\) One of the major theorems of elementary linear algebra is that such operators are diagonalizable and that there exists an orthonormal basis of eigenvectors for \(V\text{.}\)
Theorem 1.4.2.
Let \(A\) be an \(n \times n\) real (complex) matrix. Then \(A\) is diagonalizable with respect to an orthonormal basis of eigenvectors if and only if \(A\) is symmetric (hermitian).
For complex operators, one can say more. \(A\) is called normal if \(A A\ad = A\ad A\text{.}\) One reason that complex vector spaces are so much nicer than real vector spaces is that normal operators turn out to have orthonormal diagonalizations.
Theorem 1.4.3. Complex (finite) spectral theorem.
Let \(A\) be an operator on a finite dimensional Hilbert space \(V\text{.}\) Then \(A\) is normal if and only if \(A\) can be diagonalized with respect to an orthonormal basis of eigenvectors for \(V\text{.}\)
One of the goals of Hilbert space theory is to capture these kinds of structural results in the context of infinite dimensional Hilbert spaces and operators on them.
