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hanbinddd

Í­³æ (СÓÐÃûÆø)

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[ Last edited by hanbinddd on 2013-12-4 at 16:05 ]
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hank612

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¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï ¡ï
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feixiaolin: »ØÌûÖö¥ 2013-12-05 15:13:31
feixiaolin: ½ð±Ò+2, ¸ÐлӦÖú 2013-12-05 15:16:03
hanbinddd: ½ð±Ò+10, ¡ï¡ï¡ï¡ï¡ï×î¼Ñ´ð°¸ 2013-12-07 12:07:40
ÒýÓûØÌû:
6Â¥: Originally posted by hanbinddd at 2013-12-05 14:30:33
Õâ¸öûÈ˰ïæÂ𣿣¿

6.1 If N(A^p)=N(A^{p+1}), we can prove that
N(A^{p+1})=N(A^{p+2}).
Èç¹û x is in N(A^{p+2}), ¼´ A^{p+1} (Ax)=0, ´Ó¶ø A^p(Ax)=0, ËùÒÔ
x  is in N(A^{p+1}).

6.2 ÈÎÒâv= A/3 (v) - 1/3*(A-3E) (v), ¶ø A/3 (v) is in N(A-3E) and
- 1/3*(A-3E) (v) is in N(A), ËùÒÔÈ«¿Õ¼ä=N(A-3E) + N(A). Ö±ºÍÊÇÏÔÈ»µÄ£¬ÒòΪ (A+3E)w=0 ÇÒ Aw=0 ÍÆ³ö w=0.

7.1 (A+E)=(4 0 8; 3 0 6; -2 0 -4) ²»µÈÓÚÁ㣬 µ«ÊÇ(A+E)^2=0. ËùÒÔAµÄ×îС¶àÏîʽÊÇ (x+1)^2.
Im(A+E) = span( (A+E)(1 0 0) )= (4 3 -2), ËùÒÔ¿ÉÉè a2=(1 0 0),
a1=(4 3 -2), ÄÇô a3= (0 1 0) ¼´¿É¡£

7.2 Òò´ËJordan ±ê×¼ÐÍΪ (-1 1 0; 0 -1 0; 0 0 -1). ¾ÍÊÇ˵£¬ÓÐÁ½¸öJordan ¿é£¬Ò»¸öÊÇ1*1, Ò»¸öÊÇ2*2´óС¡£
We_must_know. We_will_know.
7Â¥2013-12-05 15:09:43
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû
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feixiaolin

ÈÙÓþ°æÖ÷ (ÎÄ̳¾«Ó¢)

ÓÅÐã°æÖ÷

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2Â¥2013-12-04 18:08:15
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

hanbinddd

Í­³æ (СÓÐÃûÆø)

ÒýÓûØÌû:
2Â¥: Originally posted by feixiaolin at 2013-12-04 18:08:15
½¨ÒéÂ¥Ö÷×·¼Ó5¸ö½ð±Ò¡£

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3Â¥2013-12-04 18:36:30
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hanbinddd

Í­³æ (СÓÐÃûÆø)

Çó½â´ð
4Â¥2013-12-04 20:00:47
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

hanbinddd

Í­³æ (СÓÐÃûÆø)

Çó°ïæ°¡

[ ·¢×ÔСľ³æ¿Í»§¶Ë ]
5Â¥2013-12-04 21:09:34
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hanbinddd

Í­³æ (СÓÐÃûÆø)

Õâ¸öûÈ˰ïæÂ𣿣¿

[ ·¢×ÔСľ³æ¿Í»§¶Ë ]
6Â¥2013-12-05 14:30:33
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

hanbinddd

Í­³æ (СÓÐÃûÆø)

ÒýÓûØÌû:
7Â¥: Originally posted by hank612 at 2013-12-05 15:09:43
6.1 If N(A^p)=N(A^{p+1}), we can prove that
N(A^{p+1})=N(A^{p+2}).
Èç¹û x is in N(A^{p+2}), ¼´ A^{p+1} (Ax)=0, ´Ó¶ø A^p(Ax)=0, ËùÒÔ
x  is in N(A^{p+1}).

6.2 ÈÎÒâv= A/3 (v) - 1/3*(A-3E) (v), ¶ø ...

ºÃÏñ²»¹»Í¨Ë×Ò×¶®°¡

[ ·¢×ÔСľ³æ¿Í»§¶Ë ]
8Â¥2013-12-05 23:34:09
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

feixiaolin

ÈÙÓþ°æÖ÷ (ÎÄ̳¾«Ó¢)

ÓÅÐã°æÖ÷

6.1 If N(A^p)=N(A^{p+1}), we can prove that¡¾ÎÒÃÇÄÜÖ¤Ã÷£¬=¿ÉÍÆÖª¡¿
N(A^{p+1})=N(A^{p+2}).
Èç¹û x is in¡¾ÊôÓÚ¡¿ N(A^{p+2}), ¼´ A^{p+1} (Ax)=0, ´Ó¶ø A^p(Ax)=0, ËùÒÔ
x  is in¡¾ÊôÓÚ¡¿ N(A^{p+1}).
Õâ¸öÓ¢Óï²»ÄÑ¡£
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9Â¥2013-12-06 09:46:18
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