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This is for those of you who understand the Lindstrom-Gessel-Viennot lemma. I am looking for a proof of the following identity using paths and such:

Let $A$ be an $n\times n$ matrix, and for $i,j\in\{1,\ldots,n\}$, let $A^{ij}$ denote the matrix resulting from $A$ after removing row $i$ and column $j$, then:

$$\det\left(\begin{array}{cccc}\det(A^{11})&\det(A^{12})&\cdots&\det(A^{1n})\\ \det(A^{21})&\det(A^{22})&\cdots&\det(A^{2n})\\ \vdots &\vdots &\ddots &\vdots\\ \det(A^{n1})&\det(A^{n2})&\cdots &\det(A^{nn})\end{array}\right)=\det(A)^{n-1}$$

Read this for the algebraic proof:

Is this a well known determinant identity? Are there any generalizations?

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wircho
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  • What happens algebraically if the two $n$s are distinct? – Phira Nov 07 '11 at 12:24
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    I'm not sure if I understand your comment. Do you mean if $A$ is an $m\times n$ matrix with $m\neq n$? In that case the determinant does not make sense and you would have to reformulate the equation. – wircho Nov 07 '11 at 19:30
  • There is one $n$ which is the size of the square matrix $A$ and one $n$ which is the size of the square matrix of determinants. I am sure that there is an analogue identity for the sums of subdeterminants of a square matrix, but that is another story. – Phira Nov 07 '11 at 19:32
  • You could consider the matrix of $k\times k$ minors (instead of $(n-1)\times(n-1)$ minors). This is an $\binom nk\times\binom nk$ matrix (instead of an $n\times n$ matrix). The determinant of this matrix of minors is also a power of $A$. I read that somewhere, but I don't remember the exact exponent. – wircho Nov 07 '11 at 21:55
  • The exact exponent has to be $k$, then, that follows immediately from the degree. – Phira Nov 07 '11 at 22:29
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    Actually from the degree, the exponent is $\frac kn\binom nk$, isn't it? – wircho Nov 08 '11 at 00:50
  • You are right. For some reason, I assumed that we take consecutive minors and still miscalculated. – Phira Nov 08 '11 at 07:22

1 Answers1

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I am very surprised that no solution was given yet.

Denote by $\mathrm{Com}(A)$ the comatrix of $A$. The determinant to be computed is the determinant of the product $D \times \mathrm{Com}(A) \times D$, where $D$ is the diagonal matrix with diagonal entries $((-1)^i)_{1 \le i \le n}$. Since $D^2=I_n$, this matrix is similar to $\mathrm{Com}(A)$, so it as the same determinant as $\mathrm{Com}(A)$.

Since $A \times \mathrm{Com}(A)^\top = (\det A) I_n$, we have $\det(A) \times \det(\mathrm{Com}(A)) = \det((\det A) I_n) = (\det A)^n$.

If $A$ is invertible, we derive $\det(\mathrm{Com}(A)) = (\det A)^{n-1}$.

If $A$ is not invertible and not null, then $\mathrm{Com}(A)$ cannot be invertible since $A \times \mathrm{Com}(A)^\top$ is null, so $\det(\mathrm{Com}(A))$ is null.

If $A$ is null and $n \ge 2$, then $\mathrm{Com}(A)$ is null.

If $A$ is null and $n=1$, then $\mathrm{Com}(A)$ is the $1 \times 1$ matrix with unique entry equal to $1$.

In all cases, we derive $\det(\mathrm{Com}(A)) = (\det A)^{n-1}$.

Christophe Leuridan
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