Let $A$ be an $n \times n$ matrix and fix an integer $k$ with $1 \leq k \leq n$. Define a new matrix $\text{minor}_k(A)$ whose entries are the $k \times k$ minors of $A$. This new matrix will be $\binom{n}{k} \times \binom{n}{k}$.

Theorem? Let $D$ be the determinant of $A$. The determinant of $\text{minor}_k(A)$ is $D^\binom{n-1}{k-1}$.

Is this right? Can anyone provide a reference or proof?

(If you want to be more precise in the definition of $\text{minor}_k(A)$, put an ordering on the cardinality $k$ subsets of $\{1, 2, \dots, n\}$, and index the rows and columns of $\text{minor}_k(A)$ using that ordering. The $(i,j)$ entry is then the determinant of the matrix obtained by keeping only the rows of $A$ indexed by the $i$th subset, and the columns of $A$ indexed by the $j$ subset. Changing the ordering shouldn't affect the determinant of the matrix of minors.)


1 Answer 1


This is called the Sylvester-Franke Theorem. A quick search on the web gives the following literature:

  1. Leonard Tornheim, The Sylvester-Franke Theorem, The American Mathematical Monthly, Vol. 59, No. 6 (Jun. - Jul., 1952), pp. 389-391.
  2. Harley Flanders, A Note on the Sylvester-Franke Theorem, The American Mathematical Monthly, Vol. 60, No. 8 (Oct., 1953), pp. 543-545.

You may also see Marcus and Minc, A Survey of Matrix Theory and Matrix Inequalities, Dover Publications Inc., New York, 1964, pp.16-17.

Note, however, that the theorem requires that the index sequences for producing minors are listed in lexiographic order. Otherwise the determinant may become $-D^\binom{n-1}{k-1}$.

  • $\begingroup$ If we use a different ordering for producing minors, doesn't that just mean applying the same permutation to rows and to columns? So where does the sign come from? $\endgroup$ Dec 23, 2012 at 17:11
  • $\begingroup$ You are certainly right. What I was talking about is that the row indices and column indices may have independent ordering. It doesn't hurt to be clear :-) $\endgroup$
    – user1551
    Dec 23, 2012 at 17:18

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