Geometric Interpretation of Determinant of Transpose

Question

Below are two well-known statements regarding the determinant function:

1. When $A$ is a square matrix, $\det(A)$ is the signed volume of the parallelepiped whose edges are columns of $A$.
2. When $A$ is a square matrix, $\det(A) = \det(A^T)$.

It appears to me that 1 is how $\det$ got invented in the first place. The multilinearity and alternating property are natural consequences. Cramer's rule can also be easily understood geometrically. However, 1 does not immediately yield 2.

Of course the proof of 2 can be done algebraically, and it's not wrong to say that the algebra involved does come from 1. My question is, can 2 be viewed more geometrically? Can it feed us back a better geometric interpretation?

[How would one visualize the row vectors of $A$ from the column vectors? And how would one think, geometrically, about the equality of the signed volume spanned by two visually different parallelepipeds?]

Edit: If you are going to use multiplicativity of the determinant, please provide a geometric intuition to that too. Keep reading to see what I mean.

Here is one of my failed attempts. I introduce a different interpretation:

3. $\det(A)$ is the ratio of the volume of an arbitrary parallelepiped changed by applying $A^T$ (left multiplication), viewed as a linear transformation.

The reason I say applying $A^T$ instead of applying $A$ is that I want the transpose to appear somewhere, and coincidentally the computation of $A^T u$ can be thought of geometrically in terms of columns of $A$: $A^T u$ is the vector whose components are dot products of columns of $A$ with $u$. (I'll take for granted the validity of the word arbitrary in 3 for the moment.) What is left to argue is the equality of 1 and 3, which I have not been able to do geometrically.

Is This a Duplicate Question? In a sense yes, and in a sense no. I did participate in that question a long time ago. There was no satisfactory answer. The accepted answer does not say what a "really geometric" interpretation of (*) should be. Yes you can always do certain row and column operations to put a matrix into a diagonal form without affecting the determinant (with an exception of a possible sign flip), and by induction, one can prove $\det(A) = \det(A^T)$. This, however, seems to me more algebraic than geometric even though each step can be thought of as geometric. I am aware that the property of an interpretation being "geometric" or "algebraic" is not a rigorous notion and is quite subjective. Some may say that this is as far as "geometry" can assist us. I just want to know if that is really the case. That is why I tag this as a soft question. I also secretly hope that maybe one such interpretation would serve as a simple explanation of Cauchy-Binet formula.

Answer 1

Write $A$ as a product of the form $E_1E_2\cdots E_kDF_1F_2\cdots F_\ell$, where each $E_i$ (resp. $F_j$) is an elementary matrix corresponding to row (resp. column) switching or row (resp. column) addition, and $D$ is a diagonal matrix. You are done if you can explain why

1. $\det A=\det(E_1)\cdots\det(E_k)\det(D)\det(F_1)\cdots\det(F_\ell)$ and
2. the determinant of each elementary matrix is equal to the determinant of its transpose.

But these are essentially what motivate the definition of determinant: we want to define the determinant as the volume of the parallelepiped generated by the columns of a matrix. Shears do not change the volume (thus the determinant of an elementary matrix for row/column addition is $1$) and scaling in one direction changes the volume in the same proportion (thus $\det D$ is the product of $D$'s diagonal entries). Since we want to define the volume as zero in the degenerate case, it follows that together with the previous two requirements, the volume must be signed and by flipping two columns, the sign flips too (thus the determinant of an elementary matrix for row/column switching is $-1$).

Answer 2

This is not fully geometric, but it's based on geometric considerations and I found it illuminating. It's based on the proof found in https://mathcs.clarku.edu/~ma130/determinants3.pdf. This proof is very similar to user1551's above, only presented prehaps at a more elementary level.

We want to prove that for any matrix $A$ of size $n\times n$, $det A = det A^T$. We divide the proof to two cases.

Case 1: $rank A=r<n$. In this case the Null-space of the matrix has a positive dimension, and the co-kernel (i.e. $Nul A^T$) has the same, strictly-positive, dimension. It follows that the transformation $A$ (as well as $A^T$) "flattens" n-dimensional objects into r-dimensional objects, and thus their volume drops to zero; hence $det A=det A^T=0$.

Case 2: $rank A=r=n$. In this case the transformation $A$ is reversible (it's an isomorphism between the domain and the target), and is therefor row-equivalent to the identity matrix. In other words, we can think of $A$ as being derived from $I$ by a series of expansions (row operations of multiplication by a scalar), shears (row operatios of row addition), and reflections about the line between two axes (row operations of row exchange). Each of these can be represented with an elementary matrix so that $A=E_1 E_2 ... E_k$.

The key algebraic insight needed to complete the proof is that $(AB)^T=B^T A^T$. This makes geometric sense as the transpose transformation is working in the opposite direction to the non-transposed one, but I don't know a geometric proof of this. Given this theorem, however, it is clear that one can write $A^T = E_k^T E_{k-1}^T ... E_1^T$.

The final step is to note that for any elementary operation $det E = det E^T$. This can be understood directly from the geometric interpretation of these operations, noted above. The geometric interpretation of the determinant also naturally leads to the theorem $det AB = det A det B$. Combining these, we have

$det A = (det E_1)(det E_2)...(det E_k)=(det E_k^T)...(det E_1^T)=det A^T$

So this proof isn't completely geometric, but it's very close, and it shows very clearly the geometric meaning of the theorem: the theorem holds because we can think of $A$ as a series of elementary transformations whose transpose has the same determinant, and thus the same series transposed generates $A^T$ with the same combined determinant.

Answer 3

To prove that the two (generalized) volumes are equal is going to require some algebra -- I don't see a way around that (because the transpose is defined algebraically). This proof doesn't require the use of matrices or characteristic equations or anything, though. I just use a geometric definition of the determinant and then an algebraic formula relating a linear transformation to its adjoint (transpose).

