$$\det(A^T) = \det(A)$$
Using the geometric definition of the determinant as the area spanned by the columns, could someone give a geometric interpretation of the property?
Linear Algebra – Geometric Interpretation of $\det(A^T) = \det(A)$
determinantgeometric-interpretationlinear algebramatricesvector-spaces
Related Solutions
Of course this theorem has a geometric interpretation! In a sense, it's a multidimensional analogue of «the volume of a parallelepiped is the product of the area of its base and its height».
3. Let's start with $3\times3$ case: $$ \left|\begin{matrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1&w_2&w_3\end{matrix}\right|= u_1\left|\begin{matrix}v_2&v_3\\w_2&w_3\end{matrix}\right| -u_2\left|\begin{matrix}v_1&v_3\\w_1&w_3\end{matrix}\right| +u_3\left|\begin{matrix}v_1&v_2\\w_1&w_2\end{matrix}\right|. $$ LHS is the volume of the parallelepiped spanned by three vectors, $u$, $v$ and $w$. What's the meaning of RHS? Clearly that's a scalar product of $u$ with something — namely, with the vector $$ \left(\left|\begin{matrix}v_2&v_3\\w_2&w_3\end{matrix}\right|, -\left|\begin{matrix}v_1&v_3\\w_1&w_3\end{matrix}\right|,\left|\begin{matrix}v_1&v_2\\w_1&w_2\end{matrix}\right|\right)= \left|\begin{matrix}\overrightarrow{e_1}&\overrightarrow{e_2}&\overrightarrow{e_3}\\v_1&v_2&v_3\\w_1&w_2&w_3\end{matrix}\right| $$ — i.e. with vector product of $v$ and $w$.
So the formula we get is $vol\langle u,v,w\rangle=(u,[v,w])$; now by the (geometrical) definition of scalar product it's $area\langle v,w\rangle\cdot (|u|\cdot\sin\phi)$, and the first factor is the area of the base and the second one is the height of our parallelepiped.
n. Consider the (general) case of vectors in $n$-dimensional space $V$. In RHS of the theorem we again see a scalar product of the first vector, $v$, with a vector $B$ (in coordinate-free language it really lives in $\Lambda^{n-1}V$, but let's ignore this for now) with coordinates $C_{1i}$.
The question is, what is the geometric meaning of $B$. Let me give 3 (closely related) answers.
- By the very same cofactor theorem it measures the [(n-1)-dimensional] area of projection of the base of our $n$-parallelepiped (i.e. $(n-1)$-parallelepiped spanned by all vectors but $v$) on different hyperplanes; more precisely, the area of the projection on the hyperplanes orthogonal to a unit vector $v$ is the scalar product $(B,v)$.
- Let's prove the cofactor theorem instead of using it. The function $(B,x)$ is linear in $x$. For a basis vector $x=e_i$ we have $(B,x)=C_{1i}$, which (up to sign, at least) is the area of the span of projections of our vectors on the hyperplane orthogonal to $e_i$. So $(B,x)$ is indeed the area of the projection of the base on the hyperplane orthogonal to $x$ (multiplied by $|x|$ and taken with appropriate signs).
- Even better, since everything is invariant under (special) orthogonal transforms, let's change basis to make $v$ a scalar multiple of $e_1$. Now the statement «$(B,v)$ is the $|v|$ times the area of the projection» became obvious (we literally multiply $|v|$ by the cofactor manifestly equal to this area — well, it was discussed in (2) anyway).
Now I must admit the statement we get is more like «the volume of a parallelepiped $\langle u,\text{base}\rangle$ is the product of the length of $u$ and the area of the projection of its base on the hyperplane orthogonal to $u$» — but it's of course equivalent to «the volume of a parallelepiped is the product of the area of its base and its height».
Let $P$ be the paralellogram spanned by the vectors $a$ and $b$ and let $\phi$ be the angle between $a$ and $b$. Then we have \begin{eqnarray*} (Area(P))^2 & = & |a|^2|b|^2(\sin(\phi))^2 \\ & = & |a|^2|b|^2-|a|^2|b|^2(\cos(\phi))^2 \\ & = &|a|^2|b|^2-(a\cdot b)^2 \\ & = & (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)-(a_1b_1+a_2b_2+a_3b_3)^2\\ & = &(a_2b_3-a_3b_2)^2+(a_3b_1-a_1b_3)^2+(a_1b_2-a_2b_1)^.2 \end{eqnarray*} So the area of $P$ is equal to the length of the vector $(a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1) = a \times b$. With the coordinates you gave.
Best Answer
A geometric interpretation in four intuitive steps....
The Determinant is the Volume Change Factor
Think of the matrix as a geometric transformation, mapping points (column vectors) to points: $x \mapsto Mx$. The determinant $\mbox{det}(M)$ gives the factor by which volumes change under this mapping.
For example, in the question you define the determinant as the volume of the parallelepiped whose edges are given by the matrix columns. This is exactly what the unit cube maps to, so again, the determinant is the factor by which the volume changes.
A Matrix Maps a Sphere to an Ellipsoid
Being a linear transformation, a matrix maps a sphere to an ellipsoid. The singular value decomposition makes this especially clear.
If you consider the principal axes of the ellipsoid (and their preimage in the sphere), the singular value decomposition expresses the matrix as a product of (1) a rotation that aligns the principal axes with the coordinate axes, (2) scalings in the coordinate axis directions to obtain the ellipsoidal shape, and (3) another rotation into the final position.
The Transpose Inverts the Rotation but Keeps the Scaling
The transpose of the matrix is very closely related, since the transpose of a product is the reversed product of the transposes, and the transpose of a rotation is its inverse. In this case, we see that the transpose is given by the inverse of rotation (3), the same scaling (2), and finally the inverse of rotation (1).
(This is almost the same as the inverse of the matrix, except the inverse naturally uses the inverse of the original scaling (2).)
The Transpose has the Same Determinant
Anyway, the rotations don't change the volume -- only the scaling step (2) changes the volume. Since this step is exactly the same for $M$ and $M^\top$, the determinants are the same.