Matrix determinant from the exterior algebra viewpoint
In every linear algebra course matrix determinant is a must. Often it is introduced in the following form:
Definition 1 (Determinant) Let \(A=(A^i_j)\) be a matrix and \(S_n\) be the permutation group of \(\{1, 2, \dotsc, n\}\). The determinant is defined to be the number \[ \det A = \sum_{\sigma\in S_n} \mathrm{sgn}\,\sigma \, A^{\sigma(1)}_1 A^{\sigma(2)}_2 \dotsc A^{\sigma(n)}_n \]
The definition above has a lot of advantages, but it also has an important drawback — the “why” of this construction is hidden and appears only later in a long list of its properties.
We’ll take an alternative viewpoint, which I have learned from Darling (1994, chap. 1), and is based around the exterior algebra.
All our vector spaces will be finite-dimensional.
We will assume that all vector spaces considered are over real numbers \(\mathbb R\). Mutatis mutandi everything works also over complex numbers \(\mathbb C\). However, not all theorems and exercises may work with finite fields, especially \(\mathbb F_2 = \mathbb Z/2\mathbb Z\).
Motivational examples
Consider \(V=\mathbb R^3\). For vectors \(v\) and \(w\) we can define their vector product \(v\times w\) with the following properties:
- Bilinearity: \((\lambda v+v')\times w = \lambda (v\times w) + v'\times w\) and \(v\times (\lambda w+w') = \lambda (v\times w) + v\times w'\).
- Antisymmetry: \(v\times w = -w\times v\).
Geometrically we can think of it as of a signed area of the parallelepiped spanned by \(v\) and \(w\).
For three vectors \(v, w, u\) we can form signed volume: \[\langle v, w, u\rangle = v\cdot (w\times u),\] which has similar properties:
- Trilinearity: \(\langle \lambda v+v', w, u \rangle = \lambda \langle v, w, u \rangle + \langle v', w, u\rangle\) (and similarly in \(w\) and \(u\) arguments).
- Antisymmetry: when we swap any two arguments the sign changes, e.g., \(\langle v, w, u\rangle = -\langle w, v, u\rangle = \langle w, u, v\rangle = -\langle u, w, v\rangle\).
Exterior algebra will be a generalisation of the above construction beyond the three-dimensional space \(V=\mathbb R^3\).
Exterior algebra
Let’s start with the natural definition:
Definition 2 (Antisymmetric multilinear function) Let \(V\) and \(U\) be vector spaces and \(f\colon V\times V \times \cdots \times V \to U\) be a function. We will say that it is multilinear if for all \(i = 1, 2, \dotsc, n\) it holds that \[ f(v_1, v_2, \dotsc, \lambda v_i + v_i', v_{i+1}, \dotsc, v_n) = \lambda f(v_1, \dotsc, v_i, \dotsc, v_n) + f(v_1, \dotsc, v_i', \dotsc, v_n). \] We will say that it is antisymmetric if it changes the sign whenever we swap any two arguments: \[ f(v_1, \dotsc, v_i, \dotsc, v_j, \dotsc, v_n) = -f(v_1, \dotsc, v_j, \dotsc, v_i, \dotsc, v_n). \]
As we have seen above both \((v, w)\mapsto v\times w\) and \((v, w, u)\mapsto v\cdot (w\times u)\) are antisymmetric multilinear functions.
Note that for every \(\sigma\in S_n\) it holds that \[ f(v_1, \dotsc, v_n) = \mathrm{sgn}\,\sigma \, f(v_{\sigma(1)}, \dotsc, v_{\sigma(n)}) \] as \(\mathrm{sgn}\,\sigma\) counts transpositions modulo 2.
Exercise 1 Let \(f\colon V\times V\to U\) be multilinear. Show that the following are equivalent:
- \(f\) is antisymmetric, i.e., \(f(v, w) = -f(w, v)\) for every \(v, w \in V\).
