A determinant identity

Question

The following identity involving determinants essentially appears in E.L. Ince's book on Ordinary Differential Equations:
Let $A$ be an $n times n$ matrix, $n geq 3$. Denote by $A_{j_1,ldots,j_r}^{k_1,ldots,k_r}$ the $(n-r) times (n-r)$ matrix obtained from $A$ by
erasing the $j_1$-th, ..., $j_r$-th row and the $k_1$-th, ..., $k_r$-th column. Then,
$$
left|A right| left|A_{n-1,n}^{1,n} right| = left|A_{n-1}^1 right|left|A_n^n right| - left|A_{n-1}^n right|left|A_n^1 right|.
$$
Any ideas of how to prove it? Is this a special case of a more general identity?

Steven Landsburg · Answer

I'll repeat the argument cited in the comments, adjusted to fit your situation.
A polynomial identity holds in general if it holds on an open set.   So it's enough to prove the identity for matrices $A$ in the open set where $A^{1,n}_{n-1,n}$ is invertible.  That is, we can assume $A^{1,n}_{n-1,n}$ is invertible.
Now premultiply $A$ by the inverse of the matrix.
$$pmatrix{0&1&0cr
A^{1,n}_{n-1,n}&0&0cr
0&0&1cr}$$
where the $1$s are $1times 1$ matrices with entry $1$ and the $0's$ are row or column matrices of the appropriate sizes, filled with zeros.
This does not affect the truth of the theorem and allows us to assume $A^{1,n}_{n-1,n}$ is the identity, after which row and column operations render the desired equality trivial.

A determinant identity

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