To put in context, I have just learned that the determinant can be uniquely characterized by the fact that it is multilinear, anti-symmetric, and sends the identity to 1.

I was surprised that anti-symmetric multilinear maps are uniquely characterized by where they map a single element (the identity), since conventional linear maps are unique only after deciding where each basis element is mapped to. So I wondered how might we so characterize the dot product, or inner products more generally.

It seems that they can be uniquely characterized by where they map all pairs of basis elements (hence $n\choose 2$ mappings). This information can then be packed into a positive-semidefinite symmetric matrix such that $f(x,y) = x^TAy$.

Following this idea, I wondered whether this idea of "packing the information" inside a matrix also works for determinants. It seems to work for $2x2$ matrices, where the determinant can be represented by a skew-symmetric matrix \begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix} and then $$ det(\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}) = \begin{bmatrix} a & b \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} c \\ d \end{bmatrix} = ad - bc$$

We thus see that for $2x2$ matrices, the determinant is uniquely characterized by where it maps the identity (represented by the number in the diagonal on the skew-symmetric matrix). What happens for higher dimensional determinants? I'm guessing we need a 3-tensor for $3x3$ matrices, etc. But are there such things as skew-symmetric 3-tensors? And how can I know/prove that these are uniquely characterized by a single number (only have 1 degree of freedom)?

  • $\begingroup$ You can construct "determinant coefficients", the $n$-dimensional $n \times \cdots \times n$ array $A$ of $-1$, $0$, $1$ in which $A_{\sigma(1),\dots,\sigma(n)} = 1$ if $\sigma$ is an even permutation, $-1$ if $\sigma$ is an odd permutation, and $0$ if $\sigma$ is not a permutation, but I doubt there's much insight to be found by doing so. Have you seen the usual proof that a skew-symmetric $n$-tensor is determined by its value on the identity matrix? $\endgroup$ May 7, 2017 at 15:56
  • $\begingroup$ No, but it seems intuitive. As Wikipedia puts it: "To see this it suffices to expand the determinant by multi-linearity in the columns into a (huge) linear combination of determinants of matrices in which each column is a standard basis vector." $\endgroup$
    – samlaf
    May 7, 2017 at 16:45
  • $\begingroup$ What I'm not sure though is, does this mean that the vector space of anti-symmetric multilinear maps only has dimension 1? The only thing I could find is Qiaochu Yuan in math.stackexchange.com/q/53094/438738 saying that the dimension of the the space of alternating multilinear maps $V^k \rightarrow F$ has dimension $d \choose k$, where $V$ has dimension $d$. $\endgroup$
    – samlaf
    May 7, 2017 at 16:48
  • $\begingroup$ Yes, the space of skew-symmetric $n$-multilinear functions on an $n$-dimensional vector space is one-dimensional (the $k = n$ case of Qiaochu's formula). $\endgroup$ May 7, 2017 at 17:28
  • 1
    $\begingroup$ "Skew-symmetric" ($f(v, u) = -f(u, v)$ and its generalizations) and "alternating" ($f(u, v) + f(v, u) = 0$) are synonyms except over fields of characteristic $2$. :) $\endgroup$ May 7, 2017 at 18:42


You must log in to answer this question.

Browse other questions tagged .