A complex $n$-dimensional vector space $V$ can be thought of as a real $2n$-dimensional vector space equipped with a map $J:V \to V$ with $J^2 = -I$. Complex-linear maps are then linear maps $V \to V$ which commute with $J$. One can think of $J$ as an infinitesimal rotation, so that $\exp(tJ)$ gives a family of rotations of this space, and $\mathbb R$-linear maps $V \to V$ are complex-linear if they respect this family.

From this point of view, or some other geometric point of view, is there a nice interpretation of the complex determinant $\det_{\mathbb C} L$ of a complex-linear map $L: V \to V$? Or, almost the same question, is there a geometric interpretation of the unique (up to scaling by complex numbers) antisymmetric complex-multilinear $n$-form $\operatorname{vol}_{\mathbb C}: V \times V \times ... \times V \to \mathbb C$?

The norm is fairly easy to interpret. $| \det_{\mathbb C} L |^2 = |\det_{\mathbb R} L|$. One way to see this is to look at the diagonalization of $L$ over $\mathbb C$. This also gives you a way to interpret the argument, as the total amount of rotation in all the invariant subspaces of $L$.

Is there a geometric interpretation of $\det_{\mathbb C} L$, not just its norm, which does not require one to diagonalize the matrix first?

Even the special case when $L$ is unitary is of interest.

  • 3
    $\begingroup$ This is a very interesting question, and I would be quite interested in an answer. Put a large bounty on it as to attract some more attention? $\endgroup$
    – user396323
    Commented Dec 23, 2016 at 2:44

3 Answers 3


One thing I should point out is that a general $L$ over $\mathbb C$ can only be upper-triangularized, not necessarily diagonalized. (Think about $2\times2$ matrix with $1$'s on diagonal, $0$ below, and nonzero entry above.) This is related to the phenomenon of "generalized eigenspaces", as is relevant to constant-coefficient ordinary differential equations whose "characteristic polynomial" has a repeated root in $\mathbb C$, etc. So in general one cannot decompose $V$ into a direct sum of subspaces on each of which $L$ acts by some scaling.

That being said, if one is aiming to prove some kind of "interpretation" given by a "continuous" formula then since the locus of diagonalizable $L$'s inside the space $\text{M}_n(\mathbb{C})$ is dense (just wiggle entries a bit to make the discriminant of the characteristic polynomial—some nasty but specific universal polynomial in the coefficients in the characteristic polynomial—not vanish), any such interpretation for diagonalizable $L$ should apply to general $L$ by passage to the limit with diagonalizable approximations. This is the reason that arguments of physicists only valid for diagonalizable $L$ wind up being valid for general $L$ (without the physicists realizing their argument did not apply in general). In other words, by focusing on the diagonalizable case one should not lose too much if one can find a truly geometric interpretation that does not explicitly mention diagonalizability (but maybe assumes invertibility of $L$).

The answer depends on how you understand the word "nice".

For example, let us look at the problem from the point of view of algebraic number theory.

One can start with a field $K$ and ask how the determinant of an endomorphism $T: V \to V$ of a finite-dimensional $k$-vector space $V$ behaves with respect to the field extension. In other words, let $E$ be a finite algebraic extension of $K$, and $V_E$ be the tensor product of $V$ and $E$ over $K$ ("extension of scalars" or "base change"). Then we have the corresponding endomorphism $T_E: V_E \to V_E$. Question: how are $\det T$ and $\det T_E$ related?

Your question was about $K = \mathbb{R}$ (real numbers) and and $E = \mathbb{C}$ (complex numbers), which is a degree $2$ extension of $\mathbb{R}$. Notice that from the point of view presented above there is no geometry in the problem—it is about linear algebra. Our field $K$ can be, say the finite field of $3$ elements (i.e. $\mathbb{Z}/3\mathbb{Z}$).

In general, $\det T = \text{N}_{E/K} (\det T_E)$, where $\text{N}_{E/K}$ is the norm map of field extension. In the simplest case this tells us that the (square of the) norm of a complex number is the determinant of the corresponding $2 \times 2$ matrix. Notice that the notion of norm is slightly different than the one taught in school, because the square root is not involved. But still the norm of the product is the product of norms. See here for an elementary note with the introduction to the notion of the norm and trace for extension of fields and a discussion on MathOverflow.

