Proof of Products of Real Matrices with Zero Matrices, Negatives and Differences

Computations with real numbers are those of The Real Numbers as a Type. The rules of Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes are used with sizes suited to each product. The zero matrices are the constant maps $0$ of Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations (v), the origins those of Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (v), and the negatives the pointwise negatives of Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations (iv) and Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (iv); so, by Rules for Pointwise Operations on Real-Valued Maps: a Commutative Ring, Scalar Multiples and Evaluation (iv), $0\cdot M=0$ and $(-1)M=-M$ for every real matrix or vector $M$.

Proof of (i). By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iii),

$$A0_{q\times r}=A\bigl(0\cdot0_{q\times r}\bigr)=0\cdot\bigl(A0_{q\times r}\bigr)=0_{p\times r},\qquad 0_{r\times p}A=\bigl(0\cdot0_{r\times p}\bigr)A=0\cdot\bigl(0_{r\times p}A\bigr)=0_{r\times q}.$$

By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v), $A0=0$ and $0_{p\times q}x=\bigl(0\cdot0_{p\times q}\bigr)x=0\cdot\bigl(0_{p\times q}x\bigr)=0$.

Proof of (ii). By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iii) with $-1$ for $s$,

$$(-A)B=\bigl((-1)A\bigr)B=(-1)(AB)=-(AB)=(-1)(AB)=A\bigl((-1)B\bigr)=A(-B),$$

and by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v), $(-A)x=\bigl((-1)A\bigr)x=(-1)(Ax)=-(Ax)$ and $A(-x)=A\bigl((-1)x\bigr)=(-1)(Ax)=-(Ax)$.

Proof of (iii). For real matrices $M,N$ of the same size, $M-N=M+(-N)$: by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations (iv), both have the entry $M_{ij}-N_{ij}=M_{ij}+(-N_{ij})$ in row $i$ and column $j$, so they are equal by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations (iii); likewise $u-v=u+(-v)$ for $u,v$ in a Euclidean space, by Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (iv) and Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (iii). Hence, by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (ii) and (ii),

$$A(B-B')=A\bigl(B+(-B')\bigr)=AB+A(-B')=AB+\bigl(-(AB')\bigr)=AB-AB',$$ $$(A-A')B=\bigl(A+(-A')\bigr)B=AB+(-A')B=AB+\bigl(-(A'B)\bigr)=AB-A'B .$$

The identity $A(x-y)=Ax-Ay$ is part of Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v); and by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v) together with (ii),

$$(A-A')x=\bigl(A+(-A')\bigr)x=Ax+(-A')x=Ax+\bigl(-(A'x)\bigr)=Ax-A'x .$$