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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.2011-08-03
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    Does this work for R^2 to R^3 or do you need to be moving from the same space?2011-08-03
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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)2011-08-03
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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?2011-08-03
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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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