Section 46.3 Examples
In this section.
Subsection 46.3.1 Using a “natural” basis for an operator
Example 46.3.1. Projection onto a line in \(\R^2\).
Let \(\funcdef{T}{\R^2}{\R^2}\) be the operator that orthogonally projects vectors onto the line
A “natural” basis to use to analyze the action of \(T\) on \(\R^2\) is an orthogonal basis consisting of one vector parallel to and one vector orthogonal to \(\ell\text{.}\) So let's take
From calculations
we obtain
As this is a diagonal matrix, we can say that \(T\) is a diagonalizable operator.
If we would like to obtain the matrix \(\matrixOf{T}{S}\) relative to the standard basis \(\basisfont{S}\text{,}\) we can use a change of basis matrix. First form the transition matrix
and calculate the reverse transition matrix
Then we have
Note that even though this is not a diagonal matrix, it does not contradict the fact the \(T\) is diagonalizable, as we only require that the matrix of \(T\) be diagonal relative to at least one basis of the domain space, not all bases.
Example 46.3.2. Transpose of matrices.
Consider \(\funcdef{T}{\matrixring_n(\R)}{\matrixring_n(\R)}\) by
As usual, let
represent the standard basis of \(\matrixring_n(\R)\text{,}\) where \(E_{ij}\) is the \(n \times n\) matrix with a \(1\) in the \((i,j)\) entry and \(0\) in every other entry. Then
so the matrix \(\matrixOf{T}{S}\) is not difficult to calculate.
However, we can also use these standard basis vectors to create a basis of \(\matrixring_n(\R)\) consisting of eigenvectors. Write
for the diagonal matrices in \(\basisfont{S}\text{,}\) write
for certain symmetric combinations of matrices in \(\basisfont{S}\text{,}\) and write
for certain skew-symmetric combinations of matrices in \(\basisfont{S}\text{.}\) All these matrices together form a basis \(\basisfont{B}\) of \(\matrixring_n(\R)\text{,}\) and we can verify
Relative to this basis of eigenvectors, we will indeed obtain a diagonal matrix for \(T\text{:}\)
where \(I_a,I_b\) are identity matrices of sizes
So \(T\) is a diagonalizable operator, since a basis for the domain space consisting of eigenvectors for the operator can be found.
Subsection 46.3.2 Computing determinant, eigenvalues, and eigenvectors of operators
Example 46.3.3. Computing the determinant of an operator.
Consider the “reversal” operator \(\funcdef{T}{\poly_3(\R)}{\poly_3(\R)}\) defined by
Relative to the standard basis \(\basisfont{S} = \{ x^3, x^2, x, 1 \}\text{,}\) we compute the matrix for \(T\) by first calculating the image vectors
Then
and we can compute
Using Corollary 45.5.5, we can conclude that \(T\) must be an isomorphism. (Though in this case that should have been obvious, as \(T\) is clearly its own inverse.)
Example 46.3.4. Computing eigenvalues and eigenvectors of an operator.
For \(p(x)\) in \(\poly_3(\R)\text{,}\) write \(\tilde{p}(x)\) for the “reversal” of \(p(x)\text{,}\) as in Example 46.3.3 above, but now define \(\funcdef{T}{\poly_3(\R)}{\poly_3(\R)}\) by
Compute the image vectors
from which we obtain the matrix of \(T\) relative to the standard basis \(\basisfont{S} = \{ x^3, x^2, x, 1 \}\text{:}\)
From
we can calculate
so that \(T\) has eigenvalues \(\lambda = 0, 2\text{.}\)
As usual, row reduce the matrices \(0 I - \matrixOf{T}{S}\) and \(2 I - \matrixOf{T}{S}\) to find that
To convert these from eigenvectors of \(\matrixOf{T}{S}\) into eigenvectors of \(T\text{,}\) simply convert them from coordinate vectors relative to \(\basisfont{S}\) into \(\poly_3(\R)\) vectors:
We could verify that these are eigenvectors of \(T\) directly, without appeal to properties of \(\matrixOf{T}{S}\text{.}\) For example,
as required by the eigenvalue-eigenvector pattern.