w1=5x1-x2+x3

w2= -x1+x2+7x3

w3= 2x1-4x2-x3

My solution so far: From the system I can tell that it is R3, so domain is R3, since there w1, w2, w3 I can tell that codomain is R3. Is there a way to show work? Thank you!

w1=5x1-x2+x3

w2= -x1+x2+7x3

w3= 2x1-4x2-x3

My solution so far: From the system I can tell that it is R3, so domain is R3, since there w1, w2, w3 I can tell that codomain is R3. Is there a way to show work? Thank you!

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Hi Olga,

You're right that the equations you gave define a function T from **R ^{3} **-- this is the domain, consisting of ordered triples (x

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 w_{1}=x_{1}^{2} - x_{2}x_{3}. Here, the variable w_{1} 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

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## Comments

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

linearor what avectorspaceis. 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

T(v

_{1}+v_{2})=T(v_{1})+T(v_{2})and

T(av)=aT(v)

where v, v

_{1}, and v_{2}are vectors in the vector space V andais a scalar. In your example, we're considering a function T:R^{3}->R^{3 }defined byT(x

_{1}, x_{2}, x_{3}) = <5x_{1}- x_{2}+ x_{3}, -x_{1}+ x_{2}+ 7x_{3}, 2x_{1}-4x_{2 }-x_{3}>So the vector space we're working over (the domain vector space) is

R^{3}. A vector in this space has the form<x

_{1}, x_{2}, x_{3}>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

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

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

T(av)=aT(v)

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.