Let us solve the challenge question in
Example 6.1.10 by converting it to a matrix problem.
Step 1: Choose a basis. For
\(P_2(x)\text{,}\) we use the standard basis
\(\alpha = \{1, x, x^2\}\text{.}\)
Step 2: Find the matrix of \(T\text{.}\) We compute the image of each basis element:
\begin{align*}
T(1) \amp= (x-2)(0) + 1 = 1,\\
T(x) \amp= (x-2)(1) + x = 2x - 2,\\
T(x^2) \amp= (x-2)(2x) + x^2 = 2x^2 - 4x + x^2 = 3x^2 - 4x.
\end{align*}
Writing these in terms of the basis \(\alpha\text{:}\)
\begin{align*}
[T(1)]_\alpha \amp= \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \quad [T(x)]_\alpha = \begin{pmatrix} -2 \\ 2 \\ 0 \end{pmatrix}, \quad [T(x^2)]_\alpha = \begin{pmatrix} 0 \\ -4 \\ 3 \end{pmatrix}.
\end{align*}
Therefore, the matrix of \(T\) relative to basis \(\alpha\) is
\begin{equation*}
[T]_\alpha = \begin{bmatrix} 1 \amp -2 \amp 0 \\ 0 \amp 2 \amp -4 \\ 0 \amp 0 \amp 3 \end{bmatrix}.
\end{equation*}
Step 3: Convert the problem to a matrix equation. The condition \(T(f(x)) = kf(x)\) becomes
\begin{equation*}
[T(f(x))]_\alpha = [kf(x)]_\alpha = k[f(x)]_\alpha.
\end{equation*}
By the fundamental relation between linear transformations and matrices, we have \([T(f(x))]_\alpha = [T]_\alpha [f(x)]_\alpha\text{.}\) Therefore,
\begin{equation*}
[T]_\alpha [f(x)]_\alpha = k[f(x)]_\alpha.
\end{equation*}
We are going to seek scalars \(k\) (eigenvalues) and nonzero vectors \([f(x)]_\alpha\) (eigenvectors) satisfying this equation.
Step 4: Find the values \(k\) . We rearrange the equation to get
\begin{equation*}
([T]_\alpha - kI)[f(x)]_\alpha = \mathbf{0}.
\end{equation*}
For nontrivial solutions to exist, the matrix \([T]_\alpha - kI\) must be singular, which means its determinant must be zero:
\begin{align*}
\det([T]_\alpha - kI) \amp= \det\begin{bmatrix} 1-k \amp -2 \amp 0 \\ 0 \amp 2-k \amp -4 \\ 0 \amp 0 \amp 3-k \end{bmatrix}\\
\amp= (1-k)(2-k)(3-k) = 0.
\end{align*}
So the possible values are \(k = 1, 2, 3\text{.}\)
Step 5: Find the corresponding polynomial \(f(x)\) for \(k=3\text{.}\) We need to solve \(([T]_\alpha - 3I)[f(x)]_\alpha = \mathbf{0}\text{.}\) For \(k=3\text{,}\) we have:
\begin{equation*}
[T]_\alpha - 3I = \begin{bmatrix} -2 \amp -2 \amp 0 \\ 0 \amp -1 \amp -4 \\ 0 \amp 0 \amp 0 \end{bmatrix}.
\end{equation*}
Computing the reduced row echelon form:
\begin{equation*}
\begin{bmatrix} -2 \amp -2 \amp 0 \\ 0 \amp -1 \amp -4 \\ 0 \amp 0 \amp 0 \end{bmatrix} \xrightarrow{\text{RREF}} \begin{bmatrix} 1 \amp 0 \amp -4 \\ 0 \amp 1 \amp 4 \\ 0 \amp 0 \amp 0 \end{bmatrix}.
\end{equation*}
The general solution is:
\begin{align*}
c - 4a \amp= 0 \quad \Rightarrow \quad c = 4a,\\
b + 4a \amp= 0 \quad \Rightarrow \quad b = -4a,
\end{align*}
where \(a\) is a free parameter. Since we want a monic polynomial (leading coefficient \(a = 1\)), we set \(a = 1\text{,}\) which gives:
\begin{equation*}
c = 4, \quad b = -4, \quad a = 1.
\end{equation*}
Therefore, the coordinate vector is \([f(x)]_\alpha = \begin{pmatrix} 4 \\ -4 \\ 1 \end{pmatrix}\text{,}\) which corresponds to the polynomial:
\begin{equation*}
f(x) = 4 \cdot 1 + (-4) \cdot x + 1 \cdot x^2 = x^2 - 4x + 4 = (x-2)^2.
\end{equation*}
Conclusion: The abstract question "Find all
\(k\) and monic polynomials
\(f(x)\) such that
\(T(f(x))=kf(x)\)" has been transformed into a concrete matrix problem: "Find the eigenvalues and eigenvectors of
\([T]_\alpha\text{.}\)" This illustrates the power of matrix representations — abstract questions become computational problems.
Remark: Similarly, we can find the monic polynomials for
\(k=1\) and
\(k=2\) by solving
\(([T]_\alpha - kI)[f(x)]_\alpha = \mathbf{0}\) with the constraint that the coefficient of
\(x^2\) equals 1.
