Prove a basis before reading dimension

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

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

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

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

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

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

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

\alpha b_1+\beta b_2=0

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

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

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

B\text{ independent and spans }V

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

C=((1,0))

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

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

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

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 qualifications

A basis needs both independence and spanning.

Theorem layer

Basis-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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