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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 $(g.\varphi)(v) := \varphi(g^{... 1 Complementarity problems is simply optimization problems with a special kind of constraints. Essentially orthogonality constraints between two non-negative vectors,$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$is zero ... 2 Consider$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$. Hence we see in this case indeed ... 1 Let$(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}^... 1 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 transformationT: 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$as a linear combination ... 3 This isn't actually dual; this is just the usual Yoneda lemma for functors$\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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