Determinant of -A

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diogomgf

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How can I prove that the determinant of the matrix [MATH]-A, \in M_{n}[/MATH], [MATH]det(-A)[/MATH] is positive when [MATH]n[/MATH] is even and negative when [MATH]n[/MATH] is odd?
 
Do you think that the problem has anything to do with the sign in from of the A?

What have you tried? Are you stuck somewhere? Have you even tried to see if statement is even true???? Have you considered doing it by induction? I'll ask again, have you tried to see if it is true?? Does the sign in front of the A really matter. Hint: It could be the case that matrix B = -A.
 
Jomo said:
Do you think that the problem has anything to do with the sign in from of the A?

What have you tried? Are you stuck somewhere? Have you even tried to see if statement is even true???? Have you considered doing it by induction? I'll ask again, have you tried to see if it is true?? Does the sign in front of the A really matter. Hint: It could be the case that matrix B = -A.
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I have not tried induction yet. What I'm trying to prove is actually: If [MATH]B = -A[/MATH] then [MATH]det(B) = (-1)^{n} \cdot detA[/MATH].
 
diogomgf said:
I have not tried induction yet. What I'm trying to prove is actually: If [MATH]B = -A[/MATH] then [MATH]det(B) = (-1)^{n} \cdot detA[/MATH].
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Are you saying that det(A) is always positive? Hmm, then wouldn't that make det(B) positive. But wait let(B) = (-1)ndet(A). And the right hand side changes from positive to negative based on (-1)n if we assume that set(A) is always positive. This scenario is all possible and if it is then determinant of any matrix is 0. Do you see this logic?

Why not take my hint that maybe this fact is not true at all. That is why I stated it twice. You can easily see this if you look at the determinants of diagonal matrices of various sizes with just 1's and -1's on the diagonal.
 
diogomgf said:
What I'm trying to prove is actually: If [MATH]B = -A[/MATH] then [MATH]det(B) = (-1)^{n} \cdot detA[/MATH].
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The issue here is if you do prove the above is true (and you should state that B=-A) how would that help prove your theorem which you posted.

Look, given any matrix B you can obtain a matrix -A which equals B (just factor out a -1 from B). Now I do agree that det(B) = (-1)ndet(A). My concern that the sign of det(B) is independent of the n. That is if n is odd then det(B) can be positive! But then since det(B) = (-1)ndet(A) we have that (-1)ndet(A) will be positive even though n is odd. You really do not see this???
 
diogomgf said:
How can I prove that the determinant of the matrix [MATH]-A, \in M_{n}[/MATH], [MATH]det(-A)[/MATH] is positive when [MATH]n[/MATH] is even and negative when [MATH]n[/MATH] is odd?
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If you want to do this by induction then go for it. But there is a simpler way.

Consider [math]A = \left ( \begin{matrix} 3 & 3 & 1 \\ 2 & 0 & -3 \\ 0 & -2 & 1 \end{matrix} \right )[/math]
What is det(A)?

(The negative signs in LaTeX are doing weird things. Bear with me.)
Let's define [math]B = \left ( \begin{matrix} -3 & -3 & -1 \\ 2 & 0 & -3 \\ 0 & -2 & 1 \end{matrix} \right )[/math] <-- the top line is -3, -3, -1

What is det(B)?

Let's define [math]C = \left ( \begin{matrix} -3 & -3 & -1 \\ -2 & 0 & 3 \\ 0 & -2 & 1 \end{matrix} \right )[/math] <-- the middle line is -2, 0, 3

What is det(C)?

And finally,
[math]-A = \left ( \begin{matrix} -3 & -3 & -1 \\ -2 & 0 & 3 \\ 0 & 2 & -1 \end{matrix} \right )[/math] <-- the bottom line is 0, 2, -1

What is det(-A)?

Do you see the pattern?

-Dan
 
Dan, I am sorry but I think that you are wrong in thinking that this theorem is true.

