楼主: tianyu0401
1740 1

求教一道线性代数的题目 [推广有奖]

  • 0关注
  • 0粉丝

VIP1

芰荷居士

硕士生

77%

还不是VIP/贵宾

-

威望
0
论坛币
1758 个
通用积分
0
学术水平
0 点
热心指数
0 点
信用等级
0 点
经验
754 点
帖子
55
精华
1
在线时间
184 小时
注册时间
2005-2-25
最后登录
2017-7-7

楼主
tianyu0401 发表于 2009-9-18 23:21:49 |AI写论文

+2 论坛币
k人 参与回答

经管之家送您一份

应届毕业生专属福利!

求职就业群
赵安豆老师微信:zhaoandou666

经管之家联合CDA

送您一个全额奖学金名额~ !

感谢您参与论坛问题回答

经管之家送您两个论坛币!

+2 论坛币
Suppose X is an n × K matrix with full column rank of K. Show that
X的转秩乘以X is nonsigular (invertible).
二维码

扫码加我 拉你入群

请注明:姓名-公司-职位

以便审核进群资格,未注明则拒绝

关键词:线性代数 Suppose matrix column colum 题目 线性代数 求教

回帖推荐

sherman_shi 发表于2楼  查看完整内容

Suppose A is an n × K matrix with full column rank of K. Show that A'A (K×K) is nonsigular (invertible), which means Rank(A'A)=Rank(A)=K. Prove equality of their null spaces. Null space of the A'A matrix is given by vectors x for which A'Ax=0. If this condition is fulfilled, also holds 0=x'A'Ax=0, which means Ax=0. On the other hand, if Ax=0, then A'Ax=0. So, their null spaces are same ...

本帖被以下文库推荐

沙发
sherman_shi 发表于 2009-9-19 00:08:07
Suppose A is an n × K matrix with full column rank of K. Show that
A'A (K×K) is nonsigular (invertible), which means Rank(A'A)=Rank(A)=K.

Prove equality of their null spaces. Null space of the A'A matrix is given by vectors x for which A'Ax=0. If this condition is fulfilled, also holds 0=x'A'Ax=0, which means Ax=0.

On the other hand, if Ax=0, then A'Ax=0.

So, their null spaces are same, and therefore Rank(A'A)=Rank(A)=K.
已有 1 人评分经验 论坛币 收起 理由
胖胖小龟宝 + 10 + 10 热心帮助其他会员

总评分: 经验 + 10  论坛币 + 10   查看全部评分

您需要登录后才可以回帖 登录 | 我要注册

本版微信群
加好友,备注jltj
拉您入交流群
GMT+8, 2025-12-5 22:48