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Mathematics LibreTexts

Linear Functions

( \newcommand{\kernel}{\mathrm{null}\,}\)

Definition: Linear Function

A function f is linear if f(ax+by)=af(x)+bf(y) for scalars a and b.

Or equivalently f is linear if (1.) f(ax)=af(x) and (2.) f(x+y)=f(x)+f(y)

Theorem 1

If f is linear, then f(0)=0.

 

Proof

Observe that f(0)=f(00)=0f(0)=0

Example 1

Determine if the following functions are linear or not.

1) f:RR,f(x)=2x

2) f:RR,f(x)=2x+3

Solution

The first function is linear as f(ax+by)=2(ax+by)=2ax+2by=af(x)+bf(y).

The second function is not linear as we have the counter example of when x=0 and a=2 observe that f(20)=0+3=3 but 2f(0)=2(0+3)=6 and so f(ax)af(x).

Question

When is a line f(x)=mx+b a linear function?

 

Example 2

Determine if the following function is linear, f:R2R2f((x1,x2))=(2x1,x1+x2).

Solution

Let x=(x1,x2),y=(y1,y2)
ax+by(ax1+by1,ax1,x2)+b(y1,y2)=(ax1,ax2)+(by1,by2)=f(ax1+by1,ax2+by2)=(2(ax1+by1),ax1+by1+ax2+by2)=(2ax1+2by1,ax1+ax2+by1+by2)=(2ax1,ax1+ax2)+(2by1,by1+by2)=a(2x1,x1+x2)+b(2y1,y1+y2)=af((x1,x2))+bf((y1,y2))

Thus the function is linear.

Example 3

Show that D : set of all differential functions set of all functions, D(f)=f is a linear function

Solution

Observe, D(af+bg)=(af+bg)=af+bg=aD(f)+bD(g)

Example 4

Show that Given a,b real numbers, I : set of all integrable functions on [a,b]R, I(f)=baf is a linear function.

Solution

Observe I(sf+tg)=basf+tg=sbaf+tbag= sI(f)+tI(g).

Example 5

The inverse of a linear function is linear (when the inverse exists).

Solution

Suppose f1(x)=c,f1(y)=d.
Then f(c)=x and f(d)=y and
f(ac+bd)=af(c)+bf(d)=ax+by.

Hence f1(ax+by)=ac+bd=af1(x)+bf1(y).

Example 6

Example 6.) D : set of all twice differential functions set of all functions, L(f)=af+bf+cf

Solution

L(sf+tg)=a(sf+tg)+b(sf+tg)+c(sf+tg)=saf+tag+sbf+tbg+scf+tcg=s(af+bf+cf)+t(ag+bg+cg)=sL(f)+tL(g)

Corollary 1

If ψ1,ψ2 are solutions to af+bf+ cf =0, then 3ψ1+5ψ2 is also a solution to af+bf+cf=0.

 

Proof

Since ψ1,ψ2 are solutions to af+bf+cf=0, L(ψ1)=0 and L(ψ2)=0.

Hence
L(3ψ1+5ψ2)=3L(ψ1)+5L(ψ2)=3(0)+5(0)=0

Thus 3ψ1+5ψ2 is also a solution to af+bf+cf=0

 

 

Corollary 2

If ψ1 is a solution to af+bf+cf=h and ψ2 is a solution to af+bf+cf=k, then 3ψ1+5ψ2 is a solution to af+bf+cf=3h+5k.

 

Proof

Since ψ1 is a solution to af+bf+cf=h,L(ψ1)=h.
Since ψ2 is a solution to af+bf+cf=k,L(ψ2)=k.
Hence L(3ψ1+5ψ2)=3L(ψ1)+5L(ψ2)
=3h+5k.

Thus 3ψ1+5ψ2 is also a solution to
af+bf+cf=3h+5k


This page titled Linear Functions is shared under a not declared license and was authored, remixed, and/or curated by Isabel K. Darcy.

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