How to determine which of the following transformations are linear transformations? 
Determine which of the following transformations are linear transformations
A. The transformation $T_1$ defined by $T_1(x_1,x_2,x_3)=(x_1,0,x_3)$
B. The transformation $T_2$ defined by $T_2(x_1,x_2)=(2x_1−3x_2,x_1+4,5x_2)$.
C. The transformation $T_3$ defined by $T_3(x_1,x_2,x3)=(x_1,x_2,−x_3)$
D. The transformation $T_4$ defined by $T_4(x_1,x_2,x3)=(1,x_2,x_3)$
E. The transformation $T_5$ defined by $T_5(x_1,x_2)=(4x_1−2x_2,3|x_2|)$.

I believe that it could be A and E. How can I determine this? If someone could show me one I could figure out the rest.
 A: For $T$ to be a linear transformation, two criteria need to be satisfied:


*

*$T(\mathbf{x}+\mathbf{y}) = T(\mathbf{x})+T(\mathbf{y})$

*$T(a\mathbf{x}) = aT(\mathbf{x})$ for $a$ a scalar/constant.


As an example, suppose $\mathbf{x} = (x_1, x_2, x_3)$ and $\mathbf{y} = (y_1, y_2, y_3)$. Let's start with A.
Then, $T(\mathbf{x}) = (x_1, 0, x_3)$ and $T(\mathbf{y}) = (y_1, 0, y_3)$. Now $$\mathbf{x}+\mathbf{y} = (x_1 + y_1, x_2 + y_2, x_3 + y_3)\text{.}$$ It follows that $T(\mathbf{x}+\mathbf{y}) = (x_1 + y_1, 0, x_3+y_3)$.
Is $T(\mathbf{x}+\mathbf{y}) = T(\mathbf{x})+T(\mathbf{y})$?
Now, for a constant $a$, $a\mathbf{x} = (ax_1, ax_2, ax_3)$.
We have $T(a\mathbf{x}) = (ax_1, 0, ax_3)$.
Furthermore, we know $T(\mathbf{x}) = (x_1, 0, x_3)$, so $aT(\mathbf{x}) = (ax_1, 0, ax_3)$.
Is $T(a\mathbf{x}) = aT(\mathbf{x})$?
Repeat this for all of the other $T$.
A: A linear transformation has this defintion.
$T(\mathbf x+\mathbf y) = T(\mathbf x) + T(\mathbf y)\\
T(c\mathbf x) = cT(\mathbf x)$
Show that this is (or isn't) true for each of the above.
A: To test whether T is a linear transformation, you need to check that for some vectors $a$ and $b$ and some constant $c$$$T(a + b) = T(a) +T(b)$$ 
$$T(ca) = cT(a)$$ 
$$T(0) = 0$$
So for example,
A. $T(x_1,x_2,x_3)=(x_1,0,x_3)$
$$T(x_1+y_1,x_2+y_2,x_3+y_3)=(x_1+y_1,0(x_2+y_2),x_3+y_3)=T(x_1,0,x_3)+T(y_1,0,y_2)$$$$T(cx_1,cx_2,cx_3)=T(cx_1,(c)0,cx_3)=cT(x_1,0,x_3)$$
$$T(0,0,0)=0$$
B. $T(x_1,x_2)=(2x_1−3x_2,x_1+4,5x_2)$
$$T(x_1+y_1,x_2+y_2)=(2(x_1+y_1)−3(x_2+y_2),(x_1+y_1)+4,5(x_2+y_2))=(2x_1+2y_1−3x_2-3y_2,x_1+y_1+4,5x_2+5y_2)$$$$T(x_1,x_2)+T(y_1,y_2)=(2x_1−3x_2,x_1+4,5x_2)+(2y_1−3y_2,y_1+4,5y_2)=(2x_1−3x_2+2y_1-3y_2,x_1+y_1+8,5x_2+5y_2)\not=T(x_1+y_1,x_2+y_2)$$
So B is not a linear transformation.
A: A linear transformation is defined by a set of homogeneous linear (i.e. all terms have degree $1$) polynomials, namely of the  form
$$T(x_1,x_2,x_3)=ax_1+bx_2+cx_3\quad(a,b,c\in\mathbf R).$$
A and C alone satisfy this criterion.
A: $A$. $$T(a\vec{x})=T(ax_1, ax_2, ax_3) = (ax_1, 0, ax_3) = a(x_1, 0, x_3) = aT(\vec{x})$$
$$T(\vec{x} +\vec{y})=T(x_1+y_1, x_2+y_2, x_3+y_3)\\=(x_1+y_1,0,x_3+y_3)=(x_1,0,x_3)+(y_1,0,y_3)=T(\vec{x})+T(\vec{y})$$
Therefore the first transformation is linear as you correctly guessed. Just repeat the same procedure for B-E and see if it works or not.
A: Is T in A a linear transformation?


*

*Check linearity for addition. 


Suppose $T:V \rightarrow W$ .Where $V$ and $W$ are vector spaces over $F$. Let $x_1,x_2,x_3 \in F$ and also let $x_4,x_5,x_6 \in F$. So that $(x_1,x_2,x_3) \in V$ and $(x_4,x_5,x_6) \in V$. Now need to check that $T((x_1,x_2,x_3)) + T((x_4,x_5,x_6)) = T((x_1+x_4, x_2+x_5,x_3+x_6))$. 
We have LHS $ = (x_1,0,x_3) + (x_4,0,x_6) = (x_1+x_4,0,x_3+x_6) = $RHS. Hence this holds by definition of vector addition.


*Check linearity for scalar multiplication:


Let be as above, and suppose $V$ is a vector-space over a field $F$. Then let $a\in F$. Want to prove that:
$aT(x_1,x_2,x_3)=T(ax_1,ax_2,ax_3)$. Hence we have:
LHS $ = a(x_1,0,x_3) = (ax_1,0,ax_3) = $RHS. And this holds by vector scalar multiplication and by property of zero in $\mathbb{R}$.
Hence this is a linear transformation by definition. In general you need to show that these two properties hold.
