# Unique decomposition and coverage in R3

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This is an owner-adopted natural-language teaching guide. Finite/arithmetic checks are not universal proofs; no Lean was run, and review scope is recorded per revision. Original source status labels are retained as historical data. Formulas retain exact TeX; no new semantic AST is invented.

derivation-mko-direct-sum-01 · ocme-new142-reader-adoption-20261021



## Goal and scope

R3=span{e1} direct-sum span{e2,e3}, with uniqueness and coverage separately verified.

A finite-dimensional two-summand internal sum, with many-summand and external-carrier boundaries.





## Givens and symbols


- calc


- G1 — V=R3; e1=(1,0,0),e2=(0,1,0),e3=(0,0,1); U=span{e1}, W=span{e2,e3}.


- U,W — internal subspaces
- \Phi — addition map from external to internal sum





## Rules and sources


- calc · prerequisite — Assume the real field, usual coordinate-vector arithmetic, matrix multiplication, and definitions of independence and span; elimination uses invertible elementary row operations.
- mko_definition · cited — Use complete bases and exact ranks in a finite-dimensional real ambient space. U∩W={0} ensures uniqueness in U+W, while V=U⊕W also needs U+W=V. For three or more subspaces pairwise zero intersections do not replace global uniqueness; an external direct sum uses a new product-type carrier. · mko-direct-sum





## Main derivation

S01 · impliesBefore

U=\operatorname{span}\{e_1\},\quad W=\operatorname{span}\{e_2,e_3\}

After

U=\{(a,0,0):a\in\mathbb R\},\quad W=\{(0,b,c):b,c\in\mathbb R\}

Conditions: none

Rule: calc · Uses: G1

Leaf basis: calc

Both subspaces are fully parameterized; each generating family is independent by coordinate comparison.



S02 · impliesBefore

v\in U\cap W

After

v=(a,0,0)=(0,b,c)\Rightarrow a=b=c=0

Conditions: none

Rule: calc · Uses: G1, S01

Leaf basis: calc

Comparing every coordinate makes the intersection the zero subspace.



S03 · impliesBefore

(x,y,z)\in V

After

(x,y,z)=(x,0,0)+(0,y,z)\in U+W

Conditions: none

Rule: calc · Uses: G1, S01

Leaf basis: calc

Every ambient vector has two valid summands, so U+W=V.



S04 · impliesBefore

u+w=u'+w',\quad u,u'\in U,\quad w,w'\in W

After

u-u'=w'-w\in U\cap W=\{0\}

Conditions: none

Rule: calc · Uses: G1, S01, S02

Leaf basis: calc

The difference of two decompositions lies in the intersection and therefore vanishes.



S05 · impliesBefore

u-u'=0,\quad w'-w=0

After

u=u',\quad w=w';\qquad V=U\oplus W

Conditions: none

Rule: mko_definition · Uses: G1, S03, S04

Leaf basis: mko_definition

Uniqueness combines with coverage to certify the internal direct sum of V.



S06 · impliesBefore

[e_1\ e_2\ e_3]=I_3

After

\operatorname{rank}[e_1\ e_2\ e_3]=3,\quad1+2=3

Conditions: none

Rule: calc · Uses: G1, S01, S05

Leaf basis: calc

The full concatenated basis matrix is I3; its three pivots recheck independence and coverage.



S07 · impliesBefore

L_1=\mathbb R(1,0),\ L_2=\mathbb R(0,1),\ L_3=\mathbb R(1,1)

After

L_i\cap L_j=\{0\}\ (i\ne j),\quad(1,0)+(0,1)-(1,1)=0

Conditions: none

Rule: calc · Uses: G1

Leaf basis: calc

Three R2 lines have pairwise zero intersections yet a nontrivial global relation, so the three-way sum is not direct.



S08 · impliesBefore

U\times W=\{(u,w):u\in U,w\in W\}

After

\Phi:U\times W\to V,\quad\Phi(u,w)=u+w\text{ bijective}

Conditions: none

Rule: calc · Uses: G1, S03, S04, S05

Leaf basis: calc

External-sum elements are ordered pairs; coverage gives surjectivity and uniqueness gives injectivity here.







## Substeps and lemmas

No substeps in this guide.





## Reverse audits


- S08 · premise_audit · G1, S03, S04, S05 — A bijection does not make ordered pairs literally equal to ambient vectors.
- S07 · premise_audit · G1 — Three nonzero summands add to zero, directly refuting uniqueness.
- S06 · premise_audit · G1, S01, S05 — The rank certificate concerns the same U,W, not arbitrary dimensions.
- S05 · premise_audit · G1, S03, S04 — Zero intersection alone gives uniqueness within U+W; S03 cannot be omitted.
- S04 · premise_audit · G1, S01, S02 — Closure under differences is needed to place each side in its respective subspace.
- S03 · premise_audit · G1, S01 — Zero intersection and coverage are separate; this step establishes coverage.
- S02 · premise_audit · G1, S01 — Zero belongs to both, so the intersection is not empty.
- S01 · premise_audit · G1 — Omitting e3 from W would invalidate coverage.





## Error controls and negative controls

Separate coverageZero intersection gives uniqueness; U+W=V gives coverage. Both are needed.

Global independenceA many-summand direct sum requires global independence of all basis columns.


- Two zero-intersection axes cover R3. → reject — span{e1}+span{e2} misses e3.
- Pairwise zero intersections suffice for three summands. → reject — The displayed lines have a nontrivial three-term relation.
- External-sum elements are literally u+w. → reject — The external carrier contains (u,w); Phi sends it to u+w.





## Result and limitations

R3=span{e1} direct-sum span{e2,e3}, with uniqueness and coverage separately verified.







## Verification and source binding

1 declared checks for this guide; only the listed expressions are covered.


- source commit: b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f
- adopted source SHA-256: e587d57a8edbf7ea1be841fcab7e765b95d2765b0c3bcff302efa7213d0cb9b6
- original source SHA-256: e587d57a8edbf7ea1be841fcab7e765b95d2765b0c3bcff302efa7213d0cb9b6
- Lean: NOT_RUN · independent AI: SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW

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