DLA Theory

Section 46.4 Theory

Here we will only make the connection between eigenvalues and eigenvectors of a operator versus of a matrix for that operator, and then characterize diagonalizable operators. The remainder of the story of canonical forms for linear operators is precisely the same as for matrices, as in Part III.

Theorem 46.4.1. Eigenvalue/eigenvector pairs of operator and matrix coincide.

Suppose that \(\funcdef{T}{V}{V}\) is a linear operator on a finite-dimensional vector space, \(\lambda\) is a scalar value, and \(\uvec{v}\) is a vector in the domain space \(V\text{.}\) Then the following are equivalent.

Operator \(T\) has eigenvalue/eigenvector pair \(\lambda, \uvec{v}\text{.}\)
For every choice of basis \(\basisfont{B}\) for \(V\text{,}\) the matrix \(\matrixOf{T}{B}\) has eigenvalue/eigenvector pair \(\lambda, \matrixOf{\uvec{v}}{B}\text{.}\)
For at least one choice of basis \(\basisfont{B}\) for \(V\text{,}\) the matrix \(\matrixOf{T}{B}\) has eigenvalue/eigenvector pair \(\lambda, \matrixOf{\uvec{v}}{B}\text{.}\)

Statement 1 implies Statement 2.

Suppose \(\lambda,\uvec{v}\) is an eigenvalue/eigenvector pair for \(T\text{,}\) so that

\begin{equation*} T(\uvec{v}) = \lambda \uvec{v} \text{.} \end{equation*}

Given a choice of basis \(\basisfont{B}\) for the domain space \(V\text{,}\) we have both

\begin{align*} \matrixOf{T(\uvec{v})}{B} \amp = \matrixOf{T}{B} \matrixOf{\uvec{v}}{B} \text{,} \amp \matrixOf{\lambda \uvec{v}}{B} \amp = \lambda \matrixOf{\uvec{v}}{B}\text{,} \end{align*}

where on the one hand we apply the relationship of \(\matrixOf{T}{B}\) to the input-output process for \(T\text{,}\) and on the other we apply the linearity of the coordinate map relative to \(\basisfont{B}\text{.}\) Equating these two results, we have

\begin{equation*} \matrixOf{T}{B} \matrixOf{\uvec{v}}{B} = \lambda \matrixOf{\uvec{v}}{B} \text{,} \end{equation*}

so that \(\lambda, \matrixOf{\uvec{v}}{B}\) is an eigenvalue/eigenvector pair of \(\matrixOf{T}{B}\text{.}\)

Statement 2 implies Statement 3.

This is obvious.

Statement 3 implies Statement 1.

Suppose \(\basisfont{B}\) is a basis of \(V\) for which \(\lambda, \matrixOf{\uvec{v}}{B}\) is an eigenvalue/eigenvector pair of \(\matrixOf{T}{B}\text{.}\) Then,

\begin{align*} \matrixOf{T(\uvec{v})}{B} \amp = \matrixOf{T}{B} \matrixOf{\uvec{v}}{B} \amp \amp \text{(i)} \\ \amp = \lambda \matrixOf{\uvec{v}}{B} \amp \amp \text{(ii)} \\ \amp = \matrixOf{\lambda \uvec{v}}{B} \amp \amp \text{(iii)} \text{,} \end{align*}

with justifications

the definition of \(\matrixOf{T}{B}\text{;}\)
our assumption that \(\lambda, \matrixOf{\uvec{v}}{B}\) is an eigenvalue/eigenvector pair of \(\matrixOf{T}{B}\text{;}\) and
the linearity of the coordinate map (Theorem 22.5.1).

Now, coordinate maps are isomorphisms (Corollary 44.5.17), so from

\begin{equation*} \coordmap{B}\bigl(T(\uvec{v})\bigr) = \matrixOf{T(\uvec{v})}{B} = \matrixOf{\lambda \uvec{v}}{B} = \coordmap{B}(\lambda \uvec{v}) \end{equation*}

we can conclude that

\begin{equation*} T(\uvec{v}) = \lambda \uvec{v} \text{.} \end{equation*}

In other words, \(\lambda,\uvec{v}\) is an eigenvalue/eigenvector pair for the operator \(T\text{.}\)

Corollary 46.4.2. Diagonalizable operator requires eigenvector basis.

An operator \(\funcdef{T}{V}{V}\) on a finite-dimensional space \(V\) is diagonalizable if and only if there exists a basis for \(V\) consisting of eigenvectors for \(T\text{.}\)

For \(T\) to be diagonalizable, there must exist a basis for \(V\) for which \(\matrixOf{T}{B}\) is diagonal. The eigenvectors of a diagonal matrix are precisely the standard basis vectors of \(\R^n\) (or \(\C^n\text{,}\) as appropriate), and the standard basis vectors are precisely the coordinate vectors of the basis vectors in \(\basisfont{B}\) relative to that basis. Therefore, we may apply Theorem 46.4.1 to see that the vectors in the basis \(\basisfont{B}\) must be eigenvectors for \(T\) in order for \(\matrixOf{T}{B}\) to be diagonal.

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