Proving a subspace under a linear transformation by the closure of standard addition and scalar multiplication

Question

$T(x,y,z)= (3x-2y, -2x+3y, 5z)$ be a linear transformation from $\mathbb{R}^3$ to $\mathbb{R}^3$ Show that $A= \{(u,v,z) \in \mathbb{R}^3~|~(u,v,w)=T(x,y,z)\}$ for some $(x,y,z)$ in $\mathbb{R}^3$ is a subspace of $\mathbb{R}^3$ by proving that it is closed under standard addition and scalar multiplication.

Answer 1

• Remarks

The whole exercise is equivalent to prove that the image $A:=\operatorname{im}(T)$ of a linear operator $T:V\rightarrow W$ is a linear subspace of $W$.

Following the remark by @LinearAlgebra, I will use

$$A=\{(u,v,w)\in\mathbb R^3~|~(u,v,w)=T(x,y,z)\} $$

for some $(x,y,z)\in\mathbb R^3$.

• On addition

Let $(u_1,v_1,w_1), (u_2,v_2,w_2)\in A$, with $(u_1,v_1,w_1)=T(x_1,y_1,z_1)$ and
$(u_2,v_2,w_2)=T(x_2,y_2,z_2)$. Then

$$(u,v,w):=(u_1,v_1,w_1)+(u_2,v_2,w_2)=T(x_1,y_1,z_1)+T(x_2,y_2,z_2)=\text{linearity of }T= T(x_1+x_2,y_1+y_2,z_1+z_2)\in A. $$

• On scalar multiplication

Let $(u_1,v_1,w_1)\in A$, with with $(u_1,v_1,w_1)=T(x_1,y_1,z_1)$ and $\lambda\in\mathbb R$. Then

$$\lambda(u_1,v_1,w_1)=\lambda T(x_1,y_1,z_1)=\text{linearity of }T=T(\lambda x_1,\lambda y_1,\lambda z_1)\in A.$$