Question:
Prove that for $(v_1,v_2,...,v_{d-1}) $ with $ v_i \in \mathbb{R}^d$ and $A \in \mathbb{R}^{d \times d}$ we have that: $\wedge _{1 \leq i \leq d-1} A v_i=comat(A) \wedge _{1 \leq i \leq d-1} v_i$
Rem: $ \wedge $ is the cross product $ \times $ between vectors..
Answer:
1- First remember that by definition $\wedge _{1 \leq i \leq d-1} A v_i$ is the only vector satisfying: $[Av_1,Av_2,...,Av_{d-1},w]=det[Av_1,Av_2,...,Av_{d-1},w]=\wedge _{1 \leq i \leq d-1} A v_i \cdot w $
2-We begin to assume that $A$ is an invertible matrix, so by property $A^{-1}=\frac{comat(A)^T}{det(A)}$ moreover we know that $\underline{\underline{A}}(\underline{v_1}, \underline{v_2}, ... , \underline{v_{d-1}}, \underline{\underline{A}}^{-1}\underline{w})= (\underline{\underline{A}} \underline{v_1}, \underline{\underline{A}} \underline{v_2}, ... , \underline{\underline{A}} \underline{v_{d-1}},
\underline{\underline{A}} \underline{\underline{A}}^{-1}\underline{w})$ hence $det((\underline{\underline{A}} \underline{v_1}, \underline{\underline{A}} \underline{v_2}, ... , \underline{\underline{A}} \underline{v_{d-1}},
\underline{\underline{A}} \underline{\underline{A}}^{-1}\underline{w})) = det(\underline{\underline{A}})det((\underline{v_1}, \underline{v_2}, ... , \underline{v_{d-1}}, \underline{\underline{A}}^{-1}\underline{w}))=det(A) \cdot \wedge _{1 \leq i \leq d-1} v_i \cdot (A^{-1} w) = det(A) \cdot \wedge _{1 \leq i \leq d-1} v_i (\frac{comat(A)^T}{det(A)} w) = (\wedge _{1 \leq i \leq d-1} v_i) (comat(A)^T) w \Rightarrow det((\underline{\underline{A}} \underline{v_1}, \underline{\underline{A}} \underline{v_2}, ... , \underline{\underline{A}} \underline{v_{d-1}}, \underline{w})) = (\wedge _{1 \leq i \leq d-1} v_i) (comat(A)^T) w $
-Now let use the following identity to finish $u \cdot (B w) = (B^T u) w$ and we get $ (\wedge _{1 \leq i \leq d-1} v_i) (comat(A)^T) w = comat(A) (\wedge _{1 \leq i \leq d-1} v_i) w \Rightarrow \wedge _{1 \leq i \leq d-1} A v_i=comat(A) \wedge _{1 \leq i \leq d-1} v_i $
3-Now in the case that $A$ is not an invertible matrix.
We know that $GL_n ( \mathbb{R})$ is dense in $M_n( \mathbb{R}) $. Hence $\exists (A_n)_n \in M_n( \mathbb{R}) s.t. \lim_{n \to \infty}A_n = A$. But according to "2-" for all $n$ we have that $\wedge _{1 \leq i \leq d-1} A_n v_i=comat(A_n) \wedge _{1 \leq i \leq d-1} v_i$.
-Now let return to a basic understanding of what is $\wedge _{1 \leq i \leq d-1} A v_i$ to conclude in order to "put in" the limit.
We begin by reasoning for $(v_1,...,v_{d-1})$ fixed. The mapping that to $A$ associate $\wedge _{1 \leq i \leq d-1} A v_i$ is a polynomial mapping of coefficient of $A$ hence it is continue.
In an other way the mapping that to $A$ associate $comat(A)$ is a polynomial mapping with coefficient of $A$, thus it is too continue.
In conclusion we can "put in" the limit and we get $\wedge _{1 \leq i \leq d-1} A v_i=comat(A) \wedge _{1 \leq i \leq d-1} v_i$
Is it correct, I have mostly doubt on the lats part of "3-"?
My answer is correct.
I just write this answer in order to mark this question as solved.