Is it possible to find the determinant of an $n\times n$- matrix, only given the determinant of all $p\times p$ sub-matrices in it? Here $p\leq n$ is fixed. This is obviously true if $p=1,n$. But what happens in other cases?
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2$\begingroup$ Do you mean contiguous submatrices or arbitrary? At any rate, it seems that if it's true for $n-1$ then there's an inductive argument that shows it is true for all $p$: use the information of $p\times p$ determinants to find all $(p+1)\times (p+1)$ determinants. $\endgroup$– Sam HopkinsCommented Feb 21, 2022 at 3:53
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16$\begingroup$ Over $\mathbb C$, if $p$ is not a divisor of $n$ it is impossible, because you can multiply each entry by a $p$th root of unity without affecting the $p \times p$ determinants. This should be the only obstruction - it should be possible to reconstruct the $p/\gcd(n,p)$th power of the determinant using representation theory. For $p=n-1$, this is easy - just take the determinant of the adjugate matrix. $\endgroup$– Will SawinCommented Feb 21, 2022 at 4:24
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3$\begingroup$ . . . while if $p|n$ it must be a signed sum of products of $n/p$ determinants of rows $(k-1)p+1,\ldots,kp$ and columns $P_k$ ($1 \leq k \leq n/p$) where $P_1,\ldots,P_{n/p}$ is a partition of $\{1,\ldots,n\}$ (and the sign can be fixed by testing on a permutation matrix). $\endgroup$– Noam D. ElkiesCommented Feb 21, 2022 at 5:38
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2$\begingroup$ @LSpice I'm saying it's easy to obtain a power of the determinant in the $p=n-1$ case, not the determinant itself. $\endgroup$– Will SawinCommented Feb 21, 2022 at 14:55
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2$\begingroup$ Besides the answer that I give below, I strongly suspect that there are syzygies (the $p\times p$ minors are not algebraically idenpendent). $\endgroup$– Denis SerreCommented Feb 21, 2022 at 15:30
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2 Answers
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As mentioned by Will Sawin, a necessary condition is that $p$ divides $n$. Thus let us assume that $n=pk$. Denoting $e_1,\dotsc,e_n$ the canonical basis, the knowledge of the $p\times p$ minors is the knowledge of the $p$-vectors $$(Ae_{i_1})\wedge\cdots\wedge(Ae_{i_p})\in\Lambda^p(K^n),$$ where $K$ is the field of scalars (e.g. $\mathbb C$).
Splitting $$(Ae_1)\wedge\cdots\wedge (Ae_n) = [(Ae_1)\wedge\cdots\wedge (Ae_p)]\wedge\cdots\wedge[(Ae_{n-p+1})\wedge\cdots\wedge (Ae_n)], $$ we see that $(Ae_1)\wedge\cdots\wedge (Ae_n)$ is a polynomial function in the $p\times p$ minors. Since $(Ae_1)\wedge\dotsb\wedge (Ae_n)=(\det A)e_1\wedge\dotsb\wedge e_n$, we deduce the value of $\det A$.
Let me describe how it works when $n=4$ and $p=2$. The minors are denoted $$A\binom{i\alpha}{j\beta}=a_{i\alpha}a_{j\beta}-a_{i\beta}a_{j\alpha}.$$ Then $$(Ae_\alpha)\wedge(Ae_\beta)=\sum_{i<j}A\binom{i\alpha}{j\beta}e_i\wedge e_j.$$ We thus obtain $$\det A=\sum_{\substack{\rho\in{\frak S}_4 \\ \rho(1)<\rho(2),\rho(3)<\rho(4)}}A\binom{\rho(1)1}{\rho(2)2}A\binom{\rho(3)3}{\rho(4)4}.$$
answered Feb 21, 2022 at 10:08
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2$\begingroup$ This can also be proved by induction using generalized Laplace expansion. $\endgroup$ Commented Feb 21, 2022 at 14:17
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1$\begingroup$ TeX note: In case it is of interest, you might find that $\displaystyle\sum_{\rho \in \mathfrak S_4; \rho(1) < \rho(2), \rho(3) < \rho(4)}$ looks better as $\displaystyle\sum_{\substack{\rho \in \mathfrak S_4 \\ \rho(1) < \rho(2), \rho(3) < \rho(4)}}$
\sum_{\substack{\rho \in \mathfrak S_4 \\ \rho(1) < \rho(2), \rho(3) < \rho(4)}}(or even an index of summation on three lines, trading more vertical space for horizontal space). $\endgroup$– LSpiceCommented Feb 21, 2022 at 14:56 -
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1$\begingroup$ I corrected a typo: $\Lambda^p(K)$ should be $\Lambda^p(K^n)$. $\endgroup$ Commented Feb 23, 2022 at 16:42
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$\begingroup$ @FrançoisBrunault . Thanks a lot, François ! $\endgroup$ Commented Feb 24, 2022 at 8:16
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Here is a way to see that, for all $p$ (even if $p \nmid n$), you can reconstruct the absolute value of the matrix's determinant, which was suggested by Will Sawin's comment. The condition $p \mid n$ is only needed to get the sign (or complex phase, if the ground field is $\mathbb{C}$).
If you know the determinants of all $p \times p$ submatrices of $A$, that means exactly that you know $C_p(A)$: the $p$-th compound matrix of $A$. Since $\det(C_p(A)) = (\det(A))^\binom{n-1}{p-1}$, you can compute $$ \lvert\det(A)\rvert = \lvert\det(C_p(A))\rvert^{1/\binom{n-1}{p-1}}. $$
answered Feb 21, 2022 at 17:09
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1$\begingroup$ In fact, the eigenvalues of $C_p(A)$ consist of products of $p$ of the eigenvalues of $A$. Let $M$ be the $n\times {n\choose p}$ incidence matrix between elements of $[n]=\{1,\dots,n\}$ and $p$-element subsets of $[n]$. The Smith normal form of $M$ has diagonal entries $p$ (once) and 1 ($n-1$ times), a special case of Theorem 3.6 of people.clas.ufl.edu/sin/files/snf.pdf. This implies that the $p$th powers of the eigenvalues of $A$ can be written as Laurent monomials in the eigenvalues of $C_p(A)$. $\endgroup$ Commented Feb 23, 2022 at 17:43