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I am trying to read Hoffman Kunze's book on linear algebra and I have a doubt in a particular result, (Theorem 1) of Section 5.2. Specifically, the theorem states:

Let $n > 1$ and let $D$ be an alternating $(n - 1)$-linear function on $(n - 1)\times (n - 1)$ matrices over $K$. For each $j$, $1 < j \le n$, the function $E_j$ defined by $$E_j(A) = \sum_{i=1}^n(-l)^{i+j}A_{ij}D_{ij}$$ is an alternating $n$-linear function on $n \times n$ matrices $A$. If $D$ is a determinant function, so is each $E_j$.

Here $D_{ij}=D[A(i|j)]$ where $A(i|j)$ denotes the matrix obtained by deleting the $i$th row and the $j$th column of $A$.

Now my question concerns the $n$-linear part. I understand why $D_{ij}$ is linear in every row except the $i$th row and that $D_{ij}$ is independent of the $i$th row. What I do not understand is why $D_{ij}$ is linear in the $i$th row.

For example, if $n=2$ and $D([a])=a$ then $$D_{11}\begin{pmatrix} a+a'& b+b'\\c & d\end{pmatrix}=d$$ while $$D_{11}\begin{pmatrix} a& b\\c & d\end{pmatrix}+D_{11}\begin{pmatrix} a'& b'\\c & d\end{pmatrix}=d+d=2d.$$

Yet the authors state $A_{ij}D_{ij}$ is $n$-linear.

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  • $\begingroup$ The $D_{ij}$ are not linear in the $i$-th row; they are independent of the $i$-th row. But the products $A_{ij}D_{ij}$ are linear in the $i$-th row. $\endgroup$ – darij grinberg Aug 17 at 13:56
  • $\begingroup$ @darijgrinberg Can you explain why? $\endgroup$ – Shahab Aug 17 at 13:58
  • $\begingroup$ Because $A_{ij}$ is linear in the $i$-th row, and $D_{ij}$ is just a constant as far as the $i$-th row is concerned. $\endgroup$ – darij grinberg Aug 17 at 14:01
  • $\begingroup$ Are you treating $A_{ij}$ as a function? $\endgroup$ – Shahab Aug 17 at 14:07
  • $\begingroup$ Yeah, because when you say "linear in the $i$-th row", the $i$-th row has to be regarded as being made of variables. $\endgroup$ – darij grinberg Aug 17 at 14:17

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