Proof of Additive Laws for Vectors in $\mathbb{R}^q$ and for Real $p\times q$ Matrices: Group Laws and Rules for Differences

Reduction to maps. By Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (i), 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 (v), $\mathbb{R}^{q}$ is the set $\mathbb{R}^{I}$ of maps $I\to\mathbb{R}$ for $I=[q]$, its sum, negative and difference are those of Pointwise Operations on Real-Valued Maps (i), and its origin is the constant map with value $0$ of Pointwise Operations on Real-Valued Maps (ii). By Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations (i), Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations (iv) and Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations (v), the same holds for $\mathcal{M}_{p\times q}(\mathbb{R})$ with $I=[p]\times[q]$. So it suffices to prove both clauses for a set $I$ and elements $X,Y,Z,X',Y'$ of $\mathbb{R}^{I}$, with these operations and $0$ the constant map with value $0$.

Values. By Pointwise Operations on Real-Valued Maps (i), sums, negatives and differences of elements of $\mathbb{R}^{I}$ are elements of $\mathbb{R}^{I}$, and the difference is $X-Y=X+(-Y)$; this is the sixth identity of (ii). For every $u\in I$, by Pointwise Operations on Real-Valued Maps (i), Rules for Pointwise Operations on Real-Valued Maps: a Commutative Ring, Scalar Multiples and Evaluation (ii) and Pointwise Operations on Real-Valued Maps (ii),

$$(X+Y)(u)=X(u)+Y(u),\qquad(-X)(u)=-X(u),\qquad(X-Y)(u)=X(u)-Y(u),\qquad0(u)=0 .$$

Two elements of $\mathbb{R}^{I}$ whose values agree at every $u\in I$ are equal, by Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv).

The identities. Let $u\in I$, and write $x=X(u)$, $y=Y(u)$, $z=Z(u)$, $x'=X'(u)$ and $y'=Y'(u)$, elements of $\mathbb{R}$. Applying the value rules to the parts of each side, from the inside out, the two sides of each identity of (i) and (ii) other than the sixth take at $u$ the values of the two sides of the corresponding identity in $\mathbb{R}$:

$$x+y=y+x,\quad(x+y)+z=x+(y+z),\quad x+0=x=0+x,\quad x+(-x)=0,$$ $$x-x=0,\quad x-0=x,\quad0-x=-x,\quad-(-x)=x,\quad x-y=-(y-x),\quad(x+y)-(x'+y')=(x-x')+(y-y'),\quad(x-y)+y=x,$$ $$(x-y)+(y-z)=x-z .$$

For instance, $\bigl((X+Y)-(X'+Y')\bigr)(u)=(x+y)-(x'+y')$ and $\bigl((X-X')+(Y-Y')\bigr)(u)=(x-x')+(y-y')$, and $\bigl(-(Y-X)\bigr)(u)=-(y-x)$. These real identities hold by the field laws of $\mathbb{R}$. So the two sides of each identity have the same value at every $u\in I$, and are equal. This proves (i) and the identities of (ii).

The implication. If $X-Y=0$, then for every $u\in I$, $X(u)-Y(u)=(X-Y)(u)=0(u)=0$, so $X(u)=\bigl(X(u)-Y(u)\bigr)+Y(u)=Y(u)$; hence $X=Y$. This completes the proof of (ii).