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arkadiusz-jadczyk.eu | ||
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nhigham.com
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| | | | | The trace of an $latex n\times n$ matrix is the sum of its diagonal elements: $latex \mathrm{trace}(A) = \sum_{i=1}^n a_{ii}$. The trace is linear, that is, $latex \mathrm{trace}(A+B) = \mathrm{trace}(A) + \mathrm{trace}(B)$, and $latex \mathrm{trace}(A) = \mathrm{trace}(A^T)$. A key fact is that the trace is also the sum of the eigenvalues. The proof is by... | |
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akos.ma
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| | | | | From the wonderful book by Ian Stewart, here are the equations themselves; read the book to know more about them. | |
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codethrasher.com
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| | | | | A linear mapping from a vector space to a field of scalars. In other words, a linear function which acts upon a vector resulting in a real number (scalar) \begin{equation} \alpha\,:\,\mathbf{V} \longrightarrow \mathbb{R} \end{equation} Simplistically, covectors can be thought of as "row vectors", or: \begin{equation} \begin{bmatrix} 1 & 2 \end{bmatrix} \end{equation} This might look like a standard vector, which would be true in an orthonormal basis, but it is not true generally. | |
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bomongiaitich.wordpress.com
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| | | Nh?c l?i b?t ??ng th?c Holder: Cho $latex 1\le p, p' \le \infty$ th?a mãn $latex 1/p + 1/p'=1.$ L?u ý tr??ng h?p $latex p=1, p'=\infty$ và $latex p=\infty, p'=1.$ Xét không gian ?? ?o $latex (\Omega, \mathcal B, \mu)$, và $latex f\in L^p(\Omega, \mu), g\in L^{p'}(\Omega, \mu).$ Khi ?ó $latex fg\in L^1(\Omega, \mu)$ và... | ||
