4.19 Matrix of a composition

Suppose we have two composable linear maps, S and T. The composition T∘S is still linear, as you can check. There should be a connection between the matrix of T∘S with respect to some bases and the matrices for T and S.

Theorem 4.19.1.

Let S:U→V and T:V→W be linear maps. Let

  • •

    ℬ=𝐛1,…,𝐛l be a basis of U,

  • •

    𝒞=𝐜1,…,𝐜m be a basis of V, and

  • •

    𝒟=𝐝1,…,𝐝n be a basis of W.

Then [T∘S]𝒟ℬ=[T]𝒟𝒞⁢[S]𝒞ℬ

Here is picture of this situation:

ℬ𝒞𝒟U→SV→TW

This theorem provides some justification for our definition of matrix multiplication: composition of linear maps corresponds to multiplication of matrices.

Proof.

Let [T]𝒟𝒞=(ti⁢j) and [S]𝒞ℬ=(si⁢j). We will work out [T∘S]𝒟ℬ using the definition of the matrix of a linear map. For any 1⩽c⩽l,

(T∘S)⁢(𝐛c) =T⁢(S⁢(𝐛c))
=T⁢(∑k=1msk⁢c⁢𝐜k) as ⁢[S]𝒞ℬ=(si⁢j)
=∑k=1msk⁢c⁢T⁢(𝐜k) linearity of ⁢T
=∑k=1msk⁢c⁢∑i=1nti⁢k⁢𝐝i as ⁢[T]𝒟𝒞=(ti⁢j)
=∑i=1n(∑k=1mti⁢k⁢sk⁢c)⁢𝐝i for finite sums, ⁢∑k∑i=∑i∑k

so the r,c entry of [T∘S]𝒟ℬ is ∑k=1mtr⁢k⁢sk⁢c, which is the same as the r,c entry of [T]𝒟𝒞⁢[S]𝒞ℬ by the matrix multiplication formula. ∎