Ejemplo de matriz 5x4 de rango 3
A=
(
1
2
3
4
5
0
1
1
6
2
4
5
−1
7
5
3
5
9
9
8
)
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8qrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeaabaWaaqaafaaakeaadaqadaqaauaabiqafqaaaaaabaGaaGymaaqaaiaaikdaaeaacaaIZaaabaGaaGinaaqaaiaaiwdaaeaacaaIWaaabaGaaGymaaqaaiaaigdaaeaacaaI2aaabaGaaGOmaaqaaiaaisdaaeaacaaI1aaabaGaeyOeI0IaaGymaaqaaiaaiEdaaeaacaaI1aaabaGaaG4maaqaaiaaiwdaaeaacaaI5aaabaGaaGyoaaqaaiaaiIdaaaaacaGLOaGaayzkaaaaaa@4A17@
1. Consideramos
el siguiente menor no nulo: M2=
|
1
2
5
0
|
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8qrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeaabaWaaqaafaaakeaadaabdaqaauaabeqaciaaaeaacaaIXaaabaGaaGOmaaqaaiaaiwdaaeaacaaIWaaaaaGaay5bSlaawIa7aaaa@3EC4@
.
2.
Fijamos la
fila 3.
Orlamos
con la fila 3 y la columna 3:
|
1
2
3
5
0
1
6
2
4
|=0
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8qrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeaabaWaaqaafaaakeaadaabdaqaauaabiqadmaaaeaacaaIXaaabaGaaGOmaaqaaiaaiodaaeaacaaI1aaabaGaaGimaaqaaiaaigdaaeaacaaI2aaabaGaaGOmaaqaaiaaisdaaaaacaGLhWUaayjcSdGaeyypa0JaaGimaaaa@443E@
Orlamos
con la fila 3 y la columna 4:
|
1
2
4
5
0
1
6
2
5
|=0
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8qrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeaabaWaaqaafaaakeaadaabdaqaauaabiqadmaaaeaacaaIXaaabaGaaGOmaaqaaiaaisdaaeaacaaI1aaabaGaaGimaaqaaiaaigdaaeaacaaI2aaabaGaaGOmaaqaaiaaiwdaaaaacaGLhWUaayjcSdGaeyypa0JaaGimaaaa@4440@
Como todos los menores obtenidos al usar la
fila 3 son nulos, podemos prescindir de dicha fila para cualquiera de los
cálculos siguientes.
Continuamos con otra fila:
Fijamos
la fila 4.
Orlamos
con la fila 4 y la columna 3:
|
1
2
3
5
0
1
−1
7
5
|=46
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8qrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeaabaWaaqaafaaakeaadaabdaqaauaabiqadmaaaeaacaaIXaaabaGaaGOmaaqaaiaaiodaaeaacaaI1aaabaGaaGimaaqaaiaaigdaaeaacqGHsislcaaIXaaabaGaaG4naaqaaiaaiwdaaaaacaGLhWUaayjcSdGaeyypa0JaaGinaiaaiAdaaaa@45F0@
Puesto
que este menor, M3, es no nulo, el rango de A es al menos 3.
3. Repetimos el proceso con el menor M3:
Fijamos
la fila 4 (recordar que la fila 3 ya no interviene).
Orlamos
con la fila 4 y la columna 4:
|
1
2
3
4
5
0
1
1
−1
7
5
3
5
9
9
8
|=0
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8qrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeaabaWaaqaafaaakeaadaabdaqaauaabiqaeqaaaaaabaGaaGymaaqaaiaaikdaaeaacaaIZaaabaGaaGinaaqaaiaaiwdaaeaacaaIWaaabaGaaGymaaqaaiaaigdaaeaacqGHsislcaaIXaaabaGaaG4naaqaaiaaiwdaaeaacaaIZaaabaGaaGynaaqaaiaaiMdaaeaacaaI5aaabaGaaGioaaaaaiaawEa7caGLiWoacqGH9aqpcaaIWaaaaa@4A72@
Con esto quedan agotadas las posibilidades
para orlar M3. Por tanto, concluimos que el rango de A es 3.