Basis with Respect to Which the Matrix for Linear Transformation is Diagonal

Problem 315

Let $P_1$ be the vector space of all real polynomials of degree $1$ or less. Consider the linear transformation $T: P_1 \to P_1$ defined by
\[T(ax+b)=(3a+b)x+a+3,\] for any $ax+b\in P_1$.

(a) With respect to the basis $B=\{1, x\}$, find the matrix of the linear transformation $T$.

(b) Find a basis $B’$ of the vector space $P_1$ such that the matrix of $T$ with respect to $B’$ is a diagonal matrix.

(c) Express $f(x)=5x+3$ as a linear combination of basis vectors of $B’$.

 
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(a) Find the matrix of the linear transformation $T$ with respect to the basis $B=\{1, x\}$.

The matrix $A$ for $T$ with respect to the basis $B=\{1, x\}$ is given by
\[A=[\, [T(1)]_B, [T(x)]_B\,].\] Since we have
\[T(1)=x+3 \text{ and } T(x)=3x+4,\] the coordinate vectors of these vector with respect to $B$ are
\begin{align*}
[T(1)]_B=\begin{bmatrix}
3 \\
1
\end{bmatrix} \text{ and } [T(x)]_B=\begin{bmatrix}
4 \\
3
\end{bmatrix}.
\end{align*}
Therefore, the matrix $A$ for $T$ with respect to $B$ is
\[A=\begin{bmatrix}
3 & 4\\
1& 3
\end{bmatrix}.\]

(b) Find a basis $B’$ of the vector space $P_1$ such that the matrix of $T$ with respect to $B’$ is a diagonal matrix.

We first diagonalize the matrix $A$ obtained in (a).
Solving the characteristic polynomial
\begin{align*}
p_A(t)=\det(A-tI)=\begin{bmatrix}
3-t & 4\\
1& 3-t
\end{bmatrix}\\
=t^2-6t+5=(t-1)(t-5),
\end{align*}
the eigenvalues are $1$ and $5$. Since $A$ has two distinct eigenvalues, it is diagonalizable.

We find the eigenvectors corresponding to the eigenvalue $1$.
The eigenvectors are nonzero solutions of $(A-I)\mathbf{x}=\mathbf{0}$.
From this, we find that $\begin{bmatrix}
2 \\
-1
\end{bmatrix}$ is an eigenvector.
Similarly, we find that $\begin{bmatrix}
2 \\
1
\end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $5$.

Thus, if we set
\[P=\begin{bmatrix}
2 & 2\\
-1& 1
\end{bmatrix},\] then we have the diagonalization of $A$:
\[\begin{bmatrix}
1 & 0\\
0 & 5
\end{bmatrix}=P^{-1}AP.\] It follows that $\begin{bmatrix}
1 & 0\\
0 & 5
\end{bmatrix}$ is the matrix representation for $T$ with respect to the basis
\[B’=\{\mathbf{v}_1, \mathbf{v}_2\},\] where
\[[\mathbf{v}_1]_B=\begin{bmatrix}
2 \\
-1
\end{bmatrix} \text{ and } [\mathbf{v}_2]_B=\begin{bmatrix}
2 \\
1
\end{bmatrix},\] and $P$ is the transition matrix (change of basis) from $B’$ to $B$.

(Note that
\[[\mathbf{v}_1]_B=P[\mathbf{v}_1]_{B’}=P\begin{bmatrix}
1 \\
0
\end{bmatrix}=\begin{bmatrix}
2 \\
-1
\end{bmatrix}.\] Similarly for $[\mathbf{v}_2]_B$.)

Hence, $\mathbf{v}_1=-x+2$ and $\mathbf{v}_2=x+2$ form the basis $B’=\{\mathbf{v}_1, \mathbf{v}_2\}$ with respect to which the matrix for $T$ is the diagonal matrix $\begin{bmatrix}
1 & 0\\
0 & 5
\end{bmatrix}$.

(c) Express $f(x)=5x+3$ as a linear combination of basis vectors of $B’$.

Note that since $P$ is the transition matrix from $B’$ to $B$, we have the following relation between coordinate vectors. For any $\mathbf{v}\in P_1$, we have
\[P[\mathbf{v}]_{B’}=[\mathbf{v}]_B.\] Using this, we have
\begin{align*}
[f(x)]_{B’}&=P^{-1}[f(x)]_B\\
&=\frac{1}{4}\begin{bmatrix}
1 & -2\\
1 & 2
\end{bmatrix}\begin{bmatrix}
3 \\
5
\end{bmatrix}=\begin{bmatrix}
-1 \\
4
\end{bmatrix}.
\end{align*}
Hence we obtain the linear combination of $f(x)$ with $\mathbf{v}_1, \mathbf{v}_2$:
\begin{align*}
f(x)&=\frac{-7}{4}\mathbf{v}_1 + \frac{13}{4} \mathbf{v}_2\\
\end{align*}

(You may confirm that $5x+3= \frac{-7}{4} (-x+2) + \frac{13}{4} (x+2)$.)

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