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I was trying to find the cross-product, or a vector perpendicular to vectors in $\mathbb{R}^n$

Then I recalled the formula for the cross product in $\mathbb{R}^3$ and then tried to generalize it.

So, I let the vectors be $\left(a_{1, 1}, a_{1, 2},\cdots , a_{1, n}\right), \left(a_{2, 1}, a_{2, 2},\cdots , a_{2, n}\right) \cdots \left(a_{n-1, 1}, a_{n-1, 2}, \cdots , a_{n-1, n}\right)$

Then using the formula to find the cross product of two vectors in $\mathbb{R}^3$ and generalizing it I get the perpendicular vector to be $$\det\pmatrix{e_1&e_2&\cdots &e_n\\a_{1, 1}&a_{1, 2}&\cdots & a_{1, n}\\ a_{2, 1}&a_{2, 2}&\cdots & a_{2, n}\\ \vdots&\vdots & \ddots & \vdots\\a_{n-1, 1}&a_{n-1, 2}&\cdots & a_{n-1, n}}$$where $e_1, e_2,\cdots, e_n$ is like $i, j, k...$

So, I tried this with $3$ vectors in $\mathbb{R}^4$ then took the dot product of the resultant vector with each of the original vectors and the answer came as $0$ in every case, so I did confirm that it worked, but I do not know why. Is there any proof of this?

Asv
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1 Answers1

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Consider a scalar product of

$$\mathbf{v}=\det\begin{pmatrix}\mathbf{e}_1&\cdots&\mathbf{e}_n\\ a_{1,1}&\cdots&a_{1,n}\\ \vdots&\ddots&\vdots\\ a_{n-1,1}&\cdots&a_{n-1,n}\\ \end{pmatrix} = A_1 \mathbf{e}_1 + \ldots + A_n \mathbf{e}_n \\ \text{with} \qquad \mathbf{a}_i=\begin{pmatrix}a_{i,1}\\ \vdots\\ a_{i,n}\end{pmatrix} = a_{i,1} \mathbf{e}_1 + \ldots + a_{i,n} \mathbf{e}_n $$

Since the basis $\mathbf e_i$ is orthonormal:

$$ \mathbf v\cdot \mathbf a_i = A_1a_{i,1} + \ldots+a_{i,n}A_n = \det\begin{pmatrix}a_{i,1}&\cdots&a_{i,n}\\ a_{1,1}&\cdots&a_{1,n}\\ \vdots&\ddots&\vdots\\ a_{n-1,1}&\cdots&a_{n-1,n}\\ \end{pmatrix} $$

In this matrix two rows are exactly the same, so the determinant is zero.

Vasily Mitch
  • 10,129
  • Thank you, I had gotten to the A_1a_i, but didn't think of writing it as the determinant – Asv Feb 10 '21 at 18:42