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A determinant identity

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?

MathOverflow Asked by Vassilis Papanicolaou on January 4, 2021

1 Answers

One 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.

Answered by Steven Landsburg on January 4, 2021

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