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

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In summary, 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. This is equivalent to showing that if det(A) is not equal to 0, then Ac = 0 has a unique solution, which is the trivial solution. Thus, if det(A) is not equal to 0, then we can form inv(A) and write inv(A)*A*c=inv(A)*0, leading to the trivial solution. However, if det(A) is equal to 0, then we cannot form the inverse and there must be
  • #1
mzh
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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
 
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  • #2
What you've said is equivalent to
If [itex] Ac = 0 [/itex] has multiple solutions then [itex] \det A = 0 [/itex].
To prove this, it is logically equivalent to show the contrapositive. Namely
If [itex] \det A \neq 0 [/itex] then [itex] Ac = 0 [/itex] 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.
 
  • #3
Kreizhn said:
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).
 

1. What does it mean when det(A) = 0?

When the determinant of a matrix A is equal to 0, it means that the matrix is singular and does not have an inverse. This also indicates that the matrix is not full rank, meaning that it has at least one linearly dependent row or column.

2. How does det(A) = 0 relate to the solution of the equation Ac = 0?

If det(A) = 0, it implies that the matrix A does not have an inverse. This means that the equation Ac = 0 has either no solution or infinitely many solutions. In this case, the only solution to the equation will be the trivial solution, where all the variables are equal to 0.

3. What is the significance of the trivial solution in the equation Ac = 0?

The trivial solution in the equation Ac = 0 is the solution where all the variables are equal to 0. This solution is significant because it indicates that the matrix A is singular and does not have an inverse. It also implies that the columns of A are linearly dependent, and therefore, the system of equations is either inconsistent or has infinitely many solutions.

4. Can det(A) = 0 and Ac = 0 have non-trivial solutions?

No, if det(A) = 0, it means that the only solution to the equation Ac = 0 is the trivial solution. This is because a non-trivial solution would require the matrix A to have an inverse, which is not possible when det(A) = 0.

5. How does the value of det(A) affect the solutions to the equation Ac = 0?

If det(A) = 0, it indicates that the matrix A is singular and does not have an inverse. This means that the equation Ac = 0 will either have no solution or infinitely many solutions. In both cases, the only solution will be the trivial solution, where all the variables are equal to 0. On the other hand, if det(A) ≠ 0, the matrix A is non-singular and has an inverse, and the equation Ac = 0 will have a unique non-trivial solution.

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