document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1166667>Question 1166667</A>: <I><SPAN STYLE=\"word-break: break-all;\"> : 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<BR>\n" );
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document.write( "(a) Show T is a linear transformation. (b) Show T is not one-to-one. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852695 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2138)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=852695\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> That's a great problem in linear algebra! Here's the demonstration that $T$ is a linear transformation but not one-to-one.\r<BR>\n" );
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document.write( "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 $T(p(x)) = x p'(x)$.\r<BR>\n" );
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document.write( "## (a) Show T is a Linear Transformation\r<BR>\n" );
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document.write( "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 $c \in \mathbb{R}$:\r<BR>\n" );
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document.write( "### 1. Additivity: $T(p(x) + q(x)) = T(p(x)) + T(q(x))$\r<BR>\n" );
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document.write( "Start with the left side (LHS):<BR>\n" );
document.write( "$$T(p(x) + q(x))$$\r<BR>\n" );
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document.write( "By the definition of $T$:<BR>\n" );
document.write( "$$T(p(x) + q(x)) = x \cdot \frac{d}{dx} (p(x) + q(x))$$\r<BR>\n" );
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document.write( "Using the linearity of the derivative ($\frac{d}{dx}(f+g) = \frac{df}{dx} + \frac{dg}{dx}$):<BR>\n" );
document.write( "$$T(p(x) + q(x)) = x (p'(x) + q'(x))$$\r<BR>\n" );
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document.write( "Distributing $x$:<BR>\n" );
document.write( "$$T(p(x) + q(x)) = x p'(x) + x q'(x)$$\r<BR>\n" );
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document.write( "Now, substitute back the definition of $T$:<BR>\n" );
document.write( "$$T(p(x) + q(x)) = T(p(x)) + T(q(x))$$\r<BR>\n" );
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document.write( "Since LHS = RHS, **additivity holds**.\r<BR>\n" );
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document.write( "### 2. Homogeneity: $T(c \cdot p(x)) = c \cdot T(p(x))$\r<BR>\n" );
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document.write( "Start with the left side (LHS):<BR>\n" );
document.write( "$$T(c \cdot p(x))$$\r<BR>\n" );
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document.write( "By the definition of $T$:<BR>\n" );
document.write( "$$T(c \cdot p(x)) = x \cdot \frac{d}{dx} (c \cdot p(x))$$\r<BR>\n" );
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document.write( "Using the property of the derivative ($\frac{d}{dx}(c \cdot f) = c \cdot \frac{df}{dx}$):<BR>\n" );
document.write( "$$T(c \cdot p(x)) = x (c \cdot p'(x))$$\r<BR>\n" );
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document.write( "Rearranging the terms (moving the scalar $c$ to the front):<BR>\n" );
document.write( "$$T(c \cdot p(x)) = c (x \cdot p'(x))$$\r<BR>\n" );
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document.write( "Now, substitute back the definition of $T$:<BR>\n" );
document.write( "$$T(c \cdot p(x)) = c \cdot T(p(x))$$\r<BR>\n" );
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document.write( "Since LHS = RHS, **homogeneity holds**.\r<BR>\n" );
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document.write( "Since $T$ satisfies both additivity and homogeneity, **$T$ is a linear transformation.**\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "## (b) Show T is Not One-to-One\r<BR>\n" );
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document.write( "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.\r<BR>\n" );
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document.write( "The Kernel is defined as the set of all vectors $p(x) \in V$ such that $T(p(x)) = 0$ (the zero polynomial).\r<BR>\n" );
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document.write( "$$\text{Ker}(T) = \{ p(x) \in V \mid T(p(x)) = 0 \}$$\r<BR>\n" );
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document.write( "We need to find if there is a non-zero polynomial $p(x)$ such that:<BR>\n" );
document.write( "$$T(p(x)) = x p'(x) = 0$$\r<BR>\n" );
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document.write( "If $x p'(x) = 0$ for all $x \in \mathbb{R}$, this implies that the derivative $p'(x)$ must be the zero polynomial:<BR>\n" );
document.write( "$$p'(x) = 0$$\r<BR>\n" );
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document.write( "If the derivative of a polynomial is zero, the polynomial itself must be a **constant polynomial**:<BR>\n" );
document.write( "$$p(x) = c$$<BR>\n" );
document.write( "where $c$ is any real constant.\r<BR>\n" );
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document.write( "The polynomials $p(x) = 1$, $p(x) = 5$, or $p(x) = -10$ are all non-zero vectors in $V$ (the space of all polynomials).\r<BR>\n" );
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document.write( "For example, let $p(x) = 5$.<BR>\n" );
document.write( "$$p'(x) = 0$$<BR>\n" );
document.write( "$$T(p(x)) = T(5) = x \cdot (0) = 0$$\r<BR>\n" );
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document.write( "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).\r<BR>\n" );
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document.write( "Therefore, **$T$ is not one-to-one.** </SPAN>\n" );
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