题目
证明关于向量x的方程A*Ax=A*b一定有解.
其中A是一个相应的矩阵,A*是A的转置(上标打不出T),b是一个相应的向量.求英文详细证明.
其中A是一个相应的矩阵,A*是A的转置(上标打不出T),b是一个相应的向量.求英文详细证明.
提问时间:2021-01-02
答案
Let T(A) be the transpose of the matrix A.
In order to prove that T(A) Ax = T(A) b has a non-trivial solution,we need to show that T(A) A is invertible.
[ | | | | ]
Assume A has the form:A = [ a_1 a_2 a_3 ...a_n ],a_1,a_2,a_3 ...a_n are the
[ | | | | ]
column vectors of A.
[ - a_1 - ]
Then T(A) = [ - a_2 - ]
[ .]
[ - a_n -]
thus by definition:
[ - a_1 - ]
T(A)A = [ - a_2 - ] [ | | | | ]
[ .] [ a_1 a_2 a_3 ...a_n ]
[ - a_n -] [ | | | | ]
[ a_1 * a_1 a_1 * a_2 ...a_1 * a_n ]
= [ a_2 * a_1 a_2 * a_2 ...a_2 * a_n ]
[ .]
[ a_n * a_1 a_n * a_2 ...a_n * a_n ]
If the column vectors of A are linearly independent,then there does not exist real numbers h and f such that:
a_i = h * a_j + f * a_k,for any i not equal to j and i not equal to k.
Hence,we can further investigate the entries in T(A)A:since a_i and a_j are always linearly independent,then it follows that each column of T(A)A must also be linearly independent,otherwise the column vectors of A would be linearly dependent.
This indicates that T(A)A is invertible,thus we have one unique solution.
In the other case,if the column vectors of A is linearly dependent,then by the same reason,column vectors of T(A)A is also linearly dependent,then we have a infinite many solutions.
Thus,in general the system T(A)Ax = T(A)b always have solutions.
In order to prove that T(A) Ax = T(A) b has a non-trivial solution,we need to show that T(A) A is invertible.
[ | | | | ]
Assume A has the form:A = [ a_1 a_2 a_3 ...a_n ],a_1,a_2,a_3 ...a_n are the
[ | | | | ]
column vectors of A.
[ - a_1 - ]
Then T(A) = [ - a_2 - ]
[ .]
[ - a_n -]
thus by definition:
[ - a_1 - ]
T(A)A = [ - a_2 - ] [ | | | | ]
[ .] [ a_1 a_2 a_3 ...a_n ]
[ - a_n -] [ | | | | ]
[ a_1 * a_1 a_1 * a_2 ...a_1 * a_n ]
= [ a_2 * a_1 a_2 * a_2 ...a_2 * a_n ]
[ .]
[ a_n * a_1 a_n * a_2 ...a_n * a_n ]
If the column vectors of A are linearly independent,then there does not exist real numbers h and f such that:
a_i = h * a_j + f * a_k,for any i not equal to j and i not equal to k.
Hence,we can further investigate the entries in T(A)A:since a_i and a_j are always linearly independent,then it follows that each column of T(A)A must also be linearly independent,otherwise the column vectors of A would be linearly dependent.
This indicates that T(A)A is invertible,thus we have one unique solution.
In the other case,if the column vectors of A is linearly dependent,then by the same reason,column vectors of T(A)A is also linearly dependent,then we have a infinite many solutions.
Thus,in general the system T(A)Ax = T(A)b always have solutions.
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