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I'm curious about is there any geometric property relative to negative value for determinant of matrix. $$\det{(A)} < 0$$ I knew about some of determinant of matrix properties as following, but it seems to me that it is nothing relative to negative value of determinant of matrix

$$\det{(AB)}=\det{(A)}det{(B)}(Multiplicative)$$ $$\det{(A)} = \textit{0} \iff \textit{A is singular}$$

$$ M_{2,2}=\begin{bmatrix} a&b\\ c&d\\ \end{bmatrix} $$ $$\lvert \det({M_{2 \times 2}}) \rvert = \lvert ad-bd \rvert=\textit{volumn of parallelogram} $$ $$\lvert \det({M_{n\times n})}\rvert=\prod_{j=1}^n a_{i,j}(-1)^{i+j}\det({M_{i,j}}) \quad \textit{expansion of determinant alone the }\textit{i}^{th}\textit{ row} $$

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  • $\begingroup$ the key word is signed area, and in ${\Bbb{R}}^3$ signed volume. $\endgroup$
    – janmarqz
    Feb 2, 2014 at 0:34
  • $\begingroup$ what you mean by "signed area"? $\endgroup$
    – bsdshell
    Feb 2, 2014 at 0:37
  • $\begingroup$ ok, I see, but what is the geometric meaning of negative area/volume? I think this is what I'm looking for $\endgroup$
    – bsdshell
    Feb 2, 2014 at 0:40
  • $\begingroup$ depends on the order of the factors to assign one of two orientations for the same absolute valued area $\endgroup$
    – janmarqz
    Feb 2, 2014 at 1:08

1 Answer 1

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The geometric property of $\pm 1$ signed determinant matrices is the orientation of the vectors of the matrix in its given space(for e.g $\mathbb{R}^{n}$) with respect to a fixed orientation of the basis vectors. If the vectors of your matrix $A$ has the same orientation as the basis then $sign(det(A))=+1$ otherwise its $-1$.

For an example, take the matrix $A=(e_{2},e_{1},e_{3})=((0,1,0),(1,0,0),(0,0,1))$ in $\mathbb{R}^{3}$. Then think of the orientation of the standard basis axes, $(e_{1},e_{2},e_{3})$, in $\mathbb{R}^{3}$ to be $\it{clockwise} $ such that if you drew an equilateral triangle in the $\mathbb{R}^{3}$ plane intersecting $(e_{1},e_{2},e_{3})$, then to get from $e_{1}$ to $e_{2}$ to $e_{3}$ you would have to travel around the triangle in a clockwise rotation. If you perform the same triangle exercise with the vector $(e_{2},e_{1},e_{3})$ you would see that you move $counterclockwise$ around the triangle. Thus, the matrix $A=(e_{2},e_{1},e_{3})$ has the opposite orientation as the standard basis and $sign(det(A))=-1$.

This also explains why when you swap two columns/rows of a matrix the sign of the determinant changes with each swap. Unfortunately though, this geometric idea is a bit more difficult to understand in higher dimensions.

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  • $\begingroup$ This is very elegant/simple way to expand your the idea, so I assume that relative to permutation group which is $\textbf{S_n}$, $\endgroup$
    – bsdshell
    Feb 2, 2014 at 1:35
  • $\begingroup$ Yes, the sign of the determinant is very closely related to permutations in $S_{n}$. $\endgroup$
    – h4nusGT
    Feb 2, 2014 at 10:15

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