- \(f\) is alternating, i.e., \(f(v, v) = 0\) for every \(v\in V\).
Generalise to multilinear mappings \(f\colon V\times V \times \cdots\times V\to U\).
Expand \(f(v+w, v+w)\) using multilinearity.
Now we are ready to construct (a particular) exterior algebra.
Definition 3 (Second exterior power) Let \(V\) be a vector space. Its second exterior power \(\bigwedge^2 V\) we be the vector space of expressions \[ \lambda_1 v_1\wedge w_1 + \cdots + \lambda_n v_n\wedge w_n \] with the following rules:
- The wedge \(\wedge\) operator is bilinear, i.e., \((\lambda v+v')\wedge w = \lambda v\wedge w + v'\wedge w\) and \(v\wedge (\lambda w+w') = \lambda v\wedge w + v\wedge w'\).
- \(\wedge\) is antisymmetric, i.e., \(v\wedge w = -w\wedge v\) (or, equivalently, \(v\wedge v=0\)).
- If \(e_1, \dotsc, e_n\) is a basis of \(V\), then \[\begin{align*} &e_1\wedge e_2, e_1\wedge e_3, \dotsc, e_1\wedge e_n, \\ &e_2\wedge e_3, \dotsc, e_2\wedge e_n\\ &\qquad\vdots\\ &e_{n-1}\wedge e_n \end{align*} \] is a basis of \(\bigwedge^2 V\).
Note that \(v\wedge w\) has the interpretation of a signed area of the parallelepiped spanned by \(v\) and \(w\). Such parallelepipeds can be formally added and there is a resemblance between the wedge product and the vector product in \(\mathbb R^3\).
We just need to prove that such a space actually exists (this construction can be skipped at the first reading): similarly to the tensor space, build the free vector space on the set \(V\times V\). Now quotient it by expressions like \((v, v)\), \((\lambda v, w) - (v, \lambda w)\), \((v+v', w) - (v, w) - (v', w)\) and \((v, w+w') - (v, w) - (v, w')\).
Then define \(v\wedge w\) to be the equivalence class \([(v, w)]\).
If we had introduced the determinant by other means, we could construct the exterior algebra \(\bigwedge^k V\) also as the space of antisymmetric multilinear functions \(V^*\times V^*\to \mathbb R\) (where \(V^*\) is the dual space) by
\[ (v\wedge w)(\alpha, \beta) := \det \begin{pmatrix} \alpha(v_1) & \alpha(v_2) \\ \beta(v_1) & \beta(v_2) \end{pmatrix} \]
Analogously we can construct:
Definition 4 (Exterior power) Let \(V\) be a vector space. We define \(\bigwedge^0 V = \mathbb R\), \(\bigwedge^1 V = V\) and for \(k\ge 2\) its \(k\)th exterior power \(\bigwedge^k V\) as the vector space of expressions \[ \lambda_1 a_1\wedge a_2\wedge \cdots\wedge a_k + \cdots + \lambda_n v_1\wedge v_2 \wedge \cdots\wedge v_k \] such that the wedge operator \(\wedge\) is multilinear and antisymmetric (alternating) and that if \(e_1, \dotsc, e_n\) is a basis of \(V\), then the set \[ \{ e_{i_1}\wedge e_{i_2}\wedge \cdots \wedge e_{i_k}\mid i_1 < i_2 < \cdots < i_k \} \]
is a basis of \(\bigwedge^k V\).
Exercise 2 Show that if \(\dim V = n\), then \(\dim \bigwedge^k V = \binom{n}{k}\). (And that in particular for \(k > n\) we have \(\bigwedge^k V = 0\), the trivial vector space).
The introduced space can be used to convert between antisymmetric multilinear and linear functions by the means of the universal property:
Theorem 1 (Universal property) Let \(f\colon V\times V \cdots\times V\to U\) be an antisymmetric multilinear function. Then, there exists a unique linear mapping \(\tilde f\colon \bigwedge^k V\to U\) such that for every set of vectors \(v_1, \dotsc, v_k\) \[ f(v_1, \dotsc, v_k) = \tilde f(v_1\wedge \dotsc \wedge v_k). \]
Proof. (Can be skipped at the first reading.)
