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I am new to linear algebra so I apologize beforehand for those of you who are math wizards. I need to know and understand why or why not for the following question:

T or F; $T (x,y) = (2x+5y,-x+2)$ is a linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$.

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Always my first sanity check: a linear map $A$ from $\mathbb{R}^2$ to $\mathbb{R}^2$ satisfies $A(0,0) = (0,0)$. That's because $A(0,0) = A (0 \cdot (0,0)) = 0 \cdot A(0,0) = (0,0)$. Sloppily but easier to memorize: $A(0) = 0$ (note that this ties in well with Jonas's hint "try using lots of zeros, or maybe some ones"). Now is $T(0,0) = (0,0)$? In most cases you see at a glance whether that's true or not. If not, you know that the map is not linear. –  t.b. Aug 3 '11 at 0:19
    
Does this work for R^2 to R^3 or do you need to be moving from the same space? –  Heather Aug 3 '11 at 0:32
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Oh, this always works since $A( \mathbf{0}) = A ( 0 \cdot \mathbf{0}) = 0 \cdot A(\mathbf{0}) = \mathbf{0}$ where the bold zero denotes the null vector of the appropriate size. (you can always "pull" the zero scalar out of a linear map) –  t.b. Aug 3 '11 at 0:38
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So just for clarification on the original problem because the 2 in the second part(?) doesn't have a y with it, that makes this not a linear transformation? But if there had been a y then it would have been a linear transformation? –  Heather Aug 3 '11 at 0:41
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Well, if you mean if there were $T'(x,y) = (2x+5y,-x+2y)$ then yes, this is linear. However, my first comment is inconclusive: you can't conclude from $T'(0,0) = (0,0)$ that $T'$ is linear: For instance $T''(2x+5y,-x+2y^2)$ satisfies $T''(0,0) = (0,0)$ (but it isn't linear). However, many of these examples break down already at $T(0,0) \neq (0,0)$ and you don't have to check further (and as I said, it costs no time at all to check this). –  t.b. Aug 3 '11 at 0:46
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4 Answers

We can use the definition of linear transformation (between vectors spaces like $\mathbb{R}^n$, though this can be generalized...):

$T$ is a linear transformation if for all vectors $\mathbf{u}$ and $\mathbf{v}$, $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$, and for all vectors $\mathbf{u}$ and real numbers $k$, we have $T(k\mathbf{u}) = kT(\mathbf{u})$.

Try to see if your $T$ satisfies this. Note, since the domain is $\mathbf{R}^2$, you should take $\mathbf{u} = (u_1, u_2)$, and similar for $\mathbf{v}$.

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Wizardry is not really required here. Rather, what you need to know is the definition of a linear transformation: there are two properties to check.

Do you know what these properties are? If not, go back and take a look in your text. (Or see Shaun Alt's answer.)

If you know what the properties are, did you try to check them? If so, where did you get stuck?

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If I say that u=(a,b) and v=(c,d) and do the addition property of T(u+v) to see if I get T(u) + T(v) I get lost at the second part. I have T(u+v) = T(a+c,b+d), where do I go from there? –  Heather Aug 3 '11 at 0:13
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$a+c$ is your $x$-coordinate, and $b+d$ is your $y$-coordinate, so $T(a+c,b+d) = (2(a+c) + 5(b+d),-(a+c) + 2)$. –  Alex Aug 3 '11 at 2:13
    
@Heather: Alex's comment does answer your question. In words though, you need to make the substitution $x = a+c$, $y = b+d$ in the definition of $T$, i.e., wherever in the expression defining $T$ you see an $x$, write $a+c$ instead; wherever you see a $y$, write $b+d$ instead. Then you want to do something similar (but easier) with $T(a,b) + T(c,d)$ and see if the two expressions are equal. –  Pete L. Clark Aug 3 '11 at 2:21
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Compute

$$ T((x,y) + (x',y')) = T(x+x', y+y') $$

and compare with

$$ T(x,y) + T(x',y') \ , $$

for any real numbers $x,y,x',y'$.

Then do the same with

$$ T(\lambda (x,y)) = T(\lambda x , \lambda y) $$

and

$$ \lambda T(x,y) \ , $$

for $\lambda $ any real number.

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To show that something is a linear transformation, you have to work in generality to show that it satisfies the conditions in the definition of linear. That is, you would have to check that for all $(x_1,y_1)$ and $(x_2,y_2)$ in $\mathbb R^2$ and for all $\lambda\in R$, $$T((x_1,y_1)+(x_2,y_2))=T(x_1,y_1)+T(x_2,y_2)\text{, and }$$

$$T(\lambda(x_1,x_2))=\lambda T(x_1,x_2).$$

On the other hand, to show that $T$ is not a linear transformation, you only need to show that there is one instance of a failure of a condition in the definition of linear, called a counterexample. It is a good idea to try to make counterexamples as easy as possible, and with a numerical problem like this, that means you might want to try using lots of zeros, or maybe some ones.

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$@$Jonas: in some sense I feel like the idea of searching for counterexamples is an unnecessary can of worms for this particular problem. Really we have some linear expressions and want to know whether they are equal or not. Well, if you can do the algebra, either they are or they aren't -- no searching for counterexamples is really required. (Of course I am trying to make things as simple as possible for the OP, who has led by admitting her inexperience and perplexity.) –  Pete L. Clark Aug 3 '11 at 2:26
    
Pete: Perhaps, and that is why it is good that we have these complementary answers. My hope was that (1) contrasting the ideas of proving an identity vs. finding a counterexample would be helpful for someone who is inexperienced (as similar lessons were very helpful for me to learn in my first linear algebra class) and (2) the end of the second paragraph would encourage experimentation of the simplest kind, as in Theo's first comment on the question. (I was initially tempted to make a remark similar to Theo's.) –  Jonas Meyer Aug 3 '11 at 2:53
    
$@$Jonas: fair enough. –  Pete L. Clark Aug 3 '11 at 4:07
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