Checking to see if something is a Linear Operator

Let $T\colon\mathbb{R}^2\to\mathbb{R}^2$ be given by $T(x,y) = (x + 2y, 3x + y)$.

I'm not sure if I plugged the values in the right place.

Ax. 1: $T(x + y) = T(x) + T(y)$

$[(x+2y) + 2(3x + y) + 3(x + 2y) - (3x - y)] = [(x+2y) + 2(3x + y)] + [3(x + 2y) - (3x - y)]$

Is that right?

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That's not right - what you've computed on the left hand side is T(T(x,y)). You should probably use different variables for your first axiom, because T is a function of either two variables or a vector, but you're treating $x$ and $y$ there like they were real numbers; the $x$ and $y$ of the axiom aren't the same as the $x$ and $y$ of T's defining equation, and I think you may be confusing them. I'd suggest writing the axiom as $T(\vec{a}+\vec{b}) = T(\vec{a})+T(\vec{b})$ and then use $a_x$, $a_y$ and $b_x$, $b_y$ as the two components of $\vec{a}$ and $\vec{b}$. – Steven Stadnicki Apr 15 '11 at 1:56

You can write $T(x,y)$ in matrix form: $$\begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$$ Now it's clear that it's linear.

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We can't use a matrix, we have to use the other method. – user9121 Apr 15 '11 at 1:56

In your "Ax 1", $x$ and $y$ are supposed to elements of $\mathbb{R}^2$. So $x=(x_1,x_2)$, and $y=(y_1,y_2)$. Then $x+y = (x_1+y_1, x_2+y_2)$. That is: $$T(x+y) = T(x_1+y_1,x_2+y_2) = \Bigl( (x_1+y_1)+2(x_2+y_2), 3(x_1+y_1)+(x_2+y_2)\Bigr).$$ Then you should evaluate $T(x) = T(x_1,x_2)$ and $T(y)=T(y_1,y_2)$, add the results, and compare the answers.

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HINT $\$ One easily verifies that these maps are linear: $\rm\ A\:(x,y)\ =\ (x,0)\:;\:$ $\rm\ B\:(x,y)\ =\ (y,0)\:.\:$

Thus so too is $\rm\ (A+2\:B)\:(x,y)\ :=\ A\:(x,y) + 2\:B\:(x,y)\ =\ (x,0) + (2\:y,0)\ =\ (x+2\:y,0)\:.$

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