Answer :
Yes.
Proof: Consider x, y in R^m. Then since T is linear, we have:
T(a*x + b*y) = a*T(x) + b*T(y)
But since L is linear, we have:
L(a*T(x) + b*T(y)) = a*L(T(x)) + b*L(T(y))
So:
L(T(a*x + b*y)) = a*L(T(x)) + b*L(T(y))
and the composition is linear.
Proof: Consider x, y in R^m. Then since T is linear, we have:
T(a*x + b*y) = a*T(x) + b*T(y)
But since L is linear, we have:
L(a*T(x) + b*T(y)) = a*L(T(x)) + b*L(T(y))
So:
L(T(a*x + b*y)) = a*L(T(x)) + b*L(T(y))
and the composition is linear.