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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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    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).2011-08-03

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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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    @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.2011-08-03
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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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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: fair enough.2011-08-03