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I always thought linear functions need to satisfy $$f(x+y)=f(x)+f(y).$$ I am a tad confused now, consider $f(x)=2x+3$. $f(1)=5$, $f(2)=7$, $f(1+2)=f(3)=9 \neq f(1)+f(2)$ which was what I thought linear functions should satisfy.

Could someone clarify?

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2 Answers 2

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You're confusing between two different notions.

In calculus, a linear function is a polynomial function of the form $f(x)=ax+b$.

In linear algebra and functional analysis, a linear function is a linear map. (one of the properties that it satisfies is $f(x+y)=f(x)+f(y)$, known as additivity)

The difference between the two is that the latter needs to have $f(0)=0$. Proof: $$f(0)=f(0+0)=f(0)+f(0)=2f(0)\iff f(0)=0.$$ I discuss this in more detail in my (not yet finished) note.

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    $\begingroup$ actually, $f(x) = a x + b$ is called affine in the general, and linear iff $b=0$ $\endgroup$ Feb 7, 2015 at 21:19
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    $\begingroup$ @TemplateRex In most standard calculus textbook, a linear function is defined as a function whose graph is a line. That's why I emphasized the distinction between a linear function in the context of calculus, and in the context of linear algebra and functional analysis. Although I do agree that calling it affine would have been much better. $\endgroup$
    – Workaholic
    Feb 7, 2015 at 21:26
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    $\begingroup$ Not only in calculus. If you look up the definition of a splitting field, it will probably talk about a polynomial splitting "into linear factors". The linear factors are not necessarily of the form $ax$. $\endgroup$
    – bof
    Feb 9, 2015 at 8:20
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$f(x)$ = $2x$ + $3$ isn't a linear function (from and to the set of real numbers). You can easily see that $f(x+y)$ = $2x$ + $2y$ + $3$, and that $f(x)$ + $f(y)$ = $2x$ + $2y$ + $6$. Equality obviously fails.

A linear function (as a mapping from and to the set of real numbers) should be in the from $ax$, where $a$ is a constant real number.

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    $\begingroup$ You will find a large number of textbooks and websites which would describe $f(x)=2x+3$ as a "linear function". Others may not, but that merely means that the term is ambiguous. $\endgroup$
    – Henry
    Feb 8, 2015 at 11:36

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