Linear Algebra Proof: Skew Symmetric Matrix and Odd Number Determinant

Click For Summary

Homework Help Overview

The discussion revolves around proving that the determinant of a skew symmetric matrix is zero when the matrix is of odd order. The original poster presents their understanding of the properties of determinants in relation to skew symmetric matrices.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to connect the determinant of a skew symmetric matrix with its transpose and the negative of the matrix, but expresses confusion about how these properties lead to a conclusion about the determinant being zero. Other participants suggest that the original poster has already outlined the proof but may have misunderstood the implications of the properties discussed.

Discussion Status

Participants are exploring the implications of the determinant properties of skew symmetric matrices. Some guidance has been offered, indicating that the original poster's reasoning may be on the right track, but there remains a lack of consensus on the understanding of the proof.

Contextual Notes

There is an emphasis on the properties of determinants related to skew symmetric matrices, particularly in the context of odd dimensions. The original poster expresses uncertainty about their calculations and interpretations, which may be affecting their understanding of the problem.

_Bd_
Messages
107
Reaction score
0

Homework Statement



If A is a skew symetric matrix (such that A^T = -A)
and A is an nxn matrix with n being an odd number proove that det(A) = 0


The Attempt at a Solution



all I can think of is
det(A) = det(A^T)
letting c = -1 det(cA) = c^ndet(A)

but I can't get anymore connections to proove this. . .I tried doing some random example of a 3x3 skew symetric matrix but I didnt get a det=0. . .so I have no clue how to go about this problem!
 
Physics news on Phys.org
If you took the determinant of a 3x3 skew symmetric matrix and didn't get zero, then you made a mistake. You basically just wrote down the proof. det(A)=det(A^T)=det(-A)=(-1)^n*det(A). Doesn't that show det(A)=0?
 
Dick said:
If you took the determinant of a 3x3 skew symmetric matrix and didn't get zero, then you made a mistake. You basically just wrote down the proof. det(A)=det(A^T)=det(-A)=(-1)^n*det(A). Doesn't that show det(A)=0?

i don't know. . .I don't see it

det(A)=det(A^T) which means just that if the determinant of A is 4 the determinant of A^T is 4

and the det(-A) =(-1)^n * det(A) which just doesn't mean anything . . .or I don't see it meaning anything? cause i don't know I am thinking about it as say the det of some matrix is 4 then the determinant of the negative of that matrix is 4 * (-1)^n . . .which I still don't see it as a zero. . .?
or maybe my train of thought is wrong? i don't know maybe I did make a mistake in my calculator
 
_Bd_ said:
i don't know. . .I don't see it

det(A)=det(A^T) which means just that if the determinant of A is 4 the determinant of A^T is 4

and the det(-A) =(-1)^n * det(A) which just doesn't mean anything . . .or I don't see it meaning anything? cause i don't know I am thinking about it as say the det of some matrix is 4 then the determinant of the negative of that matrix is 4 * (-1)^n . . .which I still don't see it as a zero. . .?
or maybe my train of thought is wrong? i don't know maybe I did make a mistake in my calculator

The point is that (-1)^n=(-1) if n is odd. Since you then have det(A)=(-det(A)), what's the only possible value for det(A)?
 
oooh! now I get it... thanks!
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
4K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
11K