A skew symmetric map $\phi$ if $\phi(x_1, ... ,x_p)= 0$ when $x_i = x_j$, then $\phi(x_1, ... ,x_p)= 0$ when $x_1, ... ,x_p$ are lin. dependent

Question

In the book of Linear Algebra By Werner Greub at page 100, it is given that;

Let $\phi$ a p-linear map from E to F.Then the following statements are equivalent. ... (i) $\phi$ is skew symmetric

(ii) $\phi(x_1, ... ,x_p)= 0$ whenever $x_i = x_j$ for some pair $i\not = j$

(iii) $\phi(x_1, ... ,x_p)= 0$ whenever the vectors $x_1, ... ,x_p$ are linearly dependent.

My question is if we assume that $(ii)$ true, the vectors $x_1, ... ,x_p$ have be linearly dependent since $x_i = 1_E x_j$, so (iii) is a direct conclusion of $(ii)$, but in the book, it chooses $p$ without lost of generality and writes it as a linear combination of the vectors $x_1, ... ,x_{p-1}$, but I didn't get it. I mean there is nothing wrong in that proof, but I think it is unnecessary because there is no way that (ii) is true and the vector $x_1, ... ,x_p$ are linearly independent.

Edit:

What I didn't get it is that when $ii$ is assumed, is there any way that $iii$ can be false ?

Answer 1

The problem is that the statements (ii) and (iii) have quantifiers in them : the word "whenever" must be understood as "for all $x_1,\dots,x_p$ such that...".

Condition (ii) tells you only what happens when two vectors in $x_1,\dots,x_p$ are equal, and doesn't tell you anything about what happens when $x_1,\dots,x_p$ are only linearly dependent; (ii) is only a special case of (iii), and that's why you need a proof.

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Let us make this more abstract. Let us give name to some affirmations :

• $A $ will denote the affirmation "$\phi(x_1,\dots ,x_p)=0$"
• $B$ will denote the affirmation "$x_i=x_j$ for some $i\neq j$"
• $C$ will denote the affirmation "$x_1,\dots ,x_p$ are linearly dependent".

Then $(ii)$ is the affirmation $$\forall x_1,\dots ,x_p\, (B\Rightarrow A)$$ and $(iii)$ is the affirmation $$\forall x_1,\dots ,x_p \, (C\Rightarrow A).$$ Moreover, $B\Rightarrow C$; so if $(iii)$ holds then we have for all $x_1,\dots ,x_p$ $$B\Rightarrow C\Rightarrow A,$$ and thus $(ii)$ holds (this is why I said that $(ii)$ was a special case of $(iii)$). So the proof of $(iii)\Rightarrow (ii)$ is simple.

But if you only know $(ii)$, then you have $B\Rightarrow A$ and $B\Rightarrow C$, but you can't deduce that $C\Rightarrow A$. Doing so would just be a bad syllogism, which is a frequent logical mistake.