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Question # text{Let A be an} n times n text{matrix and suppose that} L : M_{nn} rightarrow M_{nn} text{is defined by} L(x)=AX, text{for} X in M_{nn}. text{Show that L is a linear transformation.}

Transformation properties
ANSWERED $$\displaystyle\text{Let A be an}\ {n}\ \times\ {n}\ \text{matrix and suppose that}\ {L}:{M}_{{\cap}}\ \rightarrow\ {M}_{{\cap}}\ \text{is defined by}{L}{\left({x}\right)}={A}{X},\text{for}\ {X}\in{M}_{{\cap}}.\text{Show that L is a linear transformation.}$$ 2021-02-11
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.