I have an algebra final exam on tuesday, and I'm having some trouble with this kind of exercise. I tried looking for an explanation in all over the web, but couldn't find anything of this type. Of course, I also searched a lot here, but really didn't find what I need.
Ok, so: I know that, for a function to be a linear transformation, it needs to verify two properties:
1: T (u+v) = T (u) + T (v)
2: c.T (u) = T (c.u)
This is what I will need to solve in the exam, I mean, this kind of exercise:
T: R3 -> R3 / T (x; y; z) = (x+z; -2x+y+z; -3y)
The thing is, that I can't seem to find a way to verify the first property. I'm writing nonsense things or trying to do things without actually knowing what I am doing, or why I am doing it. I first tried to make two generic vectors, then tried to apply the function on it... but it looks horrible. Fun fact: I'm not having big troubles with solving R2 to R2 exercises, but I can't seem to find a way to solve this one.
How should I start to verify the first property in this kind of exercise? Note that in this case we shouldn't use matrix or anything, it should be just pure math.
Thanks in advance.
Edit 1 (12/15/19): Corrected spelling in the property no. 2 (added c instead of x), thanks for the suggestion.
Edit 2 (17/12/19): I passed the exam. I was wrong on some of the points, but was finally successfull in the linear transformation one. I didn't had to prove it, however. Just calculate the image and the nulity. Sorry for my bad english btw. Thanks you all for your help.
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$\begingroup$Let $u=(x_1,y_1,z_1)$ and $v=(x_2,y_2,z_2)$.
Can you then show $T(u+v)=T(u)+T(v)$?
$\endgroup$ 1 $\begingroup$Your vectors are in 3 dimension. When you are trying to verify $T(u + v) = T(u) + T(v)$ you just substitute $u = (u_{1}, u_{2}, u_{3})$ and $v = (v_{1}, v_{2}, v_{3})$ into the given transformation, in this case $T(w) = T(x, y, z) = (x + z, -2x + y + z, -3y)$, where $w = (x, y, z)$ in this notation, where we could have easily used $w = (w_{1}, w_{2}, w_{3})$ as a notation, but I presume your course uses as axis label $x, y, z$. Then, we have
$$\begin{align} T(u + v) &= T((u_{1}, u_{2}, u_{3}) + (v_{1}, v_{2}, v_{3})) \quad \text{by substitution of vectors} \\ &= T((u_{1} + v_{1}, u_{2} + v_{2}, u_{3} + v_{3})) \quad \text{by defintion of coordinate-wise vector addition} \\ & = (u_{1} + v_{1} + u_{3} + v_{3}, -2(u_{1} + v_{1}) + u_{2} + v_{2} + u_{3} + v_{3}, -3(u_{2} + v_{2})) \quad \text{(1)} \\ & = (u_{1} + v_{1} + u_{3} + v_{3}, -2u_{1} - 2v_{1} + u_{2} + v_{2} + u_{3} + v_{3}, -3 u_{2} - 3v_{2}) \quad \text{(2)} \\ & = (u_{1} + u_{3}, -2u_{1} + u_{2} + u_{3}, -3 u_{2}) + (v_{1} + v_{3}, - 2v_{1} + v_{2} + v_{3}, - 3v_{2}) \quad \text{(3)} \\ & = T(u_{1}, u_{2}, u_{3}) + T(u_{1}, u_{2}, u_{3}) \quad (4) \\ & = T(u) + T(v). \end{align}$$
(1) by applying the transformation (2) some vector algebra (3) recognizing what you wan to show and bringing into desired form (4) recognizing this is the transformation applied to $u$ and $v$
$\endgroup$ 2 $\begingroup$One way to do these types of problems is to show that T is equivalent to left multiplication by a matrix. Here we have
\begin{align} T(x, y, z) &= (x + z, -2x + y + z, -3y) \\ &= x(1, -2, 0) + y(0, 1, -3) + z(1, 1, 0) \\ &= \begin{bmatrix} 1 & 0 & 1 \\ -2 & 1 & 1 \\ 0 & -3 & 0 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} \\ \end{align}Because T is equivalent to left multiplication by a matrix on a vector, T must be a linear map.
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