Today I have seemingly simple question and maybe someone knows the answer without getting into messy calculations.

So we have $n$ vectors $v_1,\dots,v_n\in\mathbb{R}^n$ and let us assume for the moment that they form a basis. Furthermore we look at all vectors which are orthogonal to the hyperplane spanned by these vectors, so $$\mathrm{Ker}\begin{pmatrix}v_1-v_n\\\vdots \\ v_{n-1}-v_n\end{pmatrix}.$$

My question: Does the following equation hold for some $N\neq 0$ in the above kernel? $$\sum_{i=1}^{n}\det\begin{pmatrix}v_1 \\ \vdots \\ v_{i-1} \\ v \\ v_{i+1} \\ \vdots \\ v_n\end{pmatrix} =\langle N,v\rangle,\qquad\forall\,v\in\mathbb{R}^n $$

It's quite easy to see in $\mathbb{R}^2$, but as I said: I'm looking for a proof without messy calculations.

Thank you!



Since $\{v_1,\dotsc,v_{n-1},v_n\}$ is a basis, so too is $\{v_1-v_n,\dotsc,v_{n-1}-v_n,v_n\}$, and hence $$ \forall 1 \leq k \leq n, \quad \det(v_1-v_n \vert \cdots \vert v_{n-1}-v_n \vert v_k) = \begin{cases} 0, &\text{if $1 \leq k \leq n-1$,}\\ \det(v_1 \vert \cdots \vert v_n) \neq 0, &\text{if $k = n$.} \end{cases} $$ By multilinearity of the determinant, $v \mapsto (v_1-v_n \vert \cdots \vert v_{n-1}-v_n \vert v)$ therefore defines a non-zero linear functional on $\mathbb{R}^n$ with kernel the hyperplane $\Pi$ spanned by $\{v_1-v_n,\dotsc,v_{n-1}-v_n\}$, so that there exists a unique non-zero vector $N \in \Pi^\perp$ such that $$ \forall v \in \mathbb{R}^n, \quad \det(v_1 - v_n \vert \cdots \vert v_{n-1} - v_n \vert v) = \langle N ,v \rangle; $$ in fact, $N = (v_1 - v_n) \times \cdots \times (v_{n-1}-v_n)$ is precisely the generalised cross product of $v_1-v_n,\dotsc,v_{n-1}-v_n$. You can then use multilinearity and antisymmetry of $\det$ (and a little bit of care) to check that $$ \forall v \in \mathbb{R}^n, \quad \langle N, v\rangle = \det(v_1-v_n \vert \cdots \vert v_{n-1}-v_n \vert v) = \sum_{k=1}^{n} \det(v_1 \vert \cdots \vert v_{k-1} \vert v \vert v_{k+1} \vert \cdots \vert v_n), $$ as required.

  • $\begingroup$ great thank you! that was exactly what I was looking for! $\endgroup$ – Richard Aug 29 '14 at 12:17

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