As \(f\) is multlilinear, its values are determined by the values on the tuples \((e_{i_1}, \dotsc, e_{i_k})\), where \(\{e_1, \dotsc, e_n\}\) is a basis of \(V\).
We can use antisymmetry to show that by “sorting out” the elements such that \(i_1 \le i_2\cdots \le i_k\) and defining \(\tilde f(e_{i_1} \wedge \dotsc, \wedge e_{i_k}) = f(e_{i_1}, \dotsc, e_{i_k})\) we obtain a well-defined mapping. Linearity is easy to proof.
Now the uniqueness is proven by observing that antisymmetry and multilinearity uniquely prescribe the values at the basis elements of \(\bigwedge^k V\).
Its importance is the following: to show that a linear map \(\bigwedge^k V\to U\) is well-defined, one can construct a multilinear antisymmetric map \(V\times V\times \cdots \times V\to U\).
Determinants
Finally, we can define the determinant. Note that if \(\dim V = n\), then \(\dim \bigwedge^n V = 1\).
Definition 5 (Determinant) Let \(n=\dim V\) and \(A\colon V\to V\) be a linear mapping. We consider the mapping \[ (v_1, \dotsc, v_n) \mapsto (Av_1) \wedge \cdots \wedge (Av_n). \]
As it is antisymmetric and multilinear, we know that it induces a unique linear mapping \(\bigwedge^n V\to \bigwedge^n V\).
Because \(\bigwedge^n V\) is one-dimensional, this mapping must be multiplication by a number. Namely, we define the determinant \(\det A\) to be the number such that for every set of vectors \(v_1, \dotsc, v_n\) \[ Av_1 \wedge \cdots \wedge Av_n = \det A\, (v_1\wedge \cdots \wedge v_n). \]
In other words, determinant measures the volume stretch of the parallelepiped spanned by the vectors after they are transformed by the mapping.
I like this geometric intuition, especially that it is clear that determinant depends only on the linear map, rather than a particular matrix representation — it is independent on the chosen basis.
We can now show a number of lemmata.
Proposition 1 If \(\mathrm{id}_V\colon V\to V\) is the identity mapping, then \(\det \mathrm{id}_V = 1\).
Proof. Obvious from the definition! Similarly, it’s clear that \(\det \left(\lambda\cdot \mathrm{id}_V\right) = \lambda^{\dim V}\).
Proposition 2 For every two mappings \(A, B\colon V\to V\) it holds that \(\det (B\circ A) = \det B\cdot \det A\).
Proof. For every set of vectors we have \[ \begin{align*} \det (B\circ A) \, v_1\wedge \cdots \wedge v_n &= (BAv_1) \wedge \cdots \wedge (BAv_n) \\ &= B(Av_1) \wedge \cdots \wedge B(Av_n) \\ &= \det B \, (Av_1) \wedge \cdots \wedge (Av_n) \\ &= \det B\cdot \det A\, v_1\wedge \cdots\wedge v_n. \end{align*} \]
Proposition 3 (Only invertible matrices have non-zero determinants) A mapping is an isomorphism if and only if it has non-zero determinant.
Proof. If the mapping is invertible, then \(A\circ A^{-1} = \mathrm{id}\) and we have \(\det A \cdot \det A^{-1} = 1\), so its determinant must be non-zero.
Now assume that the mapping is non-invertible. This means that there exists a non-zero vector \(k\in \ker A\) such that \(Ak=0\). Let’s complete \(k\) to a basis \(k, e_1, \dotsc, e_{n-1}\). Then \[ \det A\, k\wedge e_1\wedge \cdots\wedge e_{n-1} = (Ak) \wedge \cdots \wedge (Ae_{n-1}) = 0, \] which means that \(\det A=0\) as \(\{k\wedge e_1\wedge \dotsc \wedge e_{n-1}\}\) is a basis of \(\bigwedge^n V\).
