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Question 16495: A Homogeneous Kernel Has Universal Closure under Real Linear Combinations

On this page
  1. Original question and givens
  2. Definition and scope
  3. Step-by-step derivation
  4. Reverse checks
  5. Result
  6. This question review and source binding

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R0031-16495 · q-20261014-advance-20261101-r01-vector-subspace-0045

Original question and givens

Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[20, -6] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).

  • G001 — Use the subset U={(x,y) in R²:a*x+b*y=0} with nonzero normal [a,b]=[20, -6] and usual real vector operations. Give (normal,zero_in?,addition_closed?,scalar_closed?,homogeneous_kernel_closed_for_all_real_scalars).

Definition and scope

Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate.

Step-by-step derivation

S001 · implies

Before

same usual real R2;fixed normal=[20, -6];U consists of ALL x with(20)*x0+(-6)*x1=0

After

zero vector:( 20)*0+(-6)*0=0;zero belongs to U

Conditions: none

The equation defines the whole set,not a finite sample.First substitute the zero vector;all later vectors and scalars stay in the same usual real vector space.

Step identifiers and dependencies

Rule: subspace_homogeneous_zero_vector_universal_parent · Uses: G001

S002 · implies

Before

ANY u,v in this U;(20)*u0+(-6)*u1=0 and(20)*v0+(-6)*v1=0

After

(20)*(u0+v0)+(-6)*(u1+v1)=((20)*u0+(-6)*u1)+((20)*v0+(-6)*v1)=0+0=0;u+v in U

Conditions: none

Distribute and regroup into the two equations already equal to zero.This proves closure for any two solutions,not just selected points.

Step identifiers and dependencies

Rule: subspace_addition_linearity_for_all_solutions · Uses: G001, S001

S003 · implies

Before

ANY real scalar r and ANY u in U;(20)*u0+(-6)*u1=0

After

(20)*(r*u0)+(-6)*(r*u1)=r*((20)*u0+(-6)*u1)=r*0=0;r*u in U

Conditions: none

Commute the fixed coefficients with an arbitrary real r and factor it out.The result is zero,including zero,negative and irrational scalars.Finitely many integer tests are not a substitute for every-real-scalar closure.

Step identifiers and dependencies

Rule: subspace_every_real_scalar_not_finite_sample_scalars · Uses: G001, S002

Reverse checks

  • S003 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S002 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.

Result

([20, -6], Yes, Yes, Yes, homogeneous_kernel_closed_for_all_real_scalars)

The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.

This question review and source binding

Review date: 2026-10-26

Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.

Source hashes and review receipts
  • input commit: 0f7be61e2c22317a4075561dc822c87d881a50db
  • original question SHA-256: 6504e8d0a601758a41ae0d124a26a1bd79884efb0c06c6577d4be6b34f0c1a22
  • candidate record SHA-256: ab866ec2a2a9266ce410e5bfbeed4d321b8589b0825cf9fcd94ba106c1337898
  • semantic review receipt SHA-256: 55bb21783e2125ecf1e67e42f83fd25bec0ccd493f911f4bb937a5e9d6f9369a
  • question-first witness SHA-256: 754ddf2a6039c5ef4aae78a538c6e9a1e9a58d4825aee194e869a27b8a4c95bb
  • Lean: NOT_RUN · human review: NOT_PERFORMED
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