TheoremBase

Let $K$ be a \reftext{def:field-c54-2026b}{field} and let $V$ be a \reftext{def:vector-space-2026a}{vector space over $K$ } with \reftext{lem:vector-space-basic-identities-2026a}{zero vector} $0_{V}$ . Let $\mathbb{N}$ be the set of \reftext{def:natural-numbers-2026a}{natural numbers} with successor map $S$ as in that definition, ordered by the relation $\le$ of \reftext{def:order-natural-numbers-2026a}{that definition}, let $n\in\mathbb{N}$ , and let $[n]$ be the \reftext{def:initial-segment-natural-numbers-2026a}{initial segment} determined by $n$ , that is, the set of natural numbers $k$ with $1\le k\le n$ . Let $u:[n]\to V$ and $v:[n]\to V$ be maps with values $u_{k}$ and $v_{k}$ , let $c:[n]\to K$ be a map with values $c_{k}$ , and let $\lambda\in K$ .

Sums of vectors are the \reftext{def:finite-sum-vector-space-2026a}{finite sums in a vector space}, and sums of scalars are the \reftext{def:finite-sum-field-2026b}{finite sums in a field}. Then the following hold.

\textbf{1. (Restriction and recursion)} If $j\in[n]$ and $v'$ denotes the restriction of $v$ to $[j]$ , then

\sum_{k=1}^{i}v'_{k}=\sum_{k=1}^{i}v_{k}\qquad\text{for every }i\in[j].

Moreover

\sum_{k=1}^{1}v_{k}=v_{1},\qquad \sum_{k=1}^{S(m)}v_{k}=\Bigl(\sum_{k=1}^{m}v_{k}\Bigr)+v_{S(m)}\quad\text{whenever }S(m)\in[n].

\textbf{2. (Additivity)}

\sum_{k=1}^{n}(u_{k}+v_{k})=\sum_{k=1}^{n}u_{k}+\sum_{k=1}^{n}v_{k}.

\textbf{3. (Homogeneity)}

\sum_{k=1}^{n}(\lambda v_{k})=\lambda\sum_{k=1}^{n}v_{k}.

\textbf{4. (Linear maps)} Let $W$ be a vector space over $K$ and let $T:V\to W$ be a \reftext{def:linear-map-2026a}{linear map}. Then

T\Bigl(\sum_{k=1}^{n}v_{k}\Bigr)=\sum_{k=1}^{n}T(v_{k}),

the sum on the right being the finite sum in $W$ .

\textbf{5. (Inner products against a finite sum)} Suppose $K$ is the field of \reftext{def:complex-numbers-2026a}{complex numbers} and $V$ together with $\langle\cdot,\cdot\rangle$ is a \reftext{def:complex-inner-product-space-2026a}{complex inner product space}, and let $w\in V$ . Then

\Bigl\langle w,\sum_{k=1}^{n}v_{k}\Bigr\rangle=\sum_{k=1}^{n}\langle w,v_{k}\rangle,\qquad \Bigl\langle \sum_{k=1}^{n}v_{k},w\Bigr\rangle=\sum_{k=1}^{n}\langle v_{k},w\rangle,

the sums on the right being finite sums in the field of complex numbers.

\textbf{6. (Inner products against a linear combination)} Under the hypotheses of claim 5, with the \reftext{def:complex-conjugate-2026a}{complex conjugate},

\Bigl\langle w,\sum_{k=1}^{n}c_{k}v_{k}\Bigr\rangle=\sum_{k=1}^{n}c_{k}\langle w,v_{k}\rangle,\qquad \Bigl\langle \sum_{k=1}^{n}c_{k}v_{k},w\Bigr\rangle=\sum_{k=1}^{n}\overline{c_{k}}\,\langle v_{k},w\rangle .

\textbf{7. (A single possibly nonzero summand)} Let $i\in[n]$ and suppose that $v_{k}=0_{V}$ for every $k\in[n]$ with $k\ne i$ . Then

\sum_{k=1}^{n}v_{k}=v_{i}.

In particular, if $v_{k}=0_{V}$ for every $k\in[n]$ , then $\sum_{k=1}^{n}v_{k}=0_{V}$ .

Properties of Finite Sums of Vectors

Properties of Finite Sums of Vectors

Dependency Graph

Prerequisites

Dependents

Dependent proofs

Authors

Citations

Comments

Proofs