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Question on Defining Linear Equation

  1. Sep 20, 2005 #1


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    I have been trying to brush up on linear algebra and I fear I still have a weak handle on some of the most basic concepts. For instance I am still trying to find a precise answer to the question, "is this function linear?"

    Most of what I have read online states, if a function has such-and-such a form then it is linear. I am hoping to form a more exact definition like, if a function has properties A and B (and perhaps C) then it is linear (or because it does not, then it is not).

    I have done some research and I think these are the two properties that must hold.

    For a function, f(x) where x can be a vector, to be linear the following properties must hold:
    A: f(u+v)=f(u)+f(v)
    B: f(c*u)=c*f(u) where c is a real scalar

    So is this correct? Am I missing any properties or have I added one where I shouldn't? (On a side note, why a real scalar? Why not allow one to restrict the range?)

    Also, using the above definition seems to cause some problems. For instance I always assumed the following equation was linear:
    But property A does not hold for this function
    y(u+v)=m(u+v)+b != y(u)+y(v) = m(u+v)+2b
    And neither does property B
    y(c*u)=m(c*u)+b != c*y(u)=c*(m*u)+c*b

    One would think a line was linear and isn't y(x)=mx+b just a re-arangement of the standard form Ay+Bx+C=0?

    Something seems amiss here and most likely it is me :)
  2. jcsd
  3. Sep 20, 2005 #2


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    Homework Helper

    I think you're confusing two interpretations of linearity.

    We call a function of the form [tex]y = ax + b[/tex] lineair because it is of the first degree. This way we call a function of the form [tex]y = ax^2 + bx + c[/tex] quadratic etc.

    The lineair you're first referring to is different. We call a map [tex]f:V \to W[/tex] lineair if the two conditions mentioned by yourself are fullfilled. This can be combined in one equivalent condition which we call the lineair combination, that is if:

    [tex]f\left( {\alpha \vec x + \beta \vec y} \right) = \alpha f\left( {\vec x} \right) + \beta f\left( {\vec y} \right)[/tex]
  4. Sep 20, 2005 #3


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    Staff Emeritus
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    "We call a function of the form y= mx+ b linear because it is of the first degree. "

    Actually, we call such a function linear precisely because its graph is a straight line!

    Yes, f(u+v)= f(u)+ f(v) and f(au)= af(u) for scalar a are precisely the conditions for a linear function, though, as TD said, they can be combined into one.

    It is NOT necessary that a be a real unless you are specifically talking about a vector space over the real numbers. In order to have a linear transformation, you must have some vector space to give you the vectors you apply the linear transformation. Any linear transformation must be "over" some number field- the rational numbers, the real numbers, the complex numbers. The "a" in f(au)= af(u) must be one of those.
    Last edited: Sep 20, 2005
  5. Sep 20, 2005 #4


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    Thank you for the replies and thanks Halls for clearing up
    that mess about the 'a'. I have been curious about that one.

    On the second page of the book I am reading equations of
    the form Ax+By+C=0 are labeled as linear and when I
    think of a linear equation this is what comes to mind.

    However when I do some simple linear rearrangement...

    y=-Ax/B-C/B=mx+b (B!=0)
    for notational convenience change to

    which violates our properties of linearity.
    y(u+v)=m(u+v)+b != y(u)+y(v) = mu+b + mv+b = m(u+v)+2b

    Now if any equation is linear it's got to be the above one.
    So where am I going wrong?
    Last edited: Sep 20, 2005
  6. Sep 21, 2005 #5


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    I have been thinking about this all day and I think I have my answer.

    I cannot solve for y in terms of x in Ax+By+C=0 because here both
    x and y are independent variables. In other words they are the inputs.
    They have a relation but solving for one in terms of the other doesn't
    really make sense in the context of the question of linearity.

    I think the correct proof for linearity of Ax+By+C=0 looks like this.

    The function is f(x,y)=Ax+By and it equals -C therefore for linearity

    Assume [tex] (x_0,y_0)[/tex] and [tex](x_1,y_1)[/tex] are points in the vector space.

    Property of addition
    [itex] \begin{align*} f(x_0+x_1,y_0+y_1)=f(x_0,y_0)+f(x_1,y_1) \\ A(x_0+x_1)+B(y_0+y_1)=Ax_0+By_0+Ax_1+By_1 \\ (Ax_0+By_0)+(Ax_1 +By_1)=(Ax_0+By_0)+(Ax_1+By_1) \\ -C-C = -C-C \end{align*} [/itex]
    and Property of scalar multiplication
    [tex] \begin{align*} af(x_0,y_0)=f(ax_0,ay_0) \\ aAx_0+aBy_0=Aax_0+Bay_0 \\ a(Ax_0+By_0)=a(Ax_0+By_0) \\ -aC=-aC \end{align*} [/tex]
    because both properties hold the equation is linear.

    Normally I don't think I would show all the algebra but I wanted to show
    that they expand and simplify in the same way.

    Is this interpretation, and proof, correct? What about my terminology?
    I am new to this stuff so I want to make sure I get it right.
  7. Sep 21, 2005 #6
    There are two conflicting definitions of linear at work here.

    A polyonomial p which satisfies p(x) = ax + b for some constants a, b is (often) called linear (or "a linear polynomial"), probably because its graph resembles a line.

    A function f: V -> W between two vector spaces V, W is called a linear transformation if f(ka) = k * f(a), etc.

    These are not the same definitions, they just happen to have similar names. If b != 0, then p is linear in the first sense, but not in the second. You shouldn't spend too much time thinking about this. If it makes you feel more comfortable, call functions of the form ax + b "affine" instead.
    Last edited: Sep 21, 2005
  8. Sep 21, 2005 #7


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    Thanks Muzza,
    I looked up affine
    and I think that word, and web-page, was exactly what I needed.

    This makes a lot of sense to me and I can see now that p(x)=mx+b is a translated linear function, aka affine.

    I think the problem with the "proof" of post #4 was simply one cannot rewrite y as y(x) because y is not a function of x in the equation Ax+By+C=0.

    Ax+By+C=0 can be parameterized to one variable (hence line, but not linear!). But in this case both x and y depend on the parameter. So even in this case y does not depend on x, and it's the parameter that defines how y and x are related.
  9. Sep 23, 2005 #8


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    Staff Emeritus
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    In terms of R2 or R3 as vector spaces, we can think of the subspaces as lines (or planes) containing the origin. The whole point of "Linear Transformations" is that the preserve those. The graph of a linear transformation on R2 or R3 must be a line (or plane) containing the origin: of the form f(x)= mx of f(x,y)= mx+ ny.

    A line (or plane) that does not contain the origin is sometimes called a "linear manifold".
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