Proof of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum

Two points of $\mathbb{R}^n$ are equal exactly when they agree at every index, so claims 1 to 3 are proved by computing coordinates. Throughout, $(Mz)_i=\sum_{j=1}^{n}M_{ij}z_j$ by Matrix-Vector Product, entries of sums, differences and scalar multiples of matrices are computed entrywise by Sum of Real Matrices, Difference of Real Matrices and Scalar Multiple of a Real Matrix, and coordinates of sums, differences and scalar multiples of points are computed coordinatewise by Sum of Points of $\mathbb{R}^n$, Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$ and Scalar Multiple of a Point of $\mathbb{R}^n$. Claims 2, 3 and 7 of Properties of Finite Sums are used as additivity, homogeneity and the single-summand rule for finite sums, and the field axioms of the field $\mathbb{R}$ for the coordinatewise computations.

Claim 1. For each $i$, distributivity gives $(P_{ij}+Q_{ij})z_j=P_{ij}z_j+Q_{ij}z_j$ for every $j$, so by additivity of finite sums

$$\bigl((P+Q)z\bigr)_i=\sum_{j=1}^{n}\bigl(P_{ij}z_j+Q_{ij}z_j\bigr)=(Pz)_i+(Qz)_i=(Pz+Qz)_i ,$$

which is the first identity. For the second, the entries of $(P-Q)+Q$ are $(P_{ij}-Q_{ij})+Q_{ij}=P_{ij}$ by associativity of addition, the additive-inverse axiom and the additive-identity axiom, so $(P-Q)+Q=P$; the first identity applied to $P-Q$ and $Q$ therefore gives $\bigl((P-Q)z\bigr)+Qz=Pz$, and adding the point $-(Qz)$ to both sides gives $(P-Q)z=Pz-Qz$. For the third, associativity of multiplication gives $(\mu P_{ij})z_j=\mu\,(P_{ij}z_j)$, and homogeneity of finite sums gives

$$\bigl((\mu P)z\bigr)_i=\sum_{j=1}^{n}\mu\,(P_{ij}z_j)=\mu\,(Pz)_i=\bigl(\mu\,(Pz)\bigr)_i .$$

Claim 2. By Identity Matrix the entry $(I_n)_{ij}$ equals $1$ if $i=j$ and $0$ otherwise. First, $0\,a=0$ for every $a\in\mathbb{R}$: indeed $0\,a+0\,a=(0+0)\,a=0\,a=0\,a+0$ by distributivity and the additive-identity axiom, so $0\,a=0$ by claim 2 of Additive Cancellation and Elementary Additive Identities in a Field. Hence the family $j\mapsto (I_n)_{ij}z_j$ takes the value $1\,z_i=z_i$ at $j=i$ and the value $0$ at every other index, so the single-summand rule gives $(I_nz)_i=z_i$.

Claim 3. For each $i$, distributivity gives $M_{ij}(z_j+z'_j)=M_{ij}z_j+M_{ij}z'_j$ for every $j$, so additivity of finite sums gives $\bigl(M(z+z')\bigr)_i=(Mz)_i+(Mz')_i$, which is the first identity. The difference identity follows from it exactly as in claim 1, using $(z-z')+z'=z$. For the last identity, commutativity and associativity of multiplication give $M_{ij}(\mu z_j)=\mu\,(M_{ij}z_j)$, and homogeneity of finite sums gives $\bigl(M(\mu z)\bigr)_i=\mu\,(Mz)_i$.

Claim 4. By Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$ and Matrix-Vector Product,

$$w\cdot(Mz)=\sum_{i=1}^{n}w_i\,(Mz)_i=\sum_{i=1}^{n}\ w_i\sum_{j=1}^{n}M_{ij}z_j .$$

For each fixed $i$, homogeneity of finite sums applied to the inner sum with the factor $w_i$, together with commutativity and associativity of multiplication, gives

$$w_i\sum_{j=1}^{n}M_{ij}z_j=\sum_{j=1}^{n}M_{ij}\,w_i\,z_j ,$$

and substituting this for every $i$ yields the asserted double sum.