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Let $A^{\#}:A_{n}(F)\rightarrow A_{n}(E)$ define by $A^{\#}f(v_{1},\cdots ,v_{n})=f(Av_{1},\cdots, Av_{n})$, where $A_{n}(F)$ is the space of the alternated n multilinear forms in F. Verify that $(\alpha A)^{\#}=\alpha (A^{\#})$, where $\alpha$ is a scalar and $A:E\rightarrow F $ is a linear transformation.

$(\alpha A)^{\#}f(v_{1},\cdots ,v_{n})=f(\alpha Av_{1},\cdots,\alpha Av_{n})=\alpha^{n}f(Av_{1},\cdots, Av_{n})\neq \alpha (A^{\#})f(v_{1},\cdots ,v_{n})$

then I showed that it's false, am I missinterpreting all?

a hint would be appreciated, thanks in advance .

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    What is $\alpha$ supposed to be? A scalar? And $A$, is it supposed to be a linear transformation $A: E\to F$?2011-06-30
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    yes $\alpha$ is a scalar. Edited it2011-06-30
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    yes I forgot, A is linear transformation from E to F2011-06-30
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    Is that the exact wording of the question? (and not something like, verify that pullbacks are linear, etc.)?2011-06-30
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    Unless the notation $\alpha(A^\sharp)$ means something special, instead of just scalar multiplication, I think what you did is correct.2011-06-30
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    Maybe it may help that all this is in the context of defining the determinant of an operator, and the text before the problem says that all linear transformations $A:E\rightarrow F$ induce a linear transformation $A^{\#}$. And the rest is the problem it also says to prove that $(A+B)^{\#}=A^{\#}+B^{\#}$, and $(BA)^{\#}=A^{\#}B^{\#}$ and $I^{\#}=I$. Well $(A+B)^{\#}=A^{\#}+B^{\#}$ also seems wierd to me but I wanted to start with $(\alpha A)^{\#}=\alpha A^{\#}$ to catch the idea2011-06-30
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    Huh, what is the text? That is really strange. determinants are not additive ($\det (A + B) \neq \det A + \det B$ in general). Now, $A^\sharp$ is a linear transformation, but the $\sharp$ "operator" is definitely not linear. (Also, @missing, [you should avail yourself the @-replies capabilities of this website](http://meta.stackexchange.com/questions/43019/how-do-comment-replies-work).)2011-07-01
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    @willie the book is "Algebra lineal" by Elon Lima, I think it's not well known in english language countries2011-07-01
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    @missing unfortunately I don't know Portuguese `:-(`2011-07-01

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