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2

Short and a little inaccurate answer: vector is one-dimensional tensor, matrix is a two-dimensional tensor. More details now: Tensors are multidimensional arrays which have certain properties. Not every multidimensional array is a tensor, check this discussion for more details. There are two types of one-dimensional tensors: vectors and co-vectors. Both ...


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Indeed, $f$ is a symmetric form, since $\omega$ and $\omega '$ are Grassmann-even: $$(\text dx \wedge \text d y)\wedge (\text d z \wedge \text d t)=(\text d z \wedge \text d t)\wedge(\text dx \wedge \text d y)$$etc.. Now, to calculate the signature, you should find a basis which diagonalizes $\omega$, the dimension of the space is $6$. A basis is given ...


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I suspect you are overwhelmed by the plethora of variables that obscure the fundamental cyclometric symmetry of the problem. You may scale all but one of them out of the symmetric matrix M, part of whose (symmetric) inverse you are seeking, really, by redefining $${\bf{M}} \equiv {\bf{k}}-\omega^2 {\bf{m}}= -k \left[\begin{matrix} 2(x-1) & 1 & 0 ...


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Write the right part of your last equation in vector form, the $T$ stands for transposition: $$\mathbf{y}= (\mathbf{{T^{-1}}})^T \cdot\mathbf{x}.$$ Take the transpose of this equation, ${(A\cdot B)}^T=B^T \cdot A^T$: $$\mathbf{y}^T= {{((\mathbf{{T^{-1}}})^T \cdot\mathbf{x})}}^T={\mathbf{x}^T}\cdot\mathbf{{\mathbf{T^{-1}}} }.$$ While the result is one ...


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I'll use a notation where I write vectors as $\vec{v}$ (because that's how \vec renders and I have no idea how to turn it into boldface in MathJax) and use primes to indicate the different basis: $v'^i$ is the components of $\vec{v}$ in the primed basis (some people use primes on the indices, which I find confusing). So given some basis ...


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I) When Ref. 1 writes let there be $\ell$ equations (17'') of constraints, it is implicitly assumed that they are independent, as also noted in Peter Diehr's answer. Obviously this implies that $\ell\leq 3N$. Moreover, the rectangular $\ell \times 3N$ matrix $$\tag{A} \left( \frac{\partial f_k}{\partial x_i}\right)_{1\leq k\leq \ell, ~1\leq i \leq ...


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The equations of constraint must be independent of each other - otherwise they would represent the same constraint; this independence is what guarantees that Lagrange's theorem is satisfied. Joos' method is correct if the conditions are met. An Introduction to Lagrange Multipliers provides a geometric analysis which is often helpful in understanding ...


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To answer the generic dimension part, in case it is not self-evident from Valter's answer: In D dimensions, the rotation matrix is the exponential of an angle θ times a matrix K, a normalized generator of the corresponding rotation group SO(D) around some unit axis D-vector k, in the vector representation, so the matrix is D×D. The eigenvectors of these ...


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Schmidt decomposition is in general a singular value decomposition (SVD) and it is applied on wave vectors and not on density matrices. While dealing with bi-partite wave vectors we use SVD because there is no restriction that the size of the two systems in question are the same. So the matrix of the wave vector coefficients can be rectangular and SVD can ...



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