If det(A)=0, Ac=0 has only trivial solution

[mzh]

Homework Statement

Show that the equation Ac = 0, where A is a NxN matrix and c is a column matrix with elements c_i, i=1..N can have a non-trivial solution (c != 0) only when det(A) = 0.

Homework Equations

inv(A) does not exist when det(A) = 0.

The Attempt at a Solution

If det(A) != 0, we can form inv(A). Then we can write
inv(A)*A*c=inv(A)*0
and so
c=0,
and so we arrive at the trivial solution. On the contraire, if det(A)=0, we can not form the inverse, and so we can not write
inv(A)*A*c=inv(A)*0, because inv(A) does not exist.
So, there must be other solutions.

Does that make sense? How do I write this in a more formal way?

Thank you for any suggestions

 

[Kreizhn]

What you've said is equivalent to

If Ac = 0 has multiple solutions then \det A = 0.

To prove this, it is logically equivalent to show the contrapositive. Namely

If \det A \neq 0 then Ac = 0 has a unique solution (namely, the trivial solution).

This is what you have done in the first part of your post, so you don't need the second part.

 

[mzh]

What you've said is equivalent to

To prove this, it is logically equivalent to show the contrapositive. Namely

This is what you have done in the first part of your post, so you don't need the second part.

Hey Kreizhn, thanks for your feedback. Ok, i see I'm saying two time the same thing. I was not sure if the first part would allow to conclude the second part. So your second argument is sufficient as a proof, i understand. And this does not require any more statements to be complete (for non-mathematicians).