Lets just look at -A which is a diagonal 3x3 matrix where the numbers on the diagonal are 1,-1 and -1. det(-A) = 1 which is positive even though n is odd.

Suppose -A is a 4x4 diagonal matrix with -1,-1,-1 and 1 on the diagonal. det(-A) = -1 which is negative even though n is even.

Am I missing something? The theorem would be true if det(A) >0 but that was not given.
 
Dan,
OK what you are showing is true. I just don't agree that when you go from A to -A you just change the signs of 1 row. I do agree that if you have a matrix A and then you multiply any row (or column) by -1 then the det of both matrices will differ by a sign. That is absolutely true.

To me, -A is the matrix which you add to A to get the zero matrix. Am I wrong????
 
Jomo said:
Dan,
OK what you are showing is true. I just don't agree that when you go from A to -A you just change the signs of 1 row. I do agree that if you have a matrix A and then you multiply any row (or column) by -1 then the det of both matrices will differ by a sign. That is absolutely true.

To me, -A is the matrix which you add to A to get the zero matrix. Am I wrong????
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My intent was to walk through the problem by demonstrating that if you change the signs on one row (or column) that we multiply the determinant by -1. So if you have a matrix with an odd rank then [math]det(-A) = (-1)^{odd power} det(A)[/math], from which the theorem follows.

-Dan
 
topsquark said:
My intent was to walk through the problem by demonstrating that if you change the signs on one row (or column) that we multiply the determinant by -1. So if you have a matrix with an odd rank then [math]det(-A) = (-1)^{odd power} det(A)[/math], from which the theorem follows.

-Dan
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Dan, I think that you are being a little sloppy. I agree that every time you change the signs of one row that the determinant changes its sign. We both agree with this. But the problem does that say that det(A) = det(-A) if n ie even and det(A) = - det(-A) if n is odd. This only works if det(-A)>=0, not when det(-A)<0
 
Jomo said:
Dan, I think that you are being a little sloppy. I agree that every time you change the signs of one row that the determinant changes its sign. We both agree with this. But the problem does that say that det(A) = det(-A) if n ie even and det(A) = - det(-A) if n is odd. This only works if det(-A)>=0, not when det(-A)<0
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Okay, you've completely lost me. Why does it matter if the determinant of A is greater than or less that 0? I'm willing to admit that I might be wrong but I have no idea where.

Sticking to n = 3, for example,
[math]| A | = \left | \left ( \begin{matrix} a & b & c \\ d & e & f \\ g & h & k \end{matrix} \right ) \right |[/math]
[math]= a(ek - fh) - b(dk - fg) + c(dh - eg)[/math]
[math] | - A | = \left | \left ( \begin{matrix} -a & -b & -c \\ -d & -e & -f \\ -g & -h & -k \end{matrix} \right ) \right |[/math] (What is with the LaTeX tonight?)

[math]= (-a)((-e)(-k) - (-f)(-h)) - (-b)((-d)(-k) - (-f)(-g)) + (-c)((-d)(-h) - (-e)(-g))[/math]
[math]= (-a)(ek - fh) - (-b)(dk - fg) + (-c)(dh - eg) = -( a(ek - fh) - b(dk - fg) + c(dh - eg)) = -|A|[/math]
Where does the condition on the positivity or negativity of |-A| come in?

-Dan
 
I might be missing something completely obvious but here goes.

You asked Where does the condition on the positivity or negativity of |-A| come in?

The OP asked how to show that det(−A) is positive when n is even and negative when n is odd.
 
Jomo said:
I might be missing something completely obvious but here goes.

You asked Where does the condition on the positivity or negativity of |-A| come in?

The OP asked how to show that det(−A) is positive when n is even and negative when n is odd.
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My bad. The original question should have been: show that If [MATH]B=−A[/MATH] then [MATH]det(B)=(−1)^{n}⋅detA.[/MATH]
 
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