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.
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Assume A has the form:A = [ a_1 a_2 a_3 ...a_n ],a_1,a_2,a_3 ...a_n are the
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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.