Beweis der Linearität der Transformationsabbildung TB

 

Ist α= T B (x) MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySdeMaeyypa0JaamivamaaBaaaleaacaWGcbaabeaakiaacIcacaWG4bGaaiykaaaa@3CBA@ und β= T B (y) MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqOSdiMaeyypa0JaamivamaaBaaaleaacaWGcbaabeaakiaacIcacaWG5bGaaiykaaaa@3CBD@ , so ist zunächst α+β,   τα∈ 𝔽 finsupp (B) MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqySdeMaey4kaSIaeqOSdiMaaiilaiaaysW7cqaHepaDcqaHXoqycqGHiiIZtuuDJXwAK1uy0HMmaeHbfv3ySLgzG0uy0HgiuD3BaGqbaiab=vi8gnaaBaaaleaacaqGMbGaaeyAaiaab6gacaqGZbGaaeyDaiaabchacaqGWbaabeaakiaacIcacaWGcbGaaiykaaaa@5603@ .

Die Darstellung

x+y = ∑ v∈B α(v)⋅v + ∑ v∈B ß(v)⋅v = ∑ v∈B (α(v)⋅v+ß(v)⋅v) = ∑ v∈B (α+ß)(v)⋅v MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@7C0A@

liefert nun: T B (x+y)= T B (x)+ T B (y) MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaBaaaleaacaWGcbaabeaakiaacIcacaWG4bGaey4kaSIaamyEaiaacMcacqGH9aqpcaWGubWaaSbaaSqaaiaadkeaaeqaaOGaaiikaiaadIhacaGGPaGaey4kaSIaamivamaaBaaaleaacaWGcbaabeaakiaacIcacaWG5bGaaiykaaaa@4636@ .

Analog ergibt sich aus

τx =τ ∑ v∈B α(v)⋅v = ∑ v∈B τ(α(v)⋅v) = ∑ v∈B (τα)(v)⋅v MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqbaeaabmGaaaqaaiabes8a0jaadIhaaeaacqGH9aqpcqaHepaDdaaeqbqaaiabeg7aHjaacIcacaWG2bGaaiykaiabgwSixlaadAhaaSqaaiaadAhacqGHiiIZcaWGcbaabeqdcqGHris5aaGcbaaabaGaeyypa0ZaaabuaeaacqaHepaDcaGGOaGaeqySdeMaaiikaiaadAhacaGGPaGaeyyXICTaamODaiaacMcaaSqaaiaadAhacqGHiiIZcaWGcbaabeqdcqGHris5aaGcbaaabaGaeyypa0ZaaabuaeaacaGGOaGaeqiXdqNaeqySdeMaaiykaiaacIcacaWG2bGaaiykaiabgwSixlaadAhaaSqaaiaadAhacqGHiiIZcaWGcbaabeqdcqGHris5aaaaaaa@69C7@

die Gleichung T B (τx)=τ T B (x) MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaBaaaleaacaWGcbaabeaakiaacIcacqaHepaDcaWG4bGaaiykaiabg2da9iabes8a0jaadsfadaWgaaWcbaGaamOqaaqabaGccaGGOaGaamiEaiaacMcaaaa@42D1@ .