I was going over the Wikipedia page for exterior products of vector spaces and we can define the determinant as the coefficient of the exterior product of vectors with respect to the standard basis when the vectors are elements in $\mathbb{R}^n$. I was wondering if there was a way to deduce the formula for the determinant of the exterior (wedge) product of two matrices from this definition.

In particular let $V$ be a finite vector space and let $\wedge^k V$ be the $k$-th exterior power of $V$ that is $T^k(V)/A^k(V)$ where $A(V)$ is the ideal generated by all $v \otimes v$ for $v \in V$ and $T^k(V) = V \otimes V \otimes \cdots \otimes V$ is tensor product of $k$ vector spaces.

Let $M$ be a square $m\times m$ matrix. Is there a known formula for $\det(M \wedge M)?$

I was thinking there must be some nice formula like $\det(M \wedge M) = \det(M)\det(M)$ but I have a feeling this does not generalize to higher powers of wedge products.

  • $\begingroup$ What does $\det(M\wedge M)$ stand for? $\endgroup$
    – anon
    Oct 26, 2011 at 5:19
  • $\begingroup$ Yes I am a little confused myself now. I am looking at an old qualifying exam problem and wondering if I just did not interpret it correctly. The original problem says to define $D_p$ to be the determinant of the square matrix $ \wedge^p M$ and give a formula for $D_p$ in terms of a determinant for $det(X)$. based on you comment does it even make since to define a determinant function for $\wedge^p M$? $\endgroup$
    – user7980
    Oct 26, 2011 at 5:30
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    $\begingroup$ The matrix $M$ is not an element of the vector space $V$. You're actually talking about $\Lambda^k M_{n\times n}$, if I understand the problem here correctly. Since I only know of $\det$ as a function of matrices, could you explain what it is as a function of wedge products of matrices? $\endgroup$
    – anon
    Oct 26, 2011 at 5:30
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    $\begingroup$ Another 'basis heavy approach' - base change to $\mathbb{C}$ and choose coordinates in which its upper triangular; if you choose the 'dictionary ordering' for $e_{i_1} \wedge e_{i_2}$, it's again upper triangular $\endgroup$ Aug 3, 2013 at 17:00

2 Answers 2


Hint: Let $\{e_1,\ldots, e_n\}$ be a basis of $V$. Then the space $\wedge^p V$ has a basis consisting of vectors of the form $e_{i_1}\wedge e_{i_2}\wedge\cdots\wedge e_{i_p}$ for some strictly increasing sequence $i_1<i_2<\ldots<i_p$ of indices. The linear mapping $\wedge^pM$ maps the vector $e_{i_1}\wedge e_{i_2}\wedge\cdots\wedge e_{i_p}$ to $M(e_{i_1})\wedge M(e_{i_2})\wedge\cdots\wedge M(e_{i_p})$. Compute the determinant of this linear mapping in the following cases:

  1. $M$ maps the basis vector $e_{i_0}$ to $\lambda e_{i_0}$ and the other basis vectors $e_i,i\neq i_0,$ to themselves.
  2. $M$ interchanges two basis vectors, $e_{i_1}$ and $e_{i_2}$, and maps the other basis vectors $e_i, i\neq i_1, i\neq i_2,$ to themselves.
  3. $M$ maps the basis vector $e_{i_0}$ to the vector $e_{i_0}+ae_{i_1}$ for some constant $a$ and $i_1\neq i_0$, and maps the other basis vectors $e_i, i\neq i_0$ to themselves.

Then keep in mind (=functoriality) that $\wedge^p(M\circ M')=\wedge^p(M) \circ \wedge^p(M')$ for all linear mappings $M,M'$ from $V$ to itself. As a further hint: This approach is a bit about elementary combinatorics. You have to count the number of changes of a given type, and remember the rule used in forming Pascal's triangle.

