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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$\begingroup$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.
$\endgroup$ 1 $\begingroup$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 $\;1\;$ .
$\endgroup$ 2 $\begingroup$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 $T(x_1+y_1,x_2+y_2,x_3+y_3)=\alpha (x_1+y_1)+\beta( x_2+y_2)+\gamma (x_3+y_3)=$
$T(x_1,x_2,x_3)+T(y_1,y_2,y_3)$
$\endgroup$ $\begingroup$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."
$\endgroup$ $\begingroup$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".)
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 $X,Y$.
$\endgroup$ $\begingroup$The lazyman's solution: every expression in the given is a linear combination. Linear combinations are well-known to be linear.
$\endgroup$ $\begingroup$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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