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# Search Results

Results tagged with Search options user 297
72 results

Questions about the properties of vector spaces and linear transformations, including linear systems in general.

A bivector is an element of $\bigwedge^2 V$, so it is dual to a $2$-form on $V$. You can think of a bi-vector as a tiny piece of area. If $V$ is three dimensional and comes with an inner product, th …
answered Nov 4 '09 by David E Speyer
I can think of (at least) two ways of interpreting this question. First: You are given some specific list of $k$ points $x_1$, $x_2$, ..., $x_k$ in $\mathbb{R}^n$, and you want to detect whether $k$ …
answered Oct 16 '15 by David E Speyer
If you have a lot of duplicate rows (and you know what they are), you can reduce to a smaller matrix. I'll start with an example, because writing out the general case will be notationally annoying. L …
answered Sep 24 '10 by David E Speyer
Another proof: For any $I$ and $J$ two subsets of $\{1,2,\ldots,n\}$ of the same cardinality, let $D(I,J)$ be the minor in rows $I$ and columns $J$. Let $Pf(I)$ be the Pfaffian $\sqrt{D(I,I)}$. We set …
answered Apr 22 '10 by David E Speyer
This isn't true if $n$ is odd. For example, if $n=3$, then your formula is $(a,b,c) = (\cos^2 \theta, 2 \sin \theta \cos \theta, \sin^2 \theta)$ and it always lies in the hyperplane $a+c=1$. More gene …
answered Mar 9 '15 by David E Speyer
What you are seeing is that the orthogonal matrices of determinant $-1$ swap the two spin representations. The first several parts of this argument will be valid for $(4n+2) \times (4n+2)$ matrices as …
answered Dec 1 '10 by David E Speyer
I'll describe a number of $n^2+n$ dimensional subvarieties. I have not yet found a point which isn't on one of them (in particular, they subsume all of your $2n$ dimensional families). If we are very …
answered Aug 23 '17 by David E Speyer
$\def\FF{\mathbb{F}_2}$ Indeed, the linear bound is close to right. I can show that we can't beat about $3.197 n$. For convenience, set $m=n/2$. Fix a constant $c$ and suppose that we can find $\geq … answered May 1 '12 by David E Speyer The inverse of a matrix is the adjoint divided by the determinant. So what you want to compute is the determinant of an$(n-1) \times (n-1)$submatrix, divided by the determinant of your original matr … answered Apr 15 '11 by David E Speyer This is false! Let $$A = \begin{bmatrix} 0&0&0&1 \\ &0&1&0 \\ &&0&0 \\ &&&0 \\ \end{bmatrix}.$$ Imposing that$XA=AX$for upper triangular$X$gives linear equations on the$10$entries of$X$. Solvi … answered Jul 10 '18 by David E Speyer Let$J_4(n)$be the$n \times n$matrix $$\begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & \cdots \\ -1 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 1 & 0 & 0 & \cdots \\ 0 & 0 & -1 & 0 & 0 & 0 & \cdots \\ 0 … answered Feb 12 '13 by David E Speyer Partial progress: It's easy to achieve n-3. Consider matrices of the form$$\begin{pmatrix} 0 & 0 & r_1 & r_2 & \cdots & r_{n-3} & 0 \\ 0 & 0 & 0 & r_1 & \cdots & r_{n-4} & r_{n-3} \\ r_1 & 0 & & … answered Feb 12 '13 by David E Speyer OK, here might be an answer to the question you are meaning to ask: Let$a_1$, ...,$a_n$be unit vectors in$\mathbb{R}^d$. Let$G$be a group acting linearly on$\mathbb{R}^d$, which permutes the$ …
answered Feb 4 '11 by David E Speyer
This is an elaboration of Thierry's answer: If $0 \leq a_0 < a_1 < \cdots < a_{d-1}$ is any sequence of nonnegative integers, then the determinant $\det \left( x_j^{a_i} \right)$ is equal to $\prod_{i … answered Apr 7 '11 by David E Speyer Assuming that the Hermitian minor has distinct eigenvalues, there is nothing we can say about the eigenvalues of$A$. Let the eigenvalues of the Hermitian minor be$\lambda_1$, ...,$\lambda_{n-1}\$, s …
answered Mar 27 '12 by David E Speyer

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