# Prove a basis before reading dimension

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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-vector-space-dimension-01 · ocme-new142-reader-adoption-20261021



## Goal and scope

B=((1,1),(1,-1)) is an ordered basis of R2, so dim R2=2.

Complete R2 coordinate families and exact rational elimination; no general invariance theorem inferred from examples.





## Givens and symbols


- calc


- G1 — V=R2; b1=(1,1), b2=(1,-1); P has b1,b2 as its ordered columns.


- P — basis-column matrix
- B — ordered candidate basis





## 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. Also explicitly assume finite-dimensional basis-cardinality invariance.
- mko_definition · cited — This slice handles complete coordinate families in R² or R³. Check exact linear independence and spanning before using basis size as dimension. An independent family supplies only a lower bound and a spanning family only an upper bound; basis-cardinality invariance invokes the finite-dimensional theorem. · mko-vector-space-dimension





## Main derivation

S01 · impliesBefore

b_1=(1,1),\quad b_2=(1,-1)

After

P=\begin{pmatrix}1&1\\1&-1\end{pmatrix}

Conditions: none

Rule: calc · Uses: G1

Leaf basis: calc

Both columns are complete; basis qualification requires independence and spanning.



S02 · impliesBefore

P=\begin{pmatrix}1&1\\1&-1\end{pmatrix}

After

R_2\leftarrow R_2-R_1:\quad\begin{pmatrix}1&1\\0&-2\end{pmatrix}

Conditions: none

Rule: calc · Uses: G1, S01

Leaf basis: calc

Subtracting row one from row two is invertible, with addition as its inverse.



S03 · impliesBefore

\begin{pmatrix}1&1\\0&-2\end{pmatrix}

After

R_2\leftarrow-\tfrac12R_2:\quad\begin{pmatrix}1&1\\0&1\end{pmatrix}\ ;\ R_1\leftarrow R_1-R_2:\quad I_2

Conditions: the pivot -2 is nonzero

Rule: calc · Uses: G1, S02

Leaf basis: calc

Scaling by -1/2 is valid because -2 is nonzero; clearing the upper entry gives the full RREF.



S04 · impliesBefore

\alpha b_1+\beta b_2=0

After

P\binom{\alpha}{\beta}=0\iff\alpha=\beta=0

Conditions: none

Rule: calc · Uses: G1, S01, S03

Leaf basis: calc

Two pivots and no free variables leave only the zero homogeneous solution, proving independence.



S05 · impliesBefore

(x,y)\in\mathbb R^2

After

(x,y)=\frac{x+y}{2}b_1+\frac{x-y}{2}b_2

Conditions: 2 is nonzero in R

Rule: calc · Uses: G1, S01

Leaf basis: calc

Explicit coefficients for arbitrary real coordinates substitute back to x,y, proving full spanning.



S06 · impliesBefore

B\text{ independent and spans }V

After

B\text{ basis};\quad\dim V=|B|=2

Conditions: none

Rule: mko_definition · Uses: G1, S04, S05

Leaf basis: mko_definition

Only after basis qualification do we read dimension using explicitly assumed finite-dimensional invariance.



S07 · impliesBefore

C=((1,0))

After

C\text{ independent},\quad\operatorname{span}C=\{(x,0):x\in\mathbb R\}\ne V

Conditions: none

Rule: calc · Uses: G1, S06

Leaf basis: calc

A single nonzero vector is independent but misses (0,1), so size 1 is not dim V.



S08 · impliesBefore

D=((1,0),(0,1),(1,1))

After

D\text{ spans }V,\quad(1,0)+(0,1)-(1,1)=0

Conditions: none

Rule: calc · Uses: G1, S06

Leaf basis: calc

The first two span V, but the three have a nontrivial relation, so list size 3 is only an upper bound.







## Substeps and lemmas

No substeps in this guide.





## Reverse audits


- S08 · premise_audit · G1, S06 — Coefficients 1,1,-1 are not all zero, so the list is not a basis.
- S07 · premise_audit · G1, S06 — This separates an independence lower bound from exact dimension.
- S06 · premise_audit · G1, S04, S05 — The dimension is 2 because the vectors form a basis, not merely because the list has length two.
- S05 · premise_audit · G1, S01 — Division by 2 is valid and x,y are not restricted to a finite grid.
- S04 · premise_audit · G1, S01, S03 — Independence gives only a lower bound of 2; spanning still needs proof.
- S03 · premise_audit · G1, S02 — Both operations are invertible, preserving homogeneous solutions and consistency.
- S02 · premise_audit · G1, S01 — The pivot -2 is exact, not a floating-point near-zero value.
- S01 · premise_audit · G1 — There are two rows and two columns in the fixed ambient R2.





## Error controls and negative controls

Two qualificationsA basis needs both independence and spanning.

Theorem layerBasis-cardinality invariance is an explicit prerequisite, not inferred universally from examples.


- One independent vector means dim R2=1. → reject — It need not span (0,1); it supplies only lower bound 1.
- A spanning list of three means dim R2=3. → reject — The displayed D is dependent.
- det P=0, hence dimension zero here. → reject — det P=-2, not zero, and the full RREF is I2.





## Result and limitations

B=((1,1),(1,-1)) is an ordered basis of R2, so dim R2=2.







## Verification and source binding

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


- source commit: b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f
- adopted source SHA-256: f627a1afb0693b4ff43791c3e3fc4f77de78ee6d8f6ae0c5183d04ab4767dfb6
- original source SHA-256: f627a1afb0693b4ff43791c3e3fc4f77de78ee6d8f6ae0c5183d04ab4767dfb6
- Lean: NOT_RUN · independent AI: SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW

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