Consider this as the geometric definition of the determinant. If $I$ is the unit $n$-cube in $\Bbb R^n$, then we can define the determinant of a linear transformation $T$ as $$\underline T(I)=\det(T)I$$ This formula tells us that the transformation $\underline T$ just scales the unit $n$-cube by its determinant.

For vectors, you might recognize that the adjoint (transpose), $\overline T$, of a linear transformation, $T$, is given by $$T(v)\cdot w = v\cdot \overline T(w)$$ However, $I$ is not really a vector, it's an object called an $n$-blade or $n$-vector. So we have to use the modified formula $$\langle\underline T(A)B\rangle = \langle A\overline T(B)\rangle$$ where $\langle M\rangle$ returns the scalar part of the multivector $M$.

So for us, it becomes $\langle\underline T(I)I\rangle = \langle I\overline T(I)\rangle$. On the LHS, we get $$\langle\underline T(I)I\rangle = \langle \det(T)I^2\rangle = \det(T)I^2$$ And on the RHS, we get $$\langle I\overline T(I)\rangle = \langle \det(\overline T)I^2\rangle = \det(\overline T)I^2$$ Therefore $$\det(T) = \det(\overline T)\ \ \ \ \square$$

Answer 4

I think there is a geometrical interpretation but it requires a bit of imagination w.r.t inner product spaces. Some notation to start:

$ f: U \rightarrow V $ is a linear map and $ f^{*} : V \rightarrow U $ is it's adjoint.

Let $ e_1 ..e_n s $ be an orthonormal basis for $U$. Considering only the non-degenerate case, we can see that $ \frac{f(e_i)}{\lVert f(e_i) \rVert} = e^{'}_i $ form a basis for $V$.

Now here is the trick: define $ e^{'}_1..e^{'}_n $ to be an orthonormal basis for $V$. From the perspective of $U$, $e^{'}_is$ may not be orthonormal, but when computing inside of $V$, we can define it to be.

Physically speaking, the space $V$ appears deformed, but only as seen from $U$ (and vice-versa).

From the definition of the adjoint, for any $u \in U$ and $ v \in V$, we have

$ f(u)\cdot v = f^{*}(v)\cdot u $

Note that the L.H.S is being computed in $V$, while the R.H.S is being computed in $U$. This equation also holds for $e_is $ and $e^{'}_js$.

$ f(e_i)\cdot e^{'}_j = f^{*}(e^{'}_j)\cdot e_i$

But the L.H.S is zero for any $ i \neq j $!

$ f^{*}(e^{'}_i)\cdot e_i = f(e_i)\cdot e^{'}_i = \lVert f(e_i) \rVert $

$ f^{*}(e^{'}_i)\cdot e_j = 0$ $ (i \neq j) $

Now we come to the crux of the argument. For this special choice of bases, $ f^{*} $ dilates the orthonormal basis vectors $e^{'}_i$ of $V$ by the same amount as $ f$ dilates the orthonormal basis vectors $ e_i$ of $U$. By keeping the order same, it also preserves the orientation.

$ \lVert f(e_i) \rVert = \lVert f^{*}(e^{'}_i) \rVert $

For this reason, $f^{*}$ maps the unit 'volume' of $V$ to the same volume as $f$ does the unit 'volume' of $U$.

Edit 1: The above treatment is admittedly not very rigorous, to start with determinants are usually defined for linear operators $ f : V \rightarrow V $ whereas I stretched the definition to $ f : U \rightarrow V $. But here is an alternate way to see the same thing:

for any $ f : V \rightarrow V $, the composite transform $ f^{*}f : V \rightarrow V $ is a symmetric linear transform (or Matrix), which has an orthornormal eigenvector basis $ e_1, e_2, .. e_n $ with real eigenvalues $ \lambda_1, \lambda_1,.. \lambda_n$

Now consider the $ f(e_i)'s $:

$ f(e_i)\cdot f(e_j) = f^{*}f(e_i)\cdot e_j = \lambda_i e_i\cdot e_j $

Therefore, $ f(e_i)s $ also form an orthornormal basis like before (except this time not by fiat). Also:

$ f(e_i).f(e_i) = \lVert f(e_i) \rVert^2 = \lambda_i $

Which is to say $f$ scales $ e_i$ by $ \sqrt{\lambda_i}$. Now set the basis $ e^{'}_i = \frac{f(e_i)}{\sqrt{\lambda_i}} $ for $ f^{*} $ and check the scaling:

$ f^{*}(e^{'}_i) = \frac{f^{*}f(e_i)}{\sqrt{\lambda_i}} = \sqrt{\lambda_i}e_i$

So again,

$ \lVert f(e_i) \rVert = \lVert f^{*}(e^{'}_i) \rVert $

Edit 2:

Another way to look at this is through the geometrical interpretation of the adjoint itself.

Every linear map, when seen in a certain special basis (as shown above), can be thought of as a composition of two maps: A positive semidefinite map ($p$) and an orthogonal map ($o$).

$p$ is a pure scaling map that scales each of the basis vectors by varying amounts. This gives a diagonal matrix.

$o$ is a re-orienting map that rotates/reflects the basis vectors but does not scale them (and keeps the angles between them). This is expressed as an orthogonal matrix with determinant $ \pm 1$ and $ o^{*} = o^{-1} $.

So what does $ f^{*} $ do?

$ f^{*} = (op)^{*} = p^{*}o^{*} = p^{*}o^{-1}$

Which can be interpreted as: $f^{*}$ un-does the re-orientation of $ f$ and then scales the basis vectors by the same factors as $f$ does.

This is really the key to why determinants are the same.