Consideremos la siguiente matriz A=( 2 1 2 1 2 2 1 2 3 1 3 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabg2da9maabmaabaqbaeqabmWaaaqaamaaL4babaqbaeqabmWaaaqaaiaaikdaaeaacaaIXaaabaaabaaabaGaaGOmaaqaaiaaigdaaeaaaeaaaeaacaaIYaaaaaaaaeaaaeaaaeaaaeaadaqjEaqaauaabeqaciaaaeaacaaIYaaabaGaaGymaaqaaaqaaiaaikdaaaaaaaqaaaqaaaqaaaqaamaaL4babaqbaeqabiGaaaqaaiaaiodaaeaacaaIXaaabaaabaGaaG4maaaaaaaaaaGaayjkaiaawMcaaaaa@4277@

formada por tres bloques de Jordan, uno de orden 3 asociado al valor propio 2, otro de orden 2 asociado al mismo valor propio, y otro de orden 2 asociado al valor propio 3:

Es sencillo calcular su polinomio característico -como el de cualquier matriz triangular:

(x−2) 5 (x−3) 2 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadIhacqGHsislcaaIYaGaaiykamaaCaaaleqabaGaaGynaaaakiaacIcacaWG4bGaeyOeI0IaaG4maiaacMcadaahaaWcbeqaaiaaikdaaaaaaa@3FCB@

Ahora su polinomio mínimo debe ser un divisor. Por tanto, será de la forma

(x−2) a (x−3) b MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadIhacqGHsislcaaIYaGaaiykamaaCaaaleqabaGaaGynaaaakiaacIcacaWG4bGaeyOeI0IaaG4maiaacMcadaahaaWcbeqaaiaaikdaaaaaaa@3FCB@ donde 0≤a≤5     0≤b≤2 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGimaiabgsMiJkaadggacqGHKjYOcaaI1aGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaaIWaGaeyizImQaamOyaiabgsMiJkaaikdaaaa@44AC@

Observa que

A−2I=( 0 1 0 1 0 0 1 0 1 1 1 )     A−3I=( −1 1 −1 1 −1 −1 1 −1 0 1 0 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@5BA9@

Sus potencias tienen siempre un 1 o un -1 en la diagonal, por lo que no pueden ser cero; en consecuencia, el polinomio mínimo no es ni (x-2)^a ni (x-3)^b Es decir, a y b son mayores que 0.

Ahora,

(A−2I) 2 =( 0 0 1 0 0 0 0 0 0 1 2 1 )   MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaafaqabeWadaaabaWaauIhaeaafaqabeWadaaabaGaaGimaaqaaiaaicdaaeaacaaIXaaabaaabaGaaGimaaqaaiaaicdaaeaaaeaaaeaacaaIWaaaaaaaaeaaaeaaaeaaaeaadaqjEaqaauaabeqaciaaaeaacaaIWaaabaGaaGimaaqaaaqaaiaaicdaaaaaaaqaaaqaaaqaaaqaamaaL4babaqbaeqabiGaaaqaaiaaigdaaeaacaaIYaaabaaabaGaaGymaaaaaaaaaaGaayjkaiaawMcaaiaabccacaqGGaaaaa@429C@ (A−2I) 3 =( 0 0 0 0 0 0 0 0 1 3 1 )     (A−3I ) 2 =( 1 −2 1 1 −2 1 1 −2 1 0 0 0 ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadgeacqGHsislcaaIYaGaamysaiaacMcadaahaaWcbeqaaiaaiodaaaGccqGH9aqpdaqadaqaauaabeqadmaaaeaadaqjEaqaauaabeqadmaaaeaacaaIWaaabaGaaGimaaqaaaqaaaqaaiaaicdaaeaacaaIWaaabaaabaaabaGaaGimaaaaaaaabaaabaaabaaabaWaauIhaeaafaqabeGacaaabaGaaGimaaqaaiaaicdaaeaaaeaacaaIWaaaaaaaaeaaaeaaaeaaaeaadaqjEaqaauaabeqaciaaaeaacaaIXaaabaGaaG4maaqaaaqaaiaaigdaaaaaaaaaaiaawIcacaGLPaaacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabIcacaWGbbGaeyOeI0IaaG4maiaadMeacaGGPaWaaWbaaSqabeaacaaIYaaaaOGaeyypa0ZaaeWaaeaafaqabeWadaaabaWaauIhaeaafaqaceWadaaabaGaaGymaaqaaiabgkHiTiaaikdaaeaacaaIXaaabaaabaGaaGymaaqaaiabgkHiTiaaikdaaeaaaeaaaeaacaaIXaaaaaaaaeaaaeaaaeaaaeaadaqjEaqaauaabiqaciaaaeaacaaIXaaabaGaeyOeI0IaaGOmaaqaaaqaaiaaigdaaaaaaaqaaaqaaaqaaaqaamaaL4babaqbaeqabiGaaaqaaiaaicdaaeaacaaIWaaabaaabaGaaGimaaaaaaaaaaGaayjkaiaawMcaaaaa@5F23@

Por tanto, el producto de matrices por bloques da

(A−2I) 3 (A−3I) 2 =(0) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadgeacqGHsislcaaIYaGaamysaiaacMcadaahaaWcbeqaaiaaiodaaaGccaGGOaGaamyqaiabgkHiTiaaiodacaWGjbGaaiykamaaCaaaleqabaGaaGOmaaaakiabg2da9iaacIcacaaIWaGaaiykaaaa@441A@

Comprobad vosotros mismos que

(A−2I) 2 (A−3I) 2 ≠(0) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadgeacqGHsislcaaIYaGaamysaiaacMcadaahaaWcbeqaaiaaikdaaaGccaGGOaGaamyqaiabgkHiTiaaiodacaWGjbGaaiykamaaCaaaleqabaGaaGOmaaaakiabgcMi5kaacIcacaaIWaGaaiykaaaa@44DA@ (A−2I) 3 (A−3I)≠(0) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadgeacqGHsislcaaIYaGaamysaiaacMcadaahaaWcbeqaaiaaiodaaaGccaGGOaGaamyqaiabgkHiTiaaiodacaWGjbGaaiykaiabgcMi5kaacIcacaaIWaGaaiykaaaa@43E8@

En consecuencia, el polinomio mínimo de A es

(x−2) 3 (x−3) 2 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadIhacqGHsislcaaIYaGaaiykamaaCaaaleqabaGaaG4maaaakiaacIcacaWG4bGaeyOeI0IaaG4maiaacMcadaahaaWcbeqaaiaaikdaaaaaaa@3FC9@