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Section 1.1 Determine if a map between vector spaces of polynomials is linear or not.

This is a generic introduction to this section.

Definition 1.1.1.

A linear transformation (also known as a linear map) is a map between vector spaces that preserves the vector space operations.

More precisely, if \(V\) and \(W\) are vector spaces, a map \(T:V\rightarrow W\) is called a linear transformation if

  1. \(T(\vec{v}+\vec{w}) = T(\vec{v})+T(\vec{w})\) for any \(\vec{v},\vec{w} \in V\text{.}\)
  2. \(T(c\vec{v}) = cT(\vec{v})\) for any \(c \in \IR,\vec{v} \in V\text{.}\)

In other words, a map is linear when vector space operations can be applied before or after the transformation without affecting the result.

Example 1.1.2.

Let \(T : \IR^3 \rightarrow \IR^2\) be given by

\begin{equation*} T\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix} \right) = \begin{bmatrix} x-z \\ 3y \end{bmatrix} \end{equation*}

To show that \(T\) is linear, we must verify...

\begin{equation*} T\left( \begin{bmatrix} x \\ y \\ z \end{bmatrix} + \begin{bmatrix} u \\ v \\ w \end{bmatrix} \right) = T\left( \begin{bmatrix} x+u \\ y+v \\ z+w \end{bmatrix} \right) = \begin{bmatrix} (x+u)-(z+w) \\ 3(y+v) \end{bmatrix} \end{equation*}
\begin{equation*} T\left( \begin{bmatrix} x \\ y \\ z \end{bmatrix} \right) + T\left( \begin{bmatrix} u \\ v \\ w \end{bmatrix} \right) = \begin{bmatrix} x-z \\ 3y \end{bmatrix} + \begin{bmatrix} u-w \\ 3v \end{bmatrix}= \begin{bmatrix} (x+u)-(z+w) \\ 3(y+v) \end{bmatrix} \end{equation*}

And also...

\begin{equation*} T\left(c\begin{bmatrix} x \\ y \\ z \end{bmatrix} \right) = T\left(\begin{bmatrix} cx \\ cy \\ cz \end{bmatrix} \right) = \begin{bmatrix} cx-cz \\ 3cy \end{bmatrix} \text{ and } cT\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix} \right) = c\begin{bmatrix} x-z \\ 3y \end{bmatrix} = \begin{bmatrix} cx-cz \\ 3cy \end{bmatrix} \end{equation*}

Therefore \(T\) is a linear transformation.

Example 1.1.3.

Let \(T : \IR^2 \rightarrow \IR^4\) be given by

\begin{equation*} T\left(\begin{bmatrix} x \\ y \end{bmatrix} \right) = \begin{bmatrix} x+y \\ x^2 \\ y+3 \\ y-2^x \end{bmatrix} \end{equation*}

To show that \(T\) is not linear, we only need to find one counterexample.

\begin{equation*} T\left( \begin{bmatrix} 0 \\ 1 \end{bmatrix} + \begin{bmatrix} 2 \\ 3 \end{bmatrix} \right) = T\left( \begin{bmatrix} 2 \\ 4 \end{bmatrix} \right) = \begin{bmatrix} 6 \\ 4 \\ 7 \\ 0 \end{bmatrix} \end{equation*}
\begin{equation*} T\left( \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right) + T\left( \begin{bmatrix} 2 \\ 3\end{bmatrix} \right) = \begin{bmatrix} 1 \\ 0 \\ 4 \\ -1 \end{bmatrix} + \begin{bmatrix} 5 \\ 4 \\ 6 \\ -5 \end{bmatrix} = \begin{bmatrix} 6 \\ 4 \\ 10 \\ -6 \end{bmatrix} \end{equation*}

Since the resulting vectors are different, \(T\) is not a linear transformation.

Here is some text I'm adding that I won't want to be in slides, but would be useful context for the student or instructor when viewing the full textbook.

Activity 1.1.1.

(~5 min)

Recall the following rules from calculus, where \(D:\Poly\to\Poly\) is the derivative map defined by \(D(f(x))=f'(x)\) for each polynomial \(f\text{.}\)

\begin{equation*} D(f+g)=f'(x)+g'(x) \end{equation*}
\begin{equation*} D(cf(x))=cf'(x) \end{equation*}

What can we conclude from these rules?

  1. \(\Poly\) is a subspace of \(\IR^3\)
  2. \(\Poly\) is not a vector space
  3. \(D\) is a linear map
  4. \(D\) is not a linear map
Activity 1.1.2.

(~10 min)

Let the polynomial maps \(S: \Poly^4 \rightarrow \Poly^3\) and \(T: \Poly^4 \rightarrow \Poly^3\) be defined by

\begin{equation*} S(f(x)) = 2f'(x)-f''(x) \hspace{3em} T(f(x)) = f'(x)+x^3 \end{equation*}

Compute \(S(x^4+x)\text{,}\) \(S(x^4)+S(x)\text{,}\) \(T(x^4+x)\text{,}\) and \(T(x^4)+T(x)\text{.}\) Which of these maps is definitely not linear?

Note 1.1.6.

Showing \(L:V\to W\) is not a linear transformation can be done by finding an example for any one of the following.

  • Show \(L(\vec z)\not=\vec z\) (where \(\vec z\) is the additive identity of \(L\) and \(W\)).
  • Find \(\vec v,\vec w\in V\) such that \(L(\vec v+\vec w)\not=L(\vec v)+L(\vec w)\text{.}\)
  • Find \(\vec v\in V\) and \(c\in \IR\) such that \(L(c\vec v)\not=cL(\vec v)\text{.}\)

Otherwise, \(L\) can be shown to be linear by proving the following in general.

  • For all \(\vec v,\vec w\in V\text{,}\) \(L(\vec v+\vec w)\not=L(\vec v)+L(\vec w)\text{.}\)
  • For all \(\vec v\in V\) and \(c\in \IR\text{,}\) \(L(c\vec v)\not=cL(\vec v)\text{.}\)

Note the similarities between this process and showing that a subset of a vector space is/isn't a subspace.

Activity 1.1.3.

(~15 min)

Continue to consider \(S: \Poly^4 \rightarrow \Poly^3\) defined by

\begin{equation*} S(f(x)) = 2f'(x)-f''(x) \end{equation*}
(a)

Verify that

\begin{equation*} S(f(x)+g(x))=2f'(x)+2g'(x)-f''(x)-g''(x) \end{equation*}

is equal to \(S(f(x))+S(g(x))\) for all polynomials \(f,g\text{.}\)

(b)

Verify that \(S(cf(x))\) is equal to \(cS(f(x))\) for all real numbers \(c\) and polynomials \(f\text{.}\)

(c)

Is \(S\) linear?

I suppose it's theoretically possible someone wants a bare paragraph or two in slides.

So let's assume that this paragraph and the previous paragraph should be on a slide together.

But let's also assume that this sentence is extra text that should only be in the textbook, not in the slides.

Activity 1.1.4.

(~20 min)

Let the polynomial maps \(S: \Poly \rightarrow \Poly\) and \(T: \Poly \rightarrow \Poly\) be defined by

\begin{equation*} S(f(x)) = (f(x))^2 \hspace{3em} T(f(x)) = 3xf(x^2) \end{equation*}
(a)

Note that \(S(0)=0\) and \(T(0)=0\text{.}\) So instead, show that \(S(x+1)\not= S(x)+S(1)\) to verify that \(S\) is not linear.

(b)

Prove that \(T\) is linear by verifying that \(T(f(x)+g(x))=T(f(x))+T(g(x))\) and \(T(cf(x))=cT(f(x))\text{.}\)

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