# I'm curious about is there any geometric property relative to negative value for determinant of matrix.

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

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

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

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

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