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Properties of Row-Vector-Matrix Multiplication

Let matrix ARm×n, let vectors x,yRm×1 and let scalars α,βR. Then, each of the following laws holds true:

  1. Distributivity: (x+y)TA=xTA+yTA
  2. Scalar Multiplication: (αx)TA=α(xTA)
  3. Linearity: (αx+βy)TA=αxTA+βyTA

Conditionality Statements

Translated into the if the antecedent P, then consequent Q structure, we write our theorem propositions

1)If ARm×n and x,yRm×1P, then (x+y)TA=xTA+yTAQ2)If ARm×n,x,yRm×1, and αRP, then (αx)TA=α(xTA)Q3)If ARm×n,x,yRm×1, and α,βRP, then (αx+βy)TA=αxT+βyTAQ

Proof Intuition

Notice:

Conjecture

Distributivity and Scalar MultiplicationLinearity

Intuitive draft: If distributivity and scalar multiplication hold, then linearity must be true.

We can translate our intuitive draft into a technical draft of conditionality as follows:

If ARm×n,x,yRm×1,α,βR and (x+y)TA=xTA+yTA and (αx)TA=α(xTA)P,then (αx+βy)TA=αxT+βyTAQ

Proofs

To prove our theorems, let's make a conjecture.

Conjecture

  1. If distributivity and scalar multiplication, then linearity.
  2. If linearity, then distributivity and scalar multiplication.

Let's show that conjecture 1 is true.
\begin{proof}
Let ARm×n,x,yRm×1, and α,βR. Suppose distributivity and scalar multiplication hold true for any vectors and scalars.

(x+y)TA=xTA+yTA and (αx)TA=α(xTA)

Consider

(αx+βy)TA=[(αx)+(βy)]TA=(αx)TA+(βy)TADistributivity=α(xTA)+β(yTA)linearityScalar multiplication

This is what we wanted to show.
\end{proof}
Now, we have to prove conjecture 2.
\begin{proof}
Let ARm×n,x,yRm×1, and α,βR. Suppose linearity holds true for any vectors and scalars. First, let's prove distributivity. Consider

(x+y)TA=(αx+βy)TAwhere α=1=β=α(xTA)+β(yTA)Linearity=1(xTA)+1(yTA)=xTA+yTA

This proves distributivity. Next, we must prove scalar multiplication. Consider

(ax)TA=(αx+βy)TAwhere β=0=α(xTA)+β(yTA)Linearity=α(xTA)+0=α(xTA)

\end{proof}
These proofs show that linearity distributivity and scalar multiplication. Finally, let's prove linearity using the 10.4 Row-Vector-Matrix Multiplication via Linear Combinations definition.
\begin{proof}
Let ARm×n,x,yRm×1, and α,βR. Consider

(αx+βy)TA=[αx1+βy1αx2+βy2αxm+βym][A(1,:)A(2,:)A(m,:)]=(αx1+βy1)A(1,:)+(αx2+βy2)A(2,:)++(αxm+βym)A(m,:)=α(x1A(1,:)+x2A(2,:)++xmA(m,:))+β(y1A(1,:)+y2A(2,:)++ymA(m,:))=α[x1x2xm][A(1,:)A(2,:)A(m,:)]+β[y1y2ym][A(1,:)A(2,:)A(m,:)]=αxTA+βyTA

This is what we wanted to show.
\end{proof}