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I have an operator

$$h: \mathbb R^4 \to \mathbb R^3\text{ given by } h(x, y, u, v) = (2x + 3y - u + 2v, x - 5y + 6v, 2y + u + v)$$

What is the easiest way to proove that this operator is linear?

I looked over on wiki etc., but I didn't really find the way to prove it mathematically.

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It is clearly linear, it follows directly from the definitions. – Listing May 3 '12 at 23:55
up vote 6 down vote accepted

You simply verify that the conditions of linearity are satisfied: for every $a\in\Bbb R$ and $\vec u,\vec v\in\Bbb R^4$, $h(\vec u+\vec v)=h(\vec u)+h(\vec v)$, and $h(a\vec u)=ah(\vec u)$. I'll do the first and leave the second for you.

Let $\vec u=\langle u_1,u_2,u_3,u_4\rangle$ and $\vec v=\langle v_1,v_2,v_3,v_4\rangle$. Then $$\begin{align*} h(\vec u+\vec y)&=h\Big(\langle u_1,u_2,u_3,u_4\rangle+\langle v_1,v_2,v_3,v_4\rangle\Big)\\ &=h\Big(\langle u_1+v_1,u_2+v_2,u_3+v_3,u_4+v_4\rangle\Big)\\ &=\Big\langle 2(u_1+v_q)+3(u_2+v_2)-(u_3+v_3)+2(u_4+v_4),\\ &\qquad\qquad(u_1+v_1)-5(u_2+v_2)+6(u_4+v_4),\\ &\qquad\qquad2(u_2+v_2)+(u_3+v_3)+(u_4+v_4)\Big\rangle\\ &=\Big\langle 2u_1+3u_2-u_3+2u_4,u_1-5u_2+6u_4,2u_2+u_3+u_4\Big\rangle\\ &\qquad\qquad+\Big\langle 2v_1+3v_2-v_3+2v_4,v_1-5v_2+6v_4,2v_2+v_3+v_4\Big\rangle\\ &=h\Big(\langle u_1,u_2,u_3,u_4\rangle\Big)+h\Big(\langle v_1,v_2,v_3,v_4\rangle\Big)\\ &=h(\vec u)+h(\vec v)\;. \end{align*}$$

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It is multiplication by the matrix, $$\pmatrix{2&3&-1&2\cr1&-5&0&6\cr0&2&1&1\cr}$$ Do you know how to prove that multiplication by a matrix is a linear transformation?

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Show that $h(x,y,u,v)+h(x',y',u',v')=h(x+x',y+y',u+u',v+v')$ and that scalars come out.

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That's only half the story - scalar multiplication counts, too. – Gerry Myerson May 3 '12 at 23:55
In his defense, he did say "and that scalars come out" :) @Gerry – Ravi May 4 '12 at 0:10
In Gerry's defense I corrected it moments after he asked it. But in my defense Gerry responded too quickly, I realized my omission within seconds of posting it. :) – user21725 May 4 '12 at 0:11

For two vectors $(x_1,y_1,u_1,v_1)$ and $h(x_1,y_1,u_1,v_1)$,

$h(x_1,y_1,u_1,v_1)+h(x_1,y_1,u_1,v_1)$ $=(2x_1 + 3y_1 - u_1 + 2v_1, x_1 - 5y_1 + 6v_1, 2y_1 + u_1 + v_1)+(2x_2 + 3y_2 - u_2 + 2v_2, x_2 - 5y_2 + 6v_2, 2y_2 + u_2 + v_2)$

and add the two vectors component wise, and you get $h(x_1+x_2,y_1+y_2,u_1+u_2,v_1+v_2)$.

Similarly, for a real number $a$, if you compute $a(h(x_1,y_1,u_1,v_1))$, and multiply the scalar and the vector, you will end up with the formula for $h(ax_1,ay_1,au_1,av_1)$.

These are the two things you need to check for a linear transformation.

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