S01 · implies
Before
A\binom{x}{y}After
x\binom{1}{2}+y\binom{2}{4}Conditions: none
Rule: calc · Uses: G1
Leaf basis: calc
Every output is a real column combination, giving inclusion in the column span.
derivation-mko-linear-map-image-01 · ocme-new142-reader-adoption-20261021
im T=span{(1,2)}; (3,6) lies in the image and (0,1) does not.
The entire real image of an exact 2-by-2 system, not a finite sampled output list.
calcG1 — T:R2 to R2, A=[[1,2],[2,4]], c1=(1,2), c2=(2,4).c_1,c_2 — matrix columnsu,v — arbitrary target coordinatesSome rule statements have reader clarifications in this revision: they are paraphrases, not verbatim quotations. Referenced original source bytes and hashes are unchanged.
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 — This slice uses exact 2-by-2 matrices and symbolic column-span certificates. Membership is exact real consistency of A*x=y. A bounded fiber search supplies witnesses or bounded nonexistence only, not global nonmembership in the real image. The image must not be conflated with the entire declared codomain or a sampled output list. · mko-linear-map-imageBefore
A\binom{x}{y}After
x\binom{1}{2}+y\binom{2}{4}Conditions: none
Rule: calc · Uses: G1
Leaf basis: calc
Every output is a real column combination, giving inclusion in the column span.
Before
z=xc_1+yc_2
After
z=T(x,y)\Rightarrow\operatorname{span}\{c_1,c_2\}\subseteq\operatorname{im}TConditions: none
Rule: calc · Uses: G1, S01
Leaf basis: calc
Any column combination is obtained by using its coefficients as input, proving reverse inclusion.
Before
c_2=2c_1
After
T(x,y)=(x+2y)c_1
Conditions: none
Rule: calc · Uses: G1, S01, S02
Leaf basis: calc
The columns are proportional, reducing every output to parameter s=x+2y.
Before
s\in\mathbb R
After
T(s,0)=(s,2s),\quad\operatorname{im}T=\{(s,2s):s\in\mathbb R\}Conditions: none
Rule: mko_definition · Uses: G1, S03
Leaf basis: mko_definition
A universal preimage shows every vector of this form is reached.
Before
A\binom{x}{y}=\binom{u}{v}After
x+2y=u,\quad 2x+4y=v\iff v=2u
Conditions: none
Rule: calc · Uses: G1, S01, S04
Leaf basis: calc
Subtracting twice equation one gives necessity; if v=2u, x=u,y=0 gives sufficiency.
Before
(u,v)=(3,6)
After
6=2\cdot3,\quad T(3,0)=(3,6)
Conditions: none
Rule: calc · Uses: G1, S05
Leaf basis: calc
Consistency holds with exact preimage (3,0), certifying membership.
Before
(u,v)=(0,1)
After
1\ne2\cdot0\Rightarrow(0,1)\notin\operatorname{im}TConditions: none
Rule: calc · Uses: G1, S05
Leaf basis: calc
Global consistency fails, proving nonmembership rather than just absence in a box.
Before
|x|,|y|\le1,\quad x,y\in\mathbb Z
After
-3\le x+2y\le3,\quad(100,200)=T(100,0)\in\operatorname{im}TConditions: none
Rule: calc · Uses: G1, S04, S05
Leaf basis: calc
The small box misses (100,200), yet it has a global preimage, illustrating the search boundary.
No substeps in this guide.
S08 · premise_audit · G1, S04, S05 — Bounded nonexistence cannot replace S05's symbolic criterion.S07 · premise_audit · G1, S05 — The contradiction 1=0 excludes all real inputs.S06 · premise_audit · G1, S05 — The preimage is not unique; adding t(-2,1) preserves output (3,6).S05 · inverse_equivalence · G1, S01, S04 — Both directions are proved without a bounded no-hit inference.S04 · premise_audit · G1, S03 — The image is a line inside R2, not the entire declared codomain.S03 · premise_audit · G1, S01, S02 — s can be any real because x=s,y=0 is available.S02 · premise_audit · G1, S01 — Both inclusions give the full image, not merely a lower bound.S01 · premise_audit · G1 — x,y range over all reals, not merely the columns themselves.Use Ax=y consistency rather than guessing surjectivity from T:R2 to R2.
A finite search's negative result applies only to its declared input box.
reject — (0,1) fails v=2u.reject — Every real multiple s(1,2) is an output, such as (3,6).reject — (100,0) is an explicit global preimage.im T=span{(1,2)}; (3,6) lies in the image and (0,1) does not.
2 declared checks for this guide; only the listed expressions are covered.
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