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2. Which types of tensors admit a representation using geometric algebra? On p.4 of this document we can see that multivectors over a given Euclidean space $\mathbb{R}^n$ do not have arbitrarily high grade/order; instead the highest order possible is $n$ (the determinant/volume element). This is because of the two products available in geometric algebra, ...


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You have the correct definition in Bogachev's book: to say that $X$ "is Gaussian with values in $C([a,b])$" is to say that $X$ is a random element of $C([a,b])$ with the property that $b^∗(X)$ is normally distributed for each $b^∗$ in the dual of $C([a,b])$.


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Indeed one has to be careful with the notion of dual module: In the context of representation theory of groups (more generally, when studying modules over Hopf algebras), the dual of a $G$-module $V$ is usually defined as having the vector space dual $V^{\ast}$ as its underlying vector space, and with the $G$-action defined by $(g.\varphi)(v) := \varphi(g^{...


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Complementarity problems is simply optimization problems with a special kind of constraints. Essentially orthogonality constraints between two non-negative vectors, $x\geq 0, y\geq 0, x^Ty = 0$. This arise, for example, in purely geometric applications (orthogonality constraints), or in situations where you want to encode either-or conditions ($x_i$ is zero ...


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Consider $S = \{\alpha_1\}, S^*=\{f_2\}$. Then $c_1f_2(\alpha_1)=0$ for any $c_1$. So your hypothesis does not hold generally. Now take any $S$ and let $S^*$ be its dual. Consider $f=\sum_{f_i\in S}c_i f_i.$ If some $c_j$ is nonzero, then the dual vector $\alpha_j$ of $f_j$ is contained in $S$, and $f(\alpha_j)=c_j\neq 0$. Hence we see in this case indeed ...


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Let $(v_1,\dots,v_n)$ be a basis for $V$, and form a dual basis $(v_1^*,\dots,v_n^*)$ for $V^*$. Similarly let $(w_1,\dots,w_m)$ be a basis for $W$, and form a dual basis $(w_1^*,\dots,w_m^*)$ for $W^*$. Suppose $Tv_j=\sum_{i=1}^ma_{ij}w_i$, so that the matrix for $T$ with respect to these bases is $[a_{ij}]$. Then \begin{align*} (T^*w_j^*)\left(\sum_{i=1}^...


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The proof that my professor used in her notes for the rank-nullity theorem didn't use either matrix representations or linear functionals. For a linear transformation $T: V \rightarrow W$, take a basis for the kernel and add on vectors to extend it to a basis for $V$. Because it is a basis for $V$, you can express a any vector in $V$ as a linear combination ...


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This isn't actually dual; this is just the usual Yoneda lemma for functors $\mathbb{D}^{\mathrm{op}}\to\mathbf{Set}$, where $\mathbb{D}=\mathbb{C}^{\mathrm{op}}$, since the functor $(C,-)$ on $\mathbb{C}$ corresponds to the functor $(-,C)$ on $\mathbb{D}^{\mathrm{op}}$. There is, regardless, a subtlety to the applying the duality principle when functors are ...



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