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 18h comment Orthogonal complement for infinite dimensional and open vector spaces. @Ranc That's correct 18h answered A set in a tensor product: 18h comment How to reconcile the existence of the least upper bound? Don't sweat computability: it doesn't mean what you think it means 18h comment If $A \ge B$ then $A[α] \ge B[α]$ @Surb also, your statement is not quite true without further assumptions on $x$ 1d revised Symmetric power subspace of tensor product deleted 3 characters in body 1d comment Symmetric power subspace of tensor product @Pedro I had planned on a second section at the time. I agree, though. 1d comment How to calculate a determinant of a general 2x2 block skew matrix? Note of course that if the matrices are real and if $L$ is odd, then neither of the matrices will be invertible. 1d comment How to calculate a determinant of a general 2x2 block skew matrix? You can express this determinant as $$\det(M) = \det(A - B^TD^{-1}B) \det(D)$$ but I don't know if this is really any faster. 1d comment Binomial theorem and the inequality $(1-e^{-x})^{\alpha}\leq (1-\alpha e^{-x})$ for $0<\alpha\le 1$ Ah, thanks sudha 1d comment Generalized “scalar product” based on multilinear form? Just an inkling, but if a nice answer to this question exists, I'm fairly sure it will have something to do with the tensor product of vector spaces (using either the symmetric or antisymmetric tensor product, probably). 1d comment Binomial theorem and the inequality $(1-e^{-x})^{\alpha}\leq (1-\alpha e^{-x})$ for $0<\alpha\le 1$ In fact, I believe the reverse inequality holds 1d comment Can there be a lottery of the natural numbers? If you want each integer to have an identical probability such that the probabilities sum to $1$, then at the very least you'll need to work over a non-archimedian field. Maybe a solution to this exists in non-standard analysis. 2d comment Is $C[0,1]$ larger than $\mathbb R$? It's equal. As a hint, consider the set of functions from $\Bbb Q\cap [0,1]$ to $\Bbb R$. 2d comment Find $D^*$ from Hoffman and Kunze Linear algebra The matrix you get for $D^*$ (just like the matrix for $D$) depends on the basis, though. 2d comment Find $D^*$ from Hoffman and Kunze Linear algebra No! $D^*$ is unique. It depends on the inner product, not the basis. 2d comment Find $D^*$ from Hoffman and Kunze Linear algebra I think your answer is probably what they're actually looking for, especially since they have specified the vector space. 2d comment Find $D^*$ from Hoffman and Kunze Linear algebra I think that $D^*$ can be defined by $D^*g= -Dg + g(0)p(t) + g(1)q(t)$ for some polynomials $p(t),q(t)$. 2d comment How to respectively give an upper bound for $\|XX^T\|_2$ and $diag (UXX^TU^T)$ with each column vector $\|x_i\|_2\leq 1$? @AlgebraicPavel ah, whoops, and thank you for the clarification. I always get those terms mixed Nov 23 comment Find $D^*$ from Hoffman and Kunze Linear algebra Well, then yes: you should have mentioned that. Even so, you need to be careful if you're using the matrix of $D$ to get the matrix of $D^*$. Note in particular that $D^*$ will only be the transpose of $D$ if you found $D$ with respect to an orthogonal (or orthonormal) basis. Nov 23 answered Let $x_0=0$. Define $x_{n+1}=\cos x_n$ for every $n\ge 0$