Compositions of linear transformations

Since linear transformations are just a special type of function, we can form compositions of them.

The composition of a linear transformation T:Rn → Rm and S:Rm → Rk is the function (S°T):Rn → Rk with the rule (S°T)(x) = S(T[x]) for all x in Rn.

We want to know two things about compositions of linear transformations:

The composition of two linear transformations is a linear transformation.

Proof: Let T:Rn → Rm and S:Rm → Rk be linear transformations and form their composition (S°T):Rn → Rk. We need to check that S°T preserves sums and scalar multiples.

For two vectors u and v in Rn

(S°T)(u + v)

= S(T[u + v])
= S(T[u] + T[v])       since T preserves sums
= S(T[u]) + S(T[v])       since S preserves sums
= (S°T)(u) + (S°T)(v).

So S°T preserves sums.

For any vector u in Rn and any scalar c,

(S°T)(cu)

= S(T[cu])
= S(cT[u])     since T preserves scalar multiples
= cS(T[u])      since S preserves scalar multiples
= c(S°T)(u)

So S°T preserves scalar multiples.

 

The standard matrix of the composition of two linear transformations is the product of the standard matrices of the individual transformations.

Proof: Suppose you have linear transformations T:Rn → Rm and S:Rm → Rk with standard matrices A and B respectively, i.e. T(u) = Au for all u in Rn and S(v) = Bv for all v in Rm. Since A is m x n and B is k x m, their product C = BA is defined and is k x m.

Suppose R:Rm → Rn is the matrix transformation given by R(u) = Cu for all u in Rn. Then

R(u)

= (BA)u
= B(Au)
= B[T(u)]
= S[T(u)]
= (S°T)(u).

This is true for all u in Rn, so R is the composition of S and T. Thus S°T has standard matrix C = BA.