SOLUTION: : Let V be the vector space of all polynomials defined on the real number line. (12 points) Suppose that T : V → V is the transformation T(p(x)) = xp′(x) where the prime denote

Question 1166667: : Let V be the vector space of all polynomials defined on the real number line. (12 points) Suppose that T : V → V is the transformation T(p(x)) = xp′(x) where the prime denotes
derivative.
(a) Show T is a linear transformation. (b) Show T is not one-to-one.
Answer by CPhill(2138) (Show Source): You can put this solution on YOUR website!
That's a great problem in linear algebra! Here's the demonstration that $T$ is a linear transformation but not one-to-one.
Let $V$ be the vector space of all polynomials defined on the real number line, and let the transformation $T: V \to V$ be defined by $T(p(x)) = x p'(x)$.
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## (a) Show T is a Linear Transformation
To show that $T$ is a **linear transformation**, we must verify two properties for any two polynomials $p(x)$ and $q(x)$ in $V$, and any scalar $c \in \mathbb{R}$:
### 1. Additivity: $T(p(x) + q(x)) = T(p(x)) + T(q(x))$
Start with the left side (LHS):
$$T(p(x) + q(x))$$
By the definition of $T$:
$$T(p(x) + q(x)) = x \cdot \frac{d}{dx} (p(x) + q(x))$$
Using the linearity of the derivative ($\frac{d}{dx}(f+g) = \frac{df}{dx} + \frac{dg}{dx}$):
$$T(p(x) + q(x)) = x (p'(x) + q'(x))$$
Distributing $x$:
$$T(p(x) + q(x)) = x p'(x) + x q'(x)$$
Now, substitute back the definition of $T$:
$$T(p(x) + q(x)) = T(p(x)) + T(q(x))$$
Since LHS = RHS, **additivity holds**.
### 2. Homogeneity: $T(c \cdot p(x)) = c \cdot T(p(x))$
Start with the left side (LHS):
$$T(c \cdot p(x))$$
By the definition of $T$:
$$T(c \cdot p(x)) = x \cdot \frac{d}{dx} (c \cdot p(x))$$
Using the property of the derivative ($\frac{d}{dx}(c \cdot f) = c \cdot \frac{df}{dx}$):
$$T(c \cdot p(x)) = x (c \cdot p'(x))$$
Rearranging the terms (moving the scalar $c$ to the front):
$$T(c \cdot p(x)) = c (x \cdot p'(x))$$
Now, substitute back the definition of $T$:
$$T(c \cdot p(x)) = c \cdot T(p(x))$$
Since LHS = RHS, **homogeneity holds**.
Since $T$ satisfies both additivity and homogeneity, **$T$ is a linear transformation.**
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## (b) Show T is Not One-to-One
A linear transformation $T$ is **not one-to-one** (or injective) if and only if its **Kernel** (or null space), $\text{Ker}(T)$, contains more than just the zero vector.
The Kernel is defined as the set of all vectors $p(x) \in V$ such that $T(p(x)) = 0$ (the zero polynomial).
$$\text{Ker}(T) = \{ p(x) \in V \mid T(p(x)) = 0 \}$$
We need to find if there is a non-zero polynomial $p(x)$ such that:
$$T(p(x)) = x p'(x) = 0$$
If $x p'(x) = 0$ for all $x \in \mathbb{R}$, this implies that the derivative $p'(x)$ must be the zero polynomial:
$$p'(x) = 0$$
If the derivative of a polynomial is zero, the polynomial itself must be a **constant polynomial**:
$$p(x) = c$$
where $c$ is any real constant.
The polynomials $p(x) = 1$, $p(x) = 5$, or $p(x) = -10$ are all non-zero vectors in $V$ (the space of all polynomials).
For example, let $p(x) = 5$.
$$p'(x) = 0$$
$$T(p(x)) = T(5) = x \cdot (0) = 0$$
Since $\text{Ker}(T)$ contains all non-zero constant polynomials (i.e., $\text{Ker}(T) = \{p(x) = c \mid c \in \mathbb{R}\}$), the **Kernel is non-trivial** (it is not just the zero vector).
Therefore, **$T$ is not one-to-one.**
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