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Let $A$ $\in$ $M_n (C)$. For each $1 \leq i \leq n$, let $A_i$ be the $(n-1)\times(n-1)$ principal submatrix of $A$ resulting from deleting the $i^\text{th}$ row and $i^\text{th}$ column. Prove that for $1 \leq k \leq n-1$, $\sum\limits_{i=1}^n E_{k}(A_{i}) = (n-k)E_{k}(A)$.

EDIT: The notation is consistent with Horn and Johnson's Matrix Analysis book. On page 40, it says that there are $\binom{n}{k}$ different k-by-k principal minors of the matrix $A=[a_{ij}]$, and the sum of these is denoted by $E_{k}(A)$

I have been trying different formulas involving the trace and the determinant, but I think I am missing some insight into this question. Any help will be appreciated, thanks.

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The definition of $E_k$ seems to be independent of $k$, which is a bit of a worry. –  Gerry Myerson Sep 15 '11 at 0:31
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up vote 3 down vote accepted

Let $A_I = A_{\{i_1, \ldots, i_k\}}$ be the principal submatrix formed by the rows/columns indexed by $I = \{i_1, \ldots, i_k\}$. If $S$ is the set of all such indexsets $I$ of size $k$, then obviously both sides of the equation can be written as $\sum_{I \in S} \lambda_{I} A_I$ for some natural numbers $\lambda_I$, as the set of principal submatrices of $A_i$ is a subset of the set of principal submatrices of $A$. So we only need to prove that each coefficient is equal on both sides, i.e. each matrix $A_I$ gets counted on both sides as many times.

Now when does $A_I$ get counted in the summation on the left hand side? Then we need that $i \notin I$, i.e. $i \in \{1, \ldots, n\} \setminus I$, which occurs $n - k$ times. On the right hand side the matrix $A_I$ is counted once, but multiplied by $n - k$. So the coefficients are the same.

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