It's enought to show that the below property got satisfied:

\(\displaystyle{\left({i}\right)}{L}{\left({x}\ +\ {y}\right)}={L}{\left({x}\right)}\ +\ {L}{\left({y}\right)}\)

\(\displaystyle{\left({i}{i}\right)}{L}{\left(\alpha{x}\right)}=\alpha{L}{\left({x}\right)}\)

Now Consider,

\(\displaystyle{L}{\left({x}\ +\ {y}\right)}={A}{\left({X}\ ++{Y}\right)}{\left(\text{use given definition}\ {L}{\left({x}\right)}={A}{X}\right)}\)

\(\displaystyle={A}{X}\ +\ {A}{Y}\)

\(\displaystyle={L}{\left({x}\right)}\ +\ {L}{\left({y}\right)}{\left(\text{By definition we get}\right)}\)

\(\displaystyle\text{Therefore},{L}{\left({x}\ +\ {y}\right)}={L}{\left({x}\right)}\ +\ {L}{\left({y}\right)}\)

Property (i) is satisfied.

Now consider,

\(\displaystyle{L}{\left({a}{x}\right)}={A}{\left({a}{X}\right)}{\left(\text{use given definition}\ {L}{\left({x}\right)}={A}{X}\right)}\)

\(\displaystyle={\left({A}\alpha\right)}{X}{\left(\text{Use associative property}\right)}\)

\(\displaystyle={\left(\alpha{A}\right)}{X}{\left(\text{Use Commutative property}\right)}\)

\(\displaystyle=\alpha{\left({A}{X}\right)}{\left(\text{Use associative property}\right)}\)

\(\displaystyle=\alpha{L}{\left({x}\right)}{\left(\text{By definition we get}\right)}\)

\(\displaystyle\text{Therefore},{L}{\left(\alpha{x}\right)}=\alpha{L}{\left({x}\right)}\)

Property (ii) is satisfied.

Therefore both properties got satisfied.

Therefore L is a linear transformation.

\(\displaystyle{\left({i}\right)}{L}{\left({x}\ +\ {y}\right)}={L}{\left({x}\right)}\ +\ {L}{\left({y}\right)}\)

\(\displaystyle{\left({i}{i}\right)}{L}{\left(\alpha{x}\right)}=\alpha{L}{\left({x}\right)}\)

Now Consider,

\(\displaystyle{L}{\left({x}\ +\ {y}\right)}={A}{\left({X}\ ++{Y}\right)}{\left(\text{use given definition}\ {L}{\left({x}\right)}={A}{X}\right)}\)

\(\displaystyle={A}{X}\ +\ {A}{Y}\)

\(\displaystyle={L}{\left({x}\right)}\ +\ {L}{\left({y}\right)}{\left(\text{By definition we get}\right)}\)

\(\displaystyle\text{Therefore},{L}{\left({x}\ +\ {y}\right)}={L}{\left({x}\right)}\ +\ {L}{\left({y}\right)}\)

Property (i) is satisfied.

Now consider,

\(\displaystyle{L}{\left({a}{x}\right)}={A}{\left({a}{X}\right)}{\left(\text{use given definition}\ {L}{\left({x}\right)}={A}{X}\right)}\)

\(\displaystyle={\left({A}\alpha\right)}{X}{\left(\text{Use associative property}\right)}\)

\(\displaystyle={\left(\alpha{A}\right)}{X}{\left(\text{Use Commutative property}\right)}\)

\(\displaystyle=\alpha{\left({A}{X}\right)}{\left(\text{Use associative property}\right)}\)

\(\displaystyle=\alpha{L}{\left({x}\right)}{\left(\text{By definition we get}\right)}\)

\(\displaystyle\text{Therefore},{L}{\left(\alpha{x}\right)}=\alpha{L}{\left({x}\right)}\)

Property (ii) is satisfied.

Therefore both properties got satisfied.

Therefore L is a linear transformation.