This pure algebraic approach, maybe, does not look "nice" from some point of view. You may ask about an approach which is specific for complex numbers, and cannot be generalized to other fields. In particular, you can observe that any, say, invertible matrix $T$ is a product of a positive matrix $P$ (the one with positive eigenvalues) and the unitary matrix $U$ (i.e. such that $UU^* = \text{Id}$, where $U^*$ is the Hermitian conjugate), $T = PU$. See here for another discussion about that classical fact.

Then the question indeed reduces to the nice geometric meaning of the determinants of the unitary matrices. Here again the question is about "nice". Unitary matrices form a Lie group $\text{U}(n)$ of transformations of a complex vector space $\mathbb{C}^n$ which preserve the standard Hermitian norm (sum of squares of absolute values of coordinates). From the point of view of the real vector space $\mathbb{R}^{2n}$ they are rotations. There is a subgroup $\text{SU}(n)$ of $\text{U}(n)$ consisting of the unitary matrices with the determinant equal to $1$. It has interesting topology. E.g. $\text{SU}(2)$ is naturally diffeomorphic to the $3$-dimensional sphere. You can study the structure of the unitary group using the notion of root of a compact simple Lie group. Then roughly speaking each unitary matrix $U$ admits a factorization $U = U_1U_2 \ldots U_k$ of some "basic" matrices. Each of the factors corresponds (roughly) to a rotation about one axis (like in the case $n = 1$, where $U = U_1 = e^{ix}$). Then from the knowledge of $\det U_k$ you can recover $\det U$ (but all the determinants can be equal to $1$, hence the factorization carries more information).


Note to moderators: I tried to edit my deleted answer, but then people would have to vote to undelete my post. So I have decided to add another answer than my deleted one. Besides, it is completely different from what I had written about 3 years ago!

Let $V = \mathbb{C}^n$, $h(-,-)$ denote the standard hermitian inner product on $V$ and denote by $\Omega \in \Lambda^n V^*$, normalized so that $$ \Omega(e_1, \ldots, e_n) = 1. $$

I claim that this data allows us to define what I would like to call the hermitian cross product of $n - 1$ vectors in $V$. Indeed, let $v_1, \ldots, v_{n-1} \in V$. Denote by $*_h(v_1 \wedge \cdots \wedge v_{n-1}) \in V$ the unique vector, say $w \in V$ for short, such that

$$ \Omega(v_1, \ldots, v_{n-1}, v) = h(v, w) $$ for all $v \in V$ (my convention is that $h(-,-)$ is antilinear in the second argument).

Taking $v = w$, we get

$$ \Omega(v_1, \ldots, v_{n-1}, w) = h(w, w) = \lVert w \rVert^2 \geq 0.$$

Thus $w$ is a vector in $V$ which is orthogonal to the span of $v_1, \ldots, v_{n-1}$ and such that the above complex determinant is actually not only real, but also non-negative. We will call $w$ the hermitian cross product of $v_1, \ldots, v_{n-1}$ (there is probably another name for it in the literature, I am sure).

In terms of coordinates, $w = (w_1, \ldots, w_n)$, it is easy to show that $$ w_i = (-1)^{i+n} \overline{m_{in}(A)} $$

where $m_{ij}(A)$ is the $ij$-th minor of $A$ and $A$ is the complex $n$ by $n-1$ matrix whose $j$-th column is $v_j$, for $j = 1, \ldots, n-1$.

If we take $v = v_n$ we then get $$ \Omega(v_1, \ldots, v_n) = h(v_n, w). $$

So we can view the complex determinant of $v_1, \ldots, v_n$ as the hermitian inner product of, say $v_n$, with a specific vector $w$, which is nothing but the hermitian cross product of the first $n - 1$ vectors ($v_1, \ldots, v_{n-1}$).

Our task is now much easier! Thinking about the real picture, orthogonaly project $v_n$, now thought of as a point in $\mathbb{R}^{2n}$, orthogonally onto the real span of $w$ and $Jw$. Note that we can identify the real span of $w$ and $Jw$ with $\mathbb{C}$, by mapping $aw + bJw$ onto $\lVert w \rVert (a + bi)$.