Let’s now connect the usual definition of the determinant to the one coming from exterior algebra:
Proposition 4 (Recovering the standard expression) Let \(e_1, \dotsc, e_n\) be a basis of \(V\) and \((A^{i}_j)\) be the matrix of coordinates, i.e., \[ Ae_k = \sum_i A^{i}_k e_i. \] Then the determinant \(\det A\) can be calculated as \[ \det A = \sum_{\sigma\in S_n} \mathrm{sgn}\,\sigma \, A^{\sigma(1)}_1 A^{\sigma(2)}_2 \dotsc A^{\sigma(n)}_n. \]
Proof. Observe that \[\begin{align*} \det A e_1 \wedge \cdots \wedge e_n &= Ae_1 \wedge \cdots \wedge Ae_n\\ &= \left( \sum_{i_1} A^{i_1}_1 e_{i_1} \right) \wedge \cdots \wedge \left( \sum_{i_n} A^{i_n}_n e_{i_n} \right)\\ &= \sum_{i_1, \dotsc, i_n} A^{i_1}A^{i_2}\cdots A^{i_n} \, e_{i_1} \wedge \cdots \wedge e_{i_n}. \end{align*}\]
Now we see that repeated indices give zero contribution to this sum, so we can only consider the indices which are permutations of \(1, 2, \dotsc, n\). We also see that \(e_{i_1} \wedge \cdots \wedge e_{i_n}\) can be then written as \(\pm 1\, e_1\wedge \dotsc \wedge e_n\), where the sign is the number of required transpositions, that is the sign of the permutation. This ends the proof.
Going just a bit further into exterior algebra we can also show that matrix transposition does not change the determinant.
To represent matrix transposition, we will use the dual mapping: if \(A\colon V\to V\) there is the dual mapping \(A^*\colon V^*\to V^*\), given as \[ (A^*\omega)(v) := \omega(Av). \]
We can therefore build the \(n\)th exterior power of \(V^*\): \(\bigwedge^n (V^*)\) and consider the determinant \(\det A^*\).
We will formally show that
Proposition 5 (Determinant of the transpose) Let \(A\colon V\to V\) be a linear map and \(A^*\colon V^*\to V^*\) be its dual. Then \[ \det A^* = \det A. \]
Proof. To do this we will need an isomorphism \[ \iota \colon {\bigwedge}^n (V^*) \to \left({\bigwedge}^n V\right)^* \] given on basis elements by \[ \iota( \omega^1 \wedge \cdots \wedge \omega^n ) (v_1\wedge \cdots \wedge v_n) = \det \big(\omega^i(v_j) \big)_{i, j = 1, \cdots, n}, \] where on the right side we use any already known formula for the determinant. It is easy to show that this mapping is well-defined and linear, as it descends from a multilinear alternating mapping.
Having this, the proof becomes straightforward calculation: \[ \begin{align*} \det A^* \iota\left( \omega^1\wedge \cdots\wedge \omega^n \right)(v_1\wedge \cdots\wedge v_n ) &= \iota\bigg( \det A^* \, \omega^1\wedge \cdots\wedge \omega^n \bigg)(v_1\wedge \cdots\wedge v_n ) \\ &=\iota\bigg( A^*\omega^1 \wedge \cdots\wedge A^*\omega^n \bigg )(v_1\wedge \cdots\wedge v_n) \\ &= \det \bigg((A^*\omega^i)(v_j)\bigg) = \det \bigg( \omega^i(Av_j ) \bigg) \\ &= \iota\left(\omega^1\wedge \cdots\wedge \omega^n \right)(Av_1\wedge\cdots\wedge Av_n) \\ &= \iota\left(\omega^1\wedge \cdots\wedge \omega^n \right)(\det A\, v_1\wedge\cdots\wedge v_n) \\ &= \det A~ \iota\left(\omega^1\wedge \cdots\wedge \omega^n\right)(v_1\wedge\cdots\wedge v_n) \end{align*} \]
Establishing such isomorphisms is quite a nice technique, which also can be used to prove
Proposition 6 (Determinant of a block-diagonal matrix) Let \(A\colon V\to V\) and \(B\colon W\to W\) be two linear mappings and \(A\oplus B\colon V\oplus W\to V\oplus W\) be the mapping given by \[ (A\oplus B)(v, w) = (Av, Bw). \]
Then \(\det (A\oplus B) = \det A\cdot \det B\).
