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I'm reading Hoffman and Kunze's linear algebra book and on page 73 in the exercise 7, they ask to verify this function

$$T(x_1,x_2,x_3)=(x_1-x_2+2x_3,2x_1+x_2,-x_1-2x_2+2x_3)$$ is a linear transformation.

This exercise is really simple, but a little bit tedious. We have to define arbitrary $u=(x_u,y_u,z_u)$ and $v=(x_v,y_v,z_v)$ elements of $F^3$ and show $T(u+v)=T(u)+T(v)$ and $T(ku)=kT(u)$ for $k\in F$. (we can see $F$ as $\mathbb R$ or $\mathbb C$)

Is there a way more elegant to prove this function is linear?

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    $\begingroup$ I don't think there is. $\endgroup$ Nov 20, 2016 at 7:32
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    $\begingroup$ It's not all that tedious, just a routine calculation. It shouldn't take all that long. $\endgroup$
    – Aweygan
    Nov 20, 2016 at 7:36
  • $\begingroup$ This ain't tedious. $\endgroup$
    – copper.hat
    Nov 20, 2016 at 8:40
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    $\begingroup$ Don't bother finding a slick way to solve this type of problems. Once you have done one or two of them you're supposed to be able to recognize immediately that "of course this is a linear transformation", without even pausing to construct a proof -- and then you'll never be expected to prove such a thing again, unless you go into teaching (or answering-questions-on-MSE). $\endgroup$ Nov 20, 2016 at 9:51
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    $\begingroup$ @HenningMakholm I know that, I'm looking for shorter solutions to show to my students. $\endgroup$
    – user42912
    Nov 20, 2016 at 10:08

8 Answers 8

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You can observe that the function is composed out of smaller linear functions. You know that $f(x) = x$ is linear and so is $f(x,y,z)=x$. Also that the sum of linear functions $f+g$ is linear, as well as a scalar multiplier $2f$. Finally, for linear $f,g,h$, the function $x \mapsto (f(x), g(x), h(x))$ is also linear.

When you know all of the above, you can just say that $F$ is made out of these linearity preserving compositions and a few obviously linear functions. That should be a good enough proof.

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    $\begingroup$ +1, nice! Alternatively add the last missing component that is function composition and strike out "good enough". $\endgroup$
    – The Vee
    Nov 20, 2016 at 10:39
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Perhaps the easiest way is to check the function acts as a matrix, and in this case (and in many more) we can do that easily:

$$T(x_1,x_2,x_3)=(x_1-x_2+2x_3,2x_1+x_2,-x_1-2x_2+2x_3)=\begin{pmatrix}1&-1&2\\2&1&0\\-1&-2&2\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$$

Observe that I simply took each coordinate in the image and "split" it in its components $\;x_1,x_2,x_3\;$ .

You can also try to prove the little more involved:

Lemma: let $\;T:\Bbb F^n\to\Bbb F^m\;,\;\;\;\Bbb F\;$ a field, be a function given componentwise:

$$T\begin{pmatrix}x_1\\x_2\\\ldots\\x_n\end{pmatrix}=\begin{pmatrix}T_1(x_1,...,x_n)\\T_2(x_1,...,x_n)\\\ldots\\T(x_1,...,x_n)\end{pmatrix}\;,\;\;\;\text{with}\;\;T_i:\Bbb F^n\to\Bbb F$$. Then $\;T\;$ is a linear map iff for all $\;i=1,...,n\;$ , the map $\;T_i\;$ is a homogeneous polynomial in $\;x_1,...,x_n\;$ of degree $\;1\;$ .

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  • $\begingroup$ Thank you for your answer. Being very rigorous, $T(x_1,x_2,x_3)$ should be an element of $F^3$ not an element of the space of the columns $3\times 1$. $\endgroup$
    – user42912
    Nov 20, 2016 at 8:09
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    $\begingroup$ @user42912 Being rigorous it is exactly (i.e., natural isomorphism) the same in these cases, and it really doesn't matter intrinsically whether you denote the vectors of $\;\Bbb F^{n}\;$ as column or row vectors. It is a matter of convention in any book. $\endgroup$
    – DonAntonio
    Nov 20, 2016 at 8:16
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You can write the above as:

$T(x_1,x_2,x_3)=x_1(1,2,-1)+x_2(-1,1,-2)+x_3(2,0,2)=\alpha x_1+\beta x_2+\gamma x_3$

where $\alpha,\beta,\gamma$ are fixed.

then $T(x_1+y_1,x_2+y_2,x_3+y_3)=\alpha (x_1+y_1)+\beta( x_2+y_2)+\gamma (x_3+y_3)=$

$T(x_1,x_2,x_3)+T(y_1,y_2,y_3)$

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I would say that, once you have been introduced to matrices, a (pharisianist ?) answer is

"The definition of $T$ : $(x_1,x_2,x_3)\mapsto(y_1,y_2,y_3)$ is equivalent to matricial equation:

$$\pmatrix{y_1\\y_2\\y_3}=\left(\begin{array}{rrr}1&-1&2\\2&1&0\\-1&-2&2\end{array}\right)\pmatrix{x_1\\x_2\\x_3}$$

which is known to be a linear operator."

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Yes. Show that all 2nd order derivatives are zero, and that the origin is a solution.
(The latter is only necessary if you're referring to the linear-algebra sense of "linear".)

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  • $\begingroup$ Good point as now I understand ( I think...) what you meant by "the origin is a solution": you meant that $\;T(0,0,0)=0\;$ , right? That wasn't clear (to me). Anyway $\endgroup$
    – DonAntonio
    Nov 21, 2016 at 8:41
  • $\begingroup$ Yes, of course. I shall delete my first comment as this explains. Thanks $\endgroup$
    – DonAntonio
    Nov 21, 2016 at 8:45
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It's only a slight improvement, but it is equivalent to show that $$T(aX + Y) = aT(X) + T(Y)$$ for all scalars $a$ and vectors $X,Y$.

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The lazyman's solution: every expression in the given is a linear combination. Linear combinations are well-known to be linear.

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For any function, if $f(ax + by) = af(x) + bf(y)$, then the function is linear.

It's easy to show (just write it out) that your T satisfies this.

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    $\begingroup$ What are your $\;a,b,x\;$ there? It looks like $\;a,b\;$ are scalars and $\;x\;$ is a vector, but then $\;a+bx\;$ makes no sense... $\endgroup$
    – DonAntonio
    Nov 20, 2016 at 17:39
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    $\begingroup$ I think you mean $ax + by$. $\endgroup$
    – asmeurer
    Nov 20, 2016 at 18:33

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