prove that the set is a vector space.

Question

Let A be a fixed 2×2 matrix, and let B be a fixed 3×3 matrix. Prove that the set $$S={\{X∈M_{2,3} |AX=XB\}}$$ is a vector space.

I don't know if my solution is correct so if someone could please help me figure it out. Thank you.
My solution is the following:

i) let L = $\begin{pmatrix}0 & 0 & 0\\\ 0 & 0 & 0\end{pmatrix}$ represent the zero matrix
substituting L into AX-XB we get:
AL = LB
A0 = 0B
0 = 0
therefore L $\in$ S

ii) if X,Y $\in$ $M_{2,3}$ then AX - XB = 0 and AY - YB = 0
Now to prove X + Y $\in$ $M_{2,3}$:
A(X+Y) - (X+Y)B
=AX + AY - XB - YB
=AX - XB + AY - YB
= 0 + 0
=0
therefore S is closed under addition
iii) if $\lambda$ $\in$ ℝ and X $\in$ $M_{2,3}$ then $\lambda$X = $\begin{pmatrix}\lambda x_1 & \lambda x_2 & \lambda x_3\\\ \lambda x_4 & \lambda x_5 & \lambda x_6\end{pmatrix}$
Now A($\lambda$X) - ($\lambda$X)B
=$\lambda$AX - $\lambda$XB
=$\lambda$(AX - XB)
=$\lambda$(0)
=0
therefore S is closed under scalar multiplication

Hence, S is a vector space

Answer 1

Your solution is correct. $ \, \! $