How to prove that the determinant of K is also zero without using eigenvalues?

[EmmaSaunders1]

Hi,

if vector b * matrix K = 0 (bK=o) what methods can one use to show that the determinant of K is therefore also zero, without using eigenvalues.

I have a feeling I am over complicating this.

Knd regards

Emma

 

[pmsrw3]

I assume b != 0. What else are you starting with? Are you allowed to use the fact that, for any square matrix K with determinant != 0, the equation Kx = y has a unique solution? If so, it is obvious that the solution to K^Tb = 0 is not unique, since K^T0 = 0 also. Thus, the determinant of K^T is 0, and since transposition doesn't change the determinant, the determinant of K must also be 0.

 

[EmmaSaunders1]

Sorry that's correct that b!=0

using for any square matrix K with determinant != 0, the equation Kx = y has a unique solution - is fine.

Im almost understand where your heading with this - would you please however clarify two steps;

you say KTb = 0 is not unique, since KT0 = 0 also. Thus, the determinant of KT is 0.

How do you know the value of the determinant from that statement alone,

Also is bK, the same as KTb, I noticed you switched the order of multiplication there.

Thanks for your help

 

[pmsrw3]

Sorry, I should have written K^Tb^T. I'm using the turnover rule for matrix multiplication: for any two matrices A and B, (AB)^T = B^TA^T. In this case b is a row vector, a 1 x n matrix, and b^T is a column vector, an n x 1 matrix. The only reason for doing it that way is that the uniqueness theorem is usually stated in terms of left multiplication and column vectors, but of course it also holds for right multiplication and row vectors.

you say KTb = 0 is not unique, since KT0 = 0 also. Thus, the determinant of KT is 0.

How do you know the value of the determinant from that statement alone

If the determinant of K (or K^T) is not 0, the solution would be unique. Since the solution is not unique, the determinant must be 0.

 

[EmmaSaunders1]

Thats great - I really appreciate your help!

 

[bpet]

It can be proved directly using Kramer's rule. Write K=[k1;...;kn] (the kj are the rows) and let Kj be the matrix K with row j replaced with 0. Obviously det(Kj)=0 by row expansion.

Also 0=b*K=b1*k1+...+bn*kn, so det(Kj)=det([k1;...;b1*k1+...+bn*kn;...;kn])=det([k1;...;bj*kj;...;kn])=bj*det(K). Choose any j such that bj<>0 and this shows that det(K)=0.