I believe that the complex determinant of $v_1, \ldots, v_n$ is actually $\lVert w \rVert$ times the orthogonal projection of $v_n$ onto the real span of $w$ and $Jw$, itself identified with $\mathbb{C}$. This in particular explains the real and imaginary part of the complex determinant and thus, in particular, its phase.

Edit 1: Here is another approach towards understanding the phase of the complex determinant, in the case of $2$ by $2$ complex matrices, using the Hopf map.

The Hopf map $h: S^3 \to S^2$ has the property that $h(e^{i\theta} \psi) = h(\psi)$, for any $\psi \in S^3 \subset \mathbb{C}^2$. So we can think of a point on $S^3$ simply as a triple consisting of $(p, v, \nu)$, where $p \in S^2$, $v$ is a unit vector in $\mathbb{R}^3$ which is orthogonal to $p$ in $\mathbb{R}^3$ but which is based at $p$. So we can think of $v \in T_p(S^2)$ as a unit vector living in the tangent plane to $S^2$ at $p$. Finally, $\nu$ is a point in the non-trivial double cover of $SO(3)$ lying above $(p, v, p \times v) \in SO(3)$.

Given $\psi_a \in S^3$ for $a = 1, 2$, which we assume to be linearly independent in $\mathbb{C}^2$ (over $\mathbb{C}$) WLOG, we then form the triples $(p_a, v_a, \nu_a)$ for $a = 1, 2$, corresponding to $\psi_a$ for $a = 1, 2$, respectively.

In this setup, how can we recover the phase of the complex determinant of the $2$ by $2$ matrix $(\psi_1, \psi_2)$?

  • $\begingroup$ The vector $w$ might be called the hermitian external product in some literature. Is there more we can say to characterize $w$ geometrically, like about its length and its orientation in the naturally oriented underlying real space, analogous to the regular cross product? $\endgroup$
    – blargoner
    Commented Jan 26, 2023 at 1:40
  • $\begingroup$ @blargoner, getting a completely geometric description of the phase of the complex determinant is still beyond my reach. Similarly, while I do know the norm of $w$ and that $w$ is orthogonal to the span of $v_1, \ldots, v_{n-1}$, yet proceeding further along a purely geometric route still seems difficult. So I guess my answer is just the beginning of an answer, possibly. I don't think it is a complete satisfactory answer yet. $\endgroup$
    – Malkoun
    Commented Jan 27, 2023 at 1:13
  • $\begingroup$ Is that really an open problem? I find it hard to believe, as it is an elementary question. In any case, I find the problem interesting. $\endgroup$
    – Malkoun
    Commented Jan 27, 2023 at 1:15
  • $\begingroup$ I agree it is interesting. I was recently thinking about this topic here. $\endgroup$
    – blargoner
    Commented Jan 27, 2023 at 1:56

I propose another answer here than my previous one.

First, it suffices to consider the case where the matrix is non-singular, for otherwise the complex determinant vanishes. But also, there is a nice orthogonalization map from $GL(n, \mathbb{C}) \to U(n)$: $$ g \mapsto (gg^*)^{-1/2}g, $$ which preserves the phase angle of the determinant (though in general not the absolute value of the determinant but, as the OP original remarked, we already know how to interpret the absolute value of the determinant geometrically).

So the problem is reduced to interpreting the determinant of an element $g \in U(n)$, i.e. of a unitary matrix.

Note that any $g \in U(n)$ can be written as $$ g = g_1 \beta, $$ where $g_1 \in SU(n)$ and $\beta \in U(1)$ is a phase factor. Note that $g_1$ is not unique and neither is $\beta$. Indeed, we can multiply $g_1$ by an $n$-th root of unity, say $\gamma$, and multiply $\beta$ by $\gamma^{-1}$, thus obtaining another similar matrix decomposition. But this is the only freedom we have, meaning that $g_1$ and $\beta$ are uniquely determined up to the above action of the group of $n$-th root of unity.

We then have $$ \operatorname{det}(g) = \operatorname{det}(g_1) \beta^n = \beta^n, $$

since $g_1 \in SU(n)$. We have thus reduced the problem of computing the determinant of $g$ to the above matrix decomposition. I am not sure how "geometric" this matrix decomposition but, as with other matrix decomposition, there probably exists a geometric way of looking at it.


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