Is the Map $T (f) (x) = f(x) – x – 1$ a Linear Transformation between Vector Spaces of Polynomials?

Linear Transformation problems and solutions

Problem 674

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 $T : \mathrm{P}_4 \rightarrow \mathrm{P}_{4}$ be the map defined by, for $f \in \mathrm{P}_4$,
\[ T (f) (x) = f(x) – x – 1.\]

Determine if $T(x)$ is a linear transformation. If it is, find the matrix representation of $T$ relative to the standard basis of $\mathrm{P}_4$.

 
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Solution.

We give two solutions.

Solution 1.

We claim that $T$ is not a linear transformation.

To see this, consider $f(x) = 1 + x$. Then
\[T(f)(x) = 1 + x – x – 1 = 0.\] On the other hand,
\[T(1) + T(x) = (1 – x – 1) + (x – x – 1) = -x – 1.\] Because $T(1+x) \neq T(1) + T(x)$, we see that $T$ is not a linear transformation.

Solution 2.

Recall the fact that any linear transformation maps the zero vector to the zero vector.
The zero vector in $\mathrm{P}_n$ is the zero polynomial $\theta(x)=0$.

Then we have
\[T(\theta)(x)=\theta(x)-x-1=-x-1 \neq \theta(x).\]

Thus, the map $T$ does not send the zero vector $\theta(x)$ to the zero vector $\theta(x)$.
Hence, $T$ is not a linear transformation.


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