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Find the domain and codomain and determine whether T is linear

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Hi Olga,

You're right that the equations you gave define a function T from R3 -- this is the domain, consisting of ordered triples (x1, x2, x3) -- to R3, the codomain, consisting of ordered triples (w1, w2, w3). (In mathematical shorthand, this is written T:R3 -> R3.) You don't need to show your work so long as you're clear about which variables are the independent or input or domain variables (in this case, the x variables) and which variables are the output or dependent or codomain variables (in this case, the w variables). If you only had two w variables, w1 and w2 for example, but still had three x variables, the domain would still be R3 but the codomain would be R2.

The title of your question also asked to determine whether the function T is linear. There are two ways to do this -- you can either do it informally or totally rigorously. Informally, this function T is linear because the w variables are just simple linear combinations of the x variables. (That means each w variable is just a bunch of multiples of the x variables added up.) An example of a non-linear function would be something like w1=x12 - x2x3. Here, the variable w1 is not just a linear combination of the x variables: this formula involves the square of one of the x variables, as well as their product.

Showing that T is linear totally rigorously, however, requires a little more work, as it always does. If you need to prove that T is linear, let me know and I can write up a more rigorous solution.

Good luck!

Matt L


Thank you very much, Matt! I have already submitted my assignment where I just explained in my own words. Your explanation made me feel better, since there is no need to actually show work. If you don't mind, I would love to learn how to show that T is linear. Thanks! 

Being totally rigorous requires having precise definitions, and I don't know if your course gave precise definitions for what it means to be linear or what a vector space is. Most introductions to multivariable calculus and linear algebra (I'm guessing you're in multivariable calculus) aren't totally rigorous; the "informal" argument I gave in my first answer would be fine.

But the rigorous definition of a linear function between two vector spaces V and W goes as follows. A function T:V -> W is linear if and only if




where v, v1, and v2 are vectors in the vector space V and a is a scalar. In your example, we're considering a function T:R3->R3 defined by

T(x1, x2, x3) = <5x1 - x2 + x3, -x1 + x2 + 7x3, 2x1 -4x-x3 >

So the vector space we're working over (the domain vector space) is R3. A vector in this space has the form

<x1, x2, x3>


So to show that T really is linear, we need to show the two conditions above hold for the equations that define T. Let's start with the first condition: given any two vectors v1=<x1, x2, x3> and v2=<y1, y2, y3>, we need to show that


What does the left-hand side (I'm going to abbreviate this as LHS) of this equation equal? Well, by the definition of vector addition, we have

v1 + v2 = <x1, x2, x3> + <y1, y2, y3> = <x1 + y1, x2 + y2, x3 + y3 >

So the LHS is equal to

T(x1+y1, x2+y2, x3+y3)=<5(x1+y1) -(x2+y2) +(x3+y3), -(x1+y1) + (x2+y2) + 7(x3+y3), 2(x1+y1) -4(x2+y2) -(x3+y3) >

Note that this just amounts to replacing x1 by (x1+y1), x2 by (x2+y2), and x3 by (x3+y3).

Now all we've done is compute the LHS of the above equation. We need to show it's equal to the right-hand side (RHS)

T(v1)+T(v2)=T(x1, x2, x3) + T(y1, y2, y3)

But what does this equal? Well, we know T(x1, x2, x3) is

<5x1 - x2 + x3, -x1 + x2 + 7x3, 2x1 -4x2 -x3 >

and T(y1, y2, y3) is

<5y1 - y2 + y3, -y1 + y2 + 7y3, 2y1 -4y2 - y3 >

So to get the RHS we just add these two vectors, which by the definition of vector addition is just

<5x1 - x2 + x3+5y1 - y2 + y3, -x1 + x2 + 7x3 -y1 + y2 + 7y3, 2x1 -4x2 -x3 2y1 -4y2 - y3>

The entire question is: is the LHS we computed above equal to this RHS? The answer is yes for a very simply reason: just distribute the coefficients from our equation for the LHS above (in bold) and you'll see you get exactly what we got in the line directly above for the RHS. This shows that

T(v1 + v2) = T(v1) + T(v2)

To finish the rigorous proof that T is linear, you'd need to show the second condition as well, namely that


But the strategy for doing this is exactly the same.

What's the point of all this work? The point is that because the w variables were defined just as sums or differences of multiples of the x variables, the distributive property will wind up guaranteeing that T(v1+v2)=T(v1)+T(v2) and T(av)=aT(v). But if the w variables had included something like the square of one of the x variables, these conditions would no longer hold. That's why the informal criterion for being linear is just making sure that the w variables are just simple linear combinations of the x variables and don't have any non-linearities like squares or products or powers.