Proof. We will use this approach: there exists an isomorphism \[ {\bigwedge}^p (V\oplus W) \simeq \bigoplus_k {\bigwedge}^k V \otimes {\bigwedge}^{p-k} W, \] so if we take \(n=\dim V\) and \(m=\dim W\) and note that \(\bigwedge^{p} V = 0\) for \(p > n\) (and similarly for \(W\)) we have \[ \iota\colon {\bigwedge}^{n+m} (V\oplus W) \simeq {\bigwedge}^n V\otimes {\bigwedge}^m W. \] If \(i\colon V\to V\oplus W\) and \(j\colon W\to V\oplus W\) are the two “canonical” inclusions, this isomorphism is given as \[ \iota\big( iv_1 \wedge \cdots \wedge iv_n\wedge jw_1 \wedge \cdots \wedge jw_m \big) = (v_1\wedge \cdots\wedge v_n) \otimes (w_1\wedge\cdots\wedge w_m). \] Now we calculate: \[\begin{align*} (A\oplus B)( iv_1 \wedge \cdots \wedge iv_n\wedge jw_1 \wedge \cdots \wedge jw_m ) &= iAv_1 \wedge \cdots \wedge iAv_n \wedge jBw_1\wedge\cdots\wedge jBw_m \\ &= \iota^{-1}\big( Av_1\wedge \cdots\wedge Av_n \otimes Bw_1\wedge\cdots\wedge Bw_m \big) \\ &= \iota^{-1}\big(\det A\cdot \det B\, v_1\wedge \cdots \wedge v_n \otimes w_1\wedge\cdots\wedge w_m) \\ &= \det A\cdot \det B \, \iota^{-1}\big( v_1\wedge \cdots \wedge v_n \otimes w_1\wedge\cdots\wedge w_m \big)\\ &= \det A\cdot \det B \, iv_1\wedge \cdots \wedge iv_n \wedge jw_1\wedge\cdots\wedge jw_m. \end{align*} \]
Proposition 7 (Determinant of an upper-triangular matrix) Let \(A\colon V\to V\) be a linear mapping and \(e_1, \dotsc, e_n\) be a basis of \(V\) such that matrix \((A^i_j)\) is upper-triangular, that is \[ \begin{align*} Ae_1 &= A^1_1 e_1\\ Ae_2 &= A^1_2 e_1 + A^2_2 e_2\\ &\vdots\\ Ae_n &= A^1_n e_1 + A^2_ne_2 + \dotsc + A^n_n e_n \end{align*} \] Then \[ \det A = \prod_{i=1}^n A^i_i. \]
Once proven, this result can also be used for lower-triangular matrices due to Proposition 5.
Proof. Recall that whenever there is \(i_j=i_k\), then \(e_{i_1}\wedge \cdots\wedge e_{i_n} = 0\). Hence, there is only one term that may be non-zero: \[ Ae_1\wedge Ae_2 \wedge \cdots \wedge Ae_n = A^1_1 e_1 \wedge \cdots \wedge A^n_n e_n = \prod_{i=1}^n A^i_i\, e_1\wedge \cdots\wedge e_n. \]
Acknowledgements
I would like to thank Adam Klukowski for helpful editing suggestions.