Check if the applications defined below are linear transformations: Check if the applications defined below are linear transformations:
a) $T: \mathbb R^2 \to \mathbb R^2$, $T(x_1, x_2) = (x_1 – 1, x_2).$
b) $T: \mathbb R^2 \to \mathbb R^2$, $T(x_1, x_2) = (x_2, x_1).$
c) $T: M_{2,2} \to \mathbb R, T (\left [ \begin{matrix}
    a & b \\
    c & d \\
   \end{matrix} \right ]) = a – 2b + c – 2d$.

My attempt:
a) Let $u = (u_1, u_2)$ e $v = (v_1, v_2)$
$T(u+v) = T(u_1+v_1, u_2+v_2) = (u_1 – 1 + v_1 – 1, u_2 + v_2) = (u_1 + v_1 – 2, u_2 + v_2) \ne T(u) + T(v)$
$T(cu) = T(cu_1, cu_2) = (c(u_1 - 1), cu_2) = (cu_1 - c, u_2) \ne cT(u)$.
No.
b)Let $u = (u_1, u_2)$ e $v = (v_1, v_2)$
$T(u+v) = T(u_1+v_1, u_2+v_2) =  (u_2 + v_2 , u_1 + v_1) = T(v) + T(u).$
$T(cu) = T(cu_1, cu_2) = (cu_2, cu_1) = c(u_2, u_1) \ne cT(u).$
No.
c)Let $u = (a, b, c, d)$ e $v = (e, f, g, h)$
$T(u+v) = T([a, b, c, d] + [e, f, g, h]) =  (a+b) – 2(b+f) + (c+g) – 2(d+h) = (a -2b +c -2d, e -2f + c -2h) = T(u) + T(v)$
$T(ku) = T(k[a, b, c, d]) = ka – k2b + kc – k2d = k(a – 2b + c – 2d) = kT(u).$
Yes.
Are my checks correct?
Thanks.
 A: a). Note that $T(0,0)=(-1,0)\neq (0,0)$. Therefore, $T$ fails to be linear map. Because if $T$ was linear then $T(0,0)=T(-1 \cdot (0,0)) = -1 T(0,0)$ and this implies that $2 T(0) =0 \implies T(0) =0$, as the underlying field is $\mathbb{R}$.
b). For $(x_1,x_2),(y_1,y_2)\in \mathbb{R}^2$ and $\alpha , \beta \in \mathbb{R}$, we have $T\big(\alpha (x_1, x_2)+\beta(y_1,y_2)\big) = T(\alpha x_1+\beta y_1, \alpha x_2+\beta y_2) = (\alpha x_2+\beta y_2, \alpha x_1+\beta y_1)=\alpha (x_2,x_1)+\beta(y_2,y_1)=\alpha T(x_1,x_2)+\beta T(y_1,y_2).$
Hence, $T$ is a linear map.
c). For $\alpha , \beta \in \mathbb{R}$, we have $\alpha \begin{pmatrix}
a & b\\
c & d 
\end{pmatrix}
+\beta \begin{pmatrix}
x & y\\
z & w 
\end{pmatrix} =  \begin{pmatrix}
\alpha a + \beta x & \alpha b + \beta y \\
\alpha c + \beta z & \alpha d + \beta w
\end{pmatrix}.\;\;$ So,
$$T \begin{pmatrix}
\alpha a + \beta x & \alpha b + \beta y \\
\alpha c + \beta z & \alpha d + \beta w
\end{pmatrix} = \alpha a + \beta x- 2(\alpha b + \beta y) + \alpha c + \beta z -2 (\alpha d + \beta w)$$ $$= \alpha (a-2b+c-2d)+\beta (x-2y+z-2w)= \alpha T\begin{pmatrix}
a & b\\
c & d 
\end{pmatrix}+\beta T\begin{pmatrix}
x & y\\
z & w 
\end{pmatrix} .$$
Hence, $T$ is a linear map.
A: Partial Answer:
For part (a) you are not applying the rule of function $T$ correctly.
$$(a+)\quad T(u_1+v_1, u_2+v_2) = (u_1   + v_1 – 1, u_2 + v_2) \not = (u_1 + v_1 – 2, u_2 + v_2).$$
$$(a\cdot)\quad\quad\quad\quad\quad T(cu) = T(cu_1, cu_2) = (cu_1 - 1, cu_2) \not= (c(u_1 - 1), cu_2).$$
"No" is however the correct conclusion. To justify it, it is better to simply note $T(0)\not=0$. (It can be shown $T(0)=0$ is a necessary condition for any linear mapping.)
For part (c+), a different sort of mistake appears in your work. You have written (with your typo fixed): $$T([a, b, c, d] + [e, f, g, h]) $$ $$=  (a+e) – 2(b+f) + (c+g) – 2(d+h) $$ $$= (a -2b +c -2d, e -2f + c -2h)$$ as though an output of $T$ is an ordered pair. But this $T$ outputs a real number (takes values in a single dimension).
A: *

*In a) you have $T(u+bv)=(u_{1}+u_{2}-1,b(v_{2}+v_{2}))\not=T(u)+bT(v)$. Hence, $T$ is not linear transformation.

*In b), $T(u+bv)=(b(v_{1}+v_{2}),u_{1}+u_{2})=T(u)+bT(v)$. Hence $T$ is a linear transformation.

*In c) \begin{align*}T(u+\beta v)&=(a_{1}+\beta a_{2})-2(b_{1}+\beta b_{2})+(c_{1}+\beta c_{2})-2(d_{1}+\beta d_{2}),\\&=T(u)+\beta T(v)\end{align*}Hence $T$ is a linear transformation.

Notice that $T$ is a linear transformation iff $T(u+bv)=T(u)+bT(v)$ for all $u,v\in V$ and $\beta\in \mathbb{F}$, with the application $T:V\to W$ and vector spaces $V$ and $W$ defined over a field $\mathbb{F}$.
In general your "checks" they have "good structure", but the problem is that you are not being careful with the arithmetic operations. For example, in a) you wrote "$T(u+v)=(u_{1}-1+u_{2}-1,v_{1}+v_{2})$" but it is not correct because $$T(u+v)=T((u_{1},u_{2})+(v_{1},v_{2}))=T(\underbrace{u_{1}+u_{2}}_{=u},\underbrace{v_{1}+v_{2}}_{=v})=(\underbrace{u_{1}+u_{2}}_{=u}-1,\underbrace{v_{1}+v_{2}}_{=v})$$
Then in 2.b) you said $T(cu)\not=cT(u)$ but you are not  writing all the operations and that is important because the RHS is just $cT(u)=c(u_{2},u_{1})$ that is the LHS so $T(cu)=cT(u)$in fact. The conclusion  of 3) is correct but in first place the transformation is from $M_{2\times 2}$ to $\mathbb{R}$ not to $\mathbb{R}^{3}$ as you write by the notation so just a small remark you should use the notation correct in the problem and for example write $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ instead of $[a,b,c,d]$. If you write all the operations so you should reduce the risk of being wrong.