  • $\begingroup$ Aha! +1, not least for making sense of the question to begin with. $\endgroup$
    – anon
    Oct 26, 2011 at 7:09
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    $\begingroup$ @anon I think you were the one who figured out what the question was :-) $\endgroup$ Oct 26, 2011 at 7:14
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    $\begingroup$ I hope that it is clear that the idea is to show that for all three types of elementary matrices $M$ a formula of the type $$\det \wedge^pM=(\det M)^{k(n,p)}$$ holds, where the exponent $k(n,p)$ is the same for all the elementary matrices. Then use the functoriality. For the sake of completeness you also need an argument to cater for the singular matrices $M$. $\endgroup$ Oct 26, 2011 at 14:51
  • $\begingroup$ This was very helpful. Even years after being answered. Thank you. $\endgroup$
    – Stiven G
    Jul 25, 2019 at 1:02

Here is an argument using SVD (which might be easier than doing the combinatorics suggested by Jyrki Lahtonen). It gets rid of all the scalings, and reduces the problem to determining orientations.

(I hope this can be done relatively easy, see comments at the end).

Put an inner product on $V$ (This induces an inner product on $\bigwedge^k V$ so we have a notion of orthogonal maps $\bigwedge^k V \to \bigwedge^k V$). Then, if $Q \in \text{SO}$, then so is $\bigwedge^k Q$. Indeed

$$ (\bigwedge^k Q)^T=(\bigwedge^k Q^T)=(\bigwedge^k Q^{-1})=(\bigwedge^k Q)^{-1}.$$

Similarly, we can convince ourselves that $\bigwedge^k Q$ preserves orientation*, so we deduce that $$\bigwedge^k Q \in \text{SO}=\text{SO}(\bigwedge^k V,\bigwedge^k V).$$

Now, given an orientation-preserving map $A:V \to V$, write $A=U\Sigma V^T$, where $U,V \in \text{SO}$. By functoriality we get $$ \bigwedge^k A=\bigwedge^k U \circ \bigwedge^k \Sigma \circ \bigwedge^k V^T, $$ so

$$ \det(\bigwedge^k A)=\det(\bigwedge^k U) \cdot \det(\bigwedge^k \Sigma) \cdot \det( \bigwedge^k V^T)=\det(\bigwedge^k \Sigma). $$

Write $\Sigma v_i=\sigma_i v_i$. ($\sigma_i$ are the singular values of $A$). Then, $$\bigwedge^k \Sigma(v_{i_1} \wedge \dots \wedge v_{i_k})=\Pi_{i=1}^k \sigma_{i_k} (v_{i_1} \wedge \dots \wedge v_{i_k}).$$

We need to multiply all this factors (over all $k$-tuples $i_1,\dots,i_k$). To find $\det (\bigwedge^k \Sigma)$, note that each $\sigma_i$ shows in exactly $\binom{d-1}{k-1}$ products, since we need to append to it $k-1$ indices out of the $d-1$ left. Hence,

$$\det (\bigwedge^k A)=\det (\bigwedge^k \Sigma)=(\Pi_{i=1}^d \sigma_{i})^{\binom{d-1}{k-1}}=(\det \Sigma)^{\binom{d-1}{k-1}}=(\det A)^{\binom{d-1}{k-1}}. $$

This finishes the proof for orientation-preserving maps.

For orientation-reversing maps, we can use SVD in a similar way, taking one of the orthogonal factors to be in $\text{O} \setminus \text{SO}$. Suppose $U \in \text{O} \setminus \text{SO}$. To repeat the argument, We only need to know if $\det(\bigwedge^k U)=1$ or $\det(\bigwedge^k U)=-1$. (We already know that $\bigwedge^k U$ is an isometry, so its determinant is $\pm 1$). In fact we need to show $\det(\bigwedge^k U)=1 \iff \binom{d-1}{k-1}$ is even.

To summarize, my proof is missing two components:

  1. $Q \in \text{SO} \Rightarrow \bigwedge^k Q \in \text{SO}.$ Equivalently, since we know $\bigwedge^k Q$ is an isometry, it suffices to show that if $A:V \to V$ is orientation-preserving, so is $\bigwedge^k A$.

  2. If $A:V \to V$ is orientation-reversing, then $\bigwedge^k A$ is orientation-preserving iff $\binom{d-1}{k-1}$ is even.

Perhaps there is an easy way to prove these two claims without too much combinatorics.


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