Linear Transformations (A1)

Section 3.1 Linear Transformations (A1)

Definition 3.1.1.

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

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

Definition 3.1.2.

Given a linear transformation \(T:V\to W\text{,}\) \(V\) is called the domain of \(T\) and \(W\) is called the co-domain of \(T\text{.}\)

\begin{tikzpicture}[x=0.2in,y=0.2in]
  \begin{scope}[shift={(0,0)}]
    \draw (0,0) -- (3,0);
    \draw (0,0) -- (0,3);
    \draw (0,0) -- (-2,-1);
    \draw[thick,-latex,blue] (0,0) -- (2,1)
          node[anchor=south west] {\(\vec v\)};
    \node[anchor=west] at (-1,-1) {domain \(\IR^3\)};
  \end{scope}
  \draw[dashed,-latex] (3,3) to [bend left=30] (7,3);
  \node[anchor=south] at (5,4) {Linear transformation \(T:\IR^3\to\IR^2\)};
  \begin{scope}[shift={(9,0.5)}]
    \draw (-2,0) -- (2,0);
    \draw (0,-2) -- (0,2);
    \draw[thick,-latex,red] (0,0) -- (-1.5,1)
          node[anchor=south east] {\(T(\vec v)\)};
    \node[anchor=west] at (0,-1) {co-domain \(\IR^2\)};
  \end{scope}
\end{tikzpicture}

Example 3.1.3.

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

\begin{equation*} T\left(\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} x-z \\ 3y \end{array}\right] \end{equation*}

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

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

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

And also...

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

Therefore \(T\) is a linear transformation.

Example 3.1.4.

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

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

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

\begin{equation*} T\left( \left[\begin{array}{c} 0 \\ 1 \end{array}\right] + \left[\begin{array}{c} 2 \\ 3 \end{array}\right] \right) = T\left( \left[\begin{array}{c} 2 \\ 4 \end{array}\right] \right) = \left[\begin{array}{c} 6 \\ 4 \\ 7 \\ 0 \end{array}\right] \end{equation*}

\begin{equation*} T\left( \left[\begin{array}{c} 0 \\ 1 \end{array}\right] \right) + T\left( \left[\begin{array}{c} 2 \\ 3\end{array}\right] \right) = \left[\begin{array}{c} 1 \\ 0 \\ 4 \\ 0 \end{array}\right] + \left[\begin{array}{c} 5 \\ 4 \\ 6 \\ -1 \end{array}\right] = \left[\begin{array}{c} 6 \\ 4 \\ 10 \\ -1 \end{array}\right] \end{equation*}

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

Fact 3.1.5.

A map between Euclidean spaces \(T:\IR^n\to\IR^m\) is linear exactly when every component of the output is a linear combination of the variables of \(\IR^n\text{.}\)

For example, the following map is definitely linear because \(x-z\) and \(3y\) are linear combinations of \(x,y,z\text{:}\)

\begin{equation*} T\left(\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} x-z \\ 3y \end{array}\right] = \left[\begin{array}{c} 1x+0y-1z \\ 0x+3y+0z \end{array}\right] \end{equation*}

But this map is not linear because \(x^2\text{,}\) \(y+3\text{,}\) and \(y-2^x\) are not linear combinations (even though \(x+y\) is):

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

Activity 3.1.1.

Recall the following rules from calculus, where \(D:\P\to\P\) 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?

\(\P\) is not a vector space
\(D\) is a linear map
\(D\) is not a linear map

Activity 3.1.2.

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

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

Fact 3.1.6.

If \(L:V\to W\) is linear, then \(L(\vec z)=L(0\vec v)=0L(\vec v)=\vec z\) where \(\vec z\) is the additive identity of the vector spaces \(V,W\text{.}\)

Put another way, an easy way to prove that a map like \(T(f(x)) = f'(x)+x^3\) can't be linear is because

\begin{equation*} T(0)=\frac{d}{dx}[0]+x^3=0+x^3=x^3\not=0. \end{equation*}

Observation 3.1.7.

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 \(\vec v,\vec w\in V\text{,}\) \(L(\vec v+\vec w)=L(\vec v)+L(\vec w)\text{.}\)
For all \(\vec v\in V\) and \(c\in \IR\text{,}\) \(L(c\vec v)=cL(\vec v)\text{.}\)

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

Activity 3.1.3.

Continue to consider \(S: \P^4 \rightarrow \P^3\) 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?

Activity 3.1.4.

Let the polynomial maps \(S: \P \rightarrow \P\) and \(T: \P \rightarrow \P\) 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 \(T(cf(x))=cT(f(x))\text{.}\)