Linear Functions
- Page ID
- 156601
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)A function \(f\) is linear if \(f(a \mathbf{x}+b \mathbf{y})=a f(\mathbf{x})+b f(\mathbf{y})\) for scalars \(a\) and \(b\).
Or equivalently \(f\) is linear if (1.) \(f(a \mathbf{x})=a f(\mathbf{x})\) and (2.) \(f(\mathbf{x}+\mathbf{y})=f(\mathbf{x})+f(\mathbf{y})\)
If \(f\) is linear, then \(f(\mathbf{0})=\mathbf{0}\).
Observe that \(f(\mathbf{0})=f(0 \cdot \mathbf{0})=0 \cdot f(\mathbf{0})=\mathbf{0}\)
Determine if the following functions are linear or not.
1) \(f: \mathbb{R} \to \mathbb{R}, f(x)=2x\)
2) \(f: \mathbb{R} \to \mathbb{R}, f(x)=2x+3\)
Solution
The first function is linear as \(f(a\mathbf{x}+b\mathbf{y}) = 2(a\mathbf{x}+b\mathbf{y}) = 2a\mathbf{x}+2b\mathbf{y} = af(\mathbf{x})+bf(\mathbf{y})\).
The second function is not linear as we have the counter example of when \(\mathbf{x}=0 \) and \(a=2\) observe that \(f(2 \cdot 0) = 0+3 = 3\) but \(2 \cdot f(0) = 2(0+3)=6\) and so \(f(a\cdot x) \neq a\cdot f(x)\).
When is a line \(f(x)=mx+b\) a linear function?
Determine if the following function is linear, \(f: R^2 \rightarrow R^2 f\left(\left(x_1, x_2\right)\right)=\left(2 x_1, x_1+x_2\right)\).
Solution
Let \(\mathbf{x}=\left(x_1, x_2\right), \mathbf{y}=\left(y_1, y_2\right)\)
\[ \notag
\begin{array}{l}
a \mathbf{x}+b \mathbf{y} \\
\left(a x_1+b y_1, a x_1, x_2\right)+b\left(y_1, y_2\right)=\left(a x_1, a x_2\right)+\left(b y_1, b y_2\right)= \\
\begin{array}{l}
f\left(a x_1+b y_1, a x_2+b y_2\right) \\
=\left(2\left(a x_1+b y_1\right), a x_1+b y_1+a x_2+b y_2\right) \\
=\left(2 a x_1+2 b y_1, a x_1+a x_2+b y_1+b y_2\right) \\
=\left(2 a x_1, a x_1+a x_2\right)+\left(2 b y_1, b y_1+b y_2\right) \\
=a\left(2 x_1, x_1+x_2\right)+b\left(2 y_1, y_1+y_2\right) \\
=a f\left(\left(x_1, x_2\right)\right)+b f\left(\left(y_1, y_2\right)\right)
\end{array}
\end{array}
\]
Thus the function is linear.
Show that \(D\) : set of all differential functions \(\rightarrow\) set of all functions, \(D(f)=f^{\prime}\) is a linear function
Solution
Observe, \(D(a f+b g)=(a f+b g)^{\prime}=a f^{\prime}+b g^{\prime}=a D(f)+b D(g)\)
Show that Given \(a, b\) real numbers, \(I\) : set of all integrable functions on \([\mathrm{a}, \mathrm{b}] \rightarrow R\), \(I(f)=\int_a^b f\) is a linear function.
Solution
Observe \(I(s f+t g)=\int_a^b s f+t g=s \int_a^b f+t \int_a^b g=\) \(s I(f)+t I(g)\).
The inverse of a linear function is linear (when the inverse exists).
Solution
Suppose \(f^{-1}(x)=c, f^{-1}(y)=d\).
Then \(f(c)=x\) and \(f(d)=y\) and
\[
f(a c+b d)=a f(c)+b f(d)=a x+b y .
\]
Hence \(f^{-1}(a x+b y)=a c+b d=a f^{-1}(x)+b f^{-1}(y)\).
Example 6.) \(D\) : set of all twice differential functions \(\rightarrow\) set of all functions, \(L(f)=a f^{\prime \prime}+b f^{\prime}+c f\)
Solution
\[ \notag
\begin{aligned}
L(s f+t g) & =a(s f+t g)^{\prime \prime}+b(s f+t g)^{\prime}+c(s f+t g) \\
& =s a f^{\prime \prime}+t a g^{\prime \prime}+s b f^{\prime}+t b g^{\prime}+s c f+t c g \\
& =s\left(a f^{\prime \prime}+b f^{\prime}+c f\right)+t\left(a g^{\prime \prime}+b g^{\prime}+c g\right) \\
& =s L(f)+t L(g)
\end{aligned}
\]
If \(\psi_1, \psi_2\) are solutions to \(a f^{\prime \prime}+b f^{\prime}+\) cf \(=0\), then \(3 \psi_1+5 \psi_2\) is also a solution to \(a f^{\prime \prime}+b f^{\prime}+c f=0\).
Since \(\psi_1, \psi_2\) are solutions to \(a f^{\prime \prime}+b f^{\prime}+c f=0\), \(L\left(\psi_1\right)=0\) and \(L\left(\psi_2\right)=0\).
Hence
\[\notag
\begin{aligned}
L\left(3 \psi_1+5 \psi_2\right) & =3 L\left(\psi_1\right)+5 L\left(\psi_2\right) \\
& =3(0)+5(0)=0
\end{aligned}
\]
Thus \(3 \psi_1+5 \psi_2\) is also a solution to \(a f^{\prime \prime}+b f^{\prime}+c f=0\)
If \(\psi_1\) is a solution to \(a f^{\prime \prime}+b f^{\prime}+c f=h\) and \(\psi_2\) is a solution to \(a f^{\prime \prime}+b f^{\prime}+c f=k\), then \(3 \psi_1+5 \psi_2\) is a solution to \(a f^{\prime \prime}+b f^{\prime}+c f=3 h+5 k\).
Since \(\psi_1\) is a solution to \(a f^{\prime \prime}+b f^{\prime}+c f=h, L\left(\psi_1\right)=h\).
Since \(\psi_2\) is a solution to \(a f^{\prime \prime}+b f^{\prime}+c f=k, L\left(\psi_2\right)=k\).
Hence \(L\left(3 \psi_1+5 \psi_2\right)=3 L\left(\psi_1\right)+5 L\left(\psi_2\right)\)
\[ \notag
=3 h+5 k .
\]
Thus \(3 \psi_1+5 \psi_2\) is also a solution to
\[ \notag
a f^{\prime \prime}+b f^{\prime}+c f=3 h+5 k
\]