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Hello
I'm studying the properties of the determinant,
on various books they say the determinant of a matrix is an alternating multilinear map.
For clarity, I will use the following notation:

${Let \ A \in Mat_{\ n \ \times \ n}(K)\ be\ a\ n \times n\ square\ matrix,\ \\ \\ A = (A^{1}|A^{2}|...|A^{n}), \\where \ A^{i}\in K^n \ is \ the \ i\!-\!eth \ column \ of \ A, \ \forall \ i \in \{1,2,...,n \} }$


They give this definition for alternating multilinear map:
${Let\ f: K^{n} \times ...\times K^{n}\longrightarrow K, \ we\ say\ f\ is\ \textbf {multilinear}\ if: \\ \small \mathsf i)\ f(A^1|...|\lambda A^j|...|A^n) = \lambda\ f(A^1|...|A^j|...|A^n) \\ \hspace{0.3mm} \mathsf i\mathsf i)\small f(A^1|...|A^i|X\! +\! Y|A^{i+2}|...|A^n)\! =\!f(A^1|...|A^i|X|A^{i+2}|...|A^n)+ f(A^1|...|A^{i}|Y|A^{i+2}|...|A^n) \\ Also\ we\ say\ f\ is\ \textbf { alternating}\ if\ whenever \ A^i=A^{i+1}\, \\we\ get\ \small f (A^1|...|A^i|A^{i+1}|...|A^n)=0\hspace{5mm} \forall i \in \{1,...,n\}. \\[4mm]}$

Then while studying the properties of the determinant I came across this proof:
${Let\ A=(A^1|...|A^{i}|A^{j}|...|A^n),\\ let\ B=(A^1|...|A^{j}|A^{i}|...|A^n)\\ then\ detB = -detA\\ \textbf{Proof}:\\ Let\ C=(A^1|...|A^{i}+A^{j}|A^{i}+A^{j}|...|A^n), we\ know\ det(C)=0.\\ Furthermore\ det(C)=det(A)+det(B)\ hence\ det(B)=-det(A).\square }$

Now since the determinant is an alternating multilinear map, I don't understand why this happens: ${Let\ A=(A^1|...|X|Y|...|A^n)\ such\ that\ det(A) \ne0, \\let \ B= (A^1|...|\underline O|Y|...|A^n),\\ let\ C = (A^1|...|X|\underline O|...|A^n),\\ det(A) \ne\ det (B)+det(C)=0}$

Can anybody help me understand how my last example is different from the approach of the proof?

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Using the approach of the proof, in your ''counter-example'' we must have: $$ B= (A^1|...|\underline O|Y|...|A^n) $$ and $$ C=(A^1|...|Y|\underline O|...|A^n) $$ so that $$ \det A= \det(A^1|...|Y|Y|...|A^n)=0 $$

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With this notation the proof of the theorem is given as:

From: $$ B= (A^1|...|X|Y|...|A^n) \quad \mbox{and}\quad C= (A^1|...|Y|X|...|A^n) $$ and $$ A= (A^1|...|X+Y|Y+X|...|A^n) $$ we have: $\det A=0$ because the determinant is an alternating form, and, from multilinearity: $$ \det A=0= \det (A^1|...|X+Y|Y+X|...|A^n)=\det(A^1|...|X|Y+X|...|A^n)+\det(A^1|...|Y|Y+X|...|A^n)= $$ $$ =\det(A^1|...|X|Y|...|A^n)+\det(A^1|...|X|X|...|A^n)+\det(A^1|...|Y|Y|...|A^n)+\det(A^1|...|Y|X|...|A^n)= $$ $$ =\det(A^1|...|X|Y|...|A^n)+0+0+\det(A^1|...|Y|X|...|A^n)= $$ $$ =\det B + \det C $$ so: $\det B=-\det C$.

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  • $\begingroup$ Hi, thank you for your time, sorry I have been unclear, I'm not trying to find a counter example to the proof, I'm confused as to wether property ii) of the definition of multilinear map can be used on multiple columns "at the same time" or not, as that is precisely the impression I get from the proof, but then again in my example I get in trouble trying to do that. $\endgroup$ – edo Feb 8 '17 at 17:53
  • $\begingroup$ I've added something. Is it useful? $\endgroup$ – Emilio Novati Feb 8 '17 at 20:08
  • $\begingroup$ yes, thank you very much! $\endgroup$ – edo Feb 9 '17 at 9:35

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