# 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 $5x+3= \frac{-7}{4} (-x+2) + \frac{13}{4} (x+2)$.)

Add to solve later

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Problem 315(a): T(x)=3x+4

1.x+0=x ; a=1, b=0;

Dear nitinbharat,

Thank you for your comment. I modified the solution accordingly.