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The calculation is lengthy, so I list steps instead. The identity is linear in $\vec p, \vec q$ respectively, so it only needs to verify the identity holds for $\vec p, \vec q$ take on the unit coordinate vectors $\vec i, \vec j, \vec k$. The rest is a length writing, but manageable this way. With the symmetry, it only needs to verify two case: $\vec p = ...


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The determinant of a $3\times3$ matrix can be viewed as the triple product of its columns (or rows): $$ \begin{align} \det\begin{bmatrix} x_1&y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3 \end{bmatrix} &= \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} \times \begin{bmatrix} y_1\\ y_2\\ y_3 \end{bmatrix} \cdot \begin{bmatrix} z_1\\ z_2\\ z_3 ...


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If you start with the definition of cross-product as $$\underline{a}\times\underline{b}=|\underline{a}||\underline{b}|\sin \theta \underline{\hat{n}},$$ where $\theta$ is the angle between $\underline{a}$ and $\underline{b}$ and $\underline{\hat{n}}$ is the unit vector perpendicular to $\underline{a}$ and $\underline{b}$ in the sense of a right-hand triad' ...


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Looking at the computation from the right angle, what you compute is a new vector in the dual space, such that $\vec a \times \vec b$ maps any vector $\vec c$ to $\mathbb R$ (or $\mathbb C$), in a way that $\det(\vec c, \vec a,\vec b)=(\vec a\times\vec b)\cdot \vec c$. You can see this by replacing $(i,j,k)$ by $(c_1, c_2, c_3)$. Everything else follows ...


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And the resulting would be your Cross Product or the coordinates of an orthogonal vector. My question is why? Why does forming it that way give you the magnitude of an orthogonal vector Your last equation can be written as $$ (a \times b)_i = \epsilon_{ijk} a_j b_k \quad (1) $$ where $\epsilon_{ijk}$ is the skew-symmetric or Levi-Civita tensor and ...


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Given a finite-dimensional oriented real inner-product space $V$ the group of symmetries preserving the inner-product and orientation is $\mathrm{SO}(V)$. The invariant tensors under $\mathrm{SO}(V)$ are $\delta_{ij}$, $\delta^{ij}$, and $\varepsilon_{ijk}$, along with the things they generate like $\delta^{ij}\varepsilon_{klm}$ and so on. The tensors ...


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1) A square matrix is non-invertible if its determinant vanishes. 2) The determinant of a matrix product of two square matrices is the product of the determinants of the two individual matrices. So, for square matrices the product is not invertible if one of the factors is not. For non-square matrices (which are non-invertible) I do not know if in some ...


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$A,B$ square: never See for example How to prove and interpret rank(AB)$\le$ min(rank(A),rank(B))? But... $$\pmatrix{1&0}\pmatrix{1\cr0} = (1).$$


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Using Cauchy-Schwarz (assuming we are talking about a Hilbert space, etc...) , $(V\cdot W)^2=V^2W^2$ iff $V$ and $W$ are parallel. I count 3 dot products, so the solution involving 1 cross product is more efficient in this sense, but the cross product is a bit more involved. If $(V\cdot W)=1$ (my interpretation of your question) and $V^2,W^2\ne1$, then at ...


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So the cross product of two vectors is actually enough. $|V \times W| = 0$ is the condition you want, for the equality $|V \times W| = |V|\cdot |W|\cdot Sin(\theta)$ where $\theta$ is the angle between V and W tells you that they are parallel.


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The fundamental problem is that there are an infinite number of solutions. Using your simultaneous equation approach gives the misleading impression that we have three equations in three unknowns. This is not quite the case as the system is not fully determined. The definition of the cross product tells us that : ...


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It's not true that the system can't have a solution if $A$ is not invertible. For example, to pick a silly example, $$ \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix} x = \begin{bmatrix}1\\0\end{bmatrix} $$ has solutions $x=(1,t)$ for every $t$. What a zero determinant means just that the solution, if it exists, will not be unique. However, instead ...


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In $\mathbb{R}^3$, we can indeed more or less identify $u\times v$ with $u\wedge v$ for two vectors $u,v \in \mathbb{R}^3$ (more accurately, $u\wedge v$ is the linear form given by scalar product with $u\times v$; or another way of putting thigs it is that $u\wedge v$ gives an "axial vector" or "pseudovector", not a vector), but this no longer works for ...


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First of all, note that the usual cross product already isn't associative! The seven-dimensional cross product also arises in this manner from the octonions. As pointed out in the latter link, the only dimensions in which cross products exist is $0,1,3,7$. This is because normed division algebras only exist in dimensions $1,2,4,8$: even though there are a ...



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