Let $A$ be an $n \times n$ matrix over a field $K$. Do the properties of the determinant of a real matrix hold for the matrix $A$? If not, in which fields do the properties of the determinant of a real matrix hold?
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What properties of the determinant of a real matrix are you talking about? – Sungjin Kim May 13 '14 at 16:36
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The basic properties of determinants, like These. – Dante May 13 '14 at 16:39
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1Every properties hold true in any fields. – Sungjin Kim May 13 '14 at 16:41
1 Answers
Let ${\bf a}_i$ be the ${\textit i}$-th column of the matrix. A $K$-valued function $D({\bf a}_1,\ldots,{\bf a}_n)$ is called an $n\times n$ determinant (or a determinant of order $n$) if it satisfies the following conditions:
I. $D$ is linear in each of its variables; i.e. for $i=1,\ldots,n$, $D({\bf a}_1,\ldots,\alpha{\bf a}_i+\beta{\bf a}_i^{\prime},\ldots,{\bf a}_n)= \alpha D({\bf a}_1,\ldots,{\bf a}_i,\ldots,{\bf a}_n)+\beta D({\bf a}_1,\ldots{\bf a}_i^{\prime},\ldots,{\bf a}_n)$ for any $\alpha,\,\beta\in K$.
II. If ${\bf a}_i={\bf a}_j$ for some $i,\,j$ ($i\neq j$), then $D({\bf a}_1,\ldots,{\bf a}_n)=0$.
III. $D({\bf e}_1,\ldots,{\bf e}_n)=1$, where ${\bf e}_i$ has a one on the ${\textit i}$-th row and zero on the others.
With this, we see that the determinant is a function that gives zero iff the columns (that can be thought as being vector in $K^n$) are linearly dependent.
Observation: I don't know if this is correct. I just made a generalization of the real case that appared to be the most intuitive for case of a general field. For references see the appendix I of the book "Elementary differential equations" by Donald L. Kreider.
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