The Matrix Representation of the Linear Transformation $T (f) (x) = ( x^2 – 2) f(x)$

Problem 673

Let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the standard basis.

Let $T : \mathrm{P}_3 \rightarrow \mathrm{P}_{5}$ be the map defined by, for $f \in \mathrm{P}_3$,
$T (f) (x) = ( x^2 – 2) f(x).$

Determine if $T(x)$ is a linear transformation. If it is, find the matrix representation of $T$ relative to the standard basis of $\mathrm{P}_3$ and $\mathrm{P}_{5}$.

Solution.

We must check that $T$ satisfies the two axioms for linear transformations. Suppose $f, g \in \mathrm{P}_3$. Then
$T(f+g)(x) = (x^2-2) (f+g)(x) = (x^2 – 2) f(x) + (x^2 – 2) g(x) = T(f)(x) + T(g)(x).$

Second, if $c \in \mathbb{R}$ then
$T( cf )(x) = (x^2 – 2) ( cf) (x) = c (x^2 – 2) f(x) = c T(f)(x).$

The two axioms are satisfied, and so $T$ is a linear transformation.

Now we find its matrix representation relative to the standard basis $B = \{ 1 , x , x^2 , x^3 \}$ of $\mathrm{P}_3$ and the standard basis $C = \{ 1 , x , x^2 , x^3 , x^4 , x^5 \}$ . To do this, for every $f \in B$ we calculate the coordinate vector of $T(f)$ relative to the basis $C$.

For example, we find that $T(1) = x^2 – 2$. The coordinate vector for this element, relative to $C$, is
$[ x^2 – 2 ]_{C} = \begin{bmatrix} -2 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}.$ Similarly, we find that $T(x) = x^3 – 2x$. The coordinate vector for this element, relative to $C$,
$[ x^3 – 2x ]_{C} = \begin{bmatrix} 0 \\ -2 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}.$

Performing this process for the rest of the elements of $B$, we get
$[T(x^2)]_{C} = [ x^4 – 2x^2 ]_{C} = \begin{bmatrix} 0 \\ 0 \\ -2 \\ 0 \\ 1 \\ 0 \end{bmatrix} \quad [T(x^3)]_{C} = [ x^5 – 2x^3 ]_{C} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ -2 \\ 0 \\ 1 \end{bmatrix}.$ And now putting these four column vectors in order, we get the matrix representation for $T$:
$[T]_{B}^{C} = \begin{bmatrix} -2 & 0 & 0 & 0 \\ 0 & -2 & 0 & 0 \\ 1 & 0 & -2 & 0 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}.$

2 Responses

in case of T(x),
You have one extra row.
instead of 6 there are 7.
one 0 is extra.

• Yu says:

The Range and Nullspace of the Linear Transformation $T (f) (x) = x f(x)$
For an integer $n > 0$, let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The...