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matheuscmss.wordpress.com | ||
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lucatrevisan.wordpress.com
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| | | | | The spectral norm of the infinite $latex {d}&fg=000000$-regular tree is $latex {2 \sqrt {d-1}}&fg=000000$. We will see what this means and how to prove it. When talking about the expansion of random graphs, abobut the construction of Ramanujan expanders, as well as about sparsifiers, community detection, and several other problems, the number $latex {2 \sqrt{d-1}}&fg=000000$... | |
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nhigham.com
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| | | | | The spectral radius $latex \rho(A)$ of a square matrix $latex A\in\mathbb{C}^{n\times n}$ is the largest absolute value of any eigenvalue of $LATEX A$: $latex \notag \rho(A) = \max\{\, |\lambda|: \lambda~ \mbox{is an eigenvalue of}~ A\,\}. $ For Hermitian matrices (or more generally normal matrices, those satisfying $LATEX AA^* = A^*A$) the spectral radius is just... | |
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djalil.chafai.net
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| | | | | Let $X$ be an $n\times n$ complex matrix. The eigenvalues $\lambda_1(X), \ldots, \lambda_n(X)$ of $X$ are the roots in $\mathbb{C}$ of its characteristic polynomial. We label them in such a way that $\displaystyle |\lambda_1(X)|\geq\cdots\geq|\lambda_n(X)|$ with growing phases. The spectral radius of $X$ is $\rho(X):=|\lambda_1(X)|$. The singular values $\displaystyle s_1(X)\geq\cdots\geq s_n(X)$ of $X$ are the eigenvalues of the positive semi-definite Hermitian... | |
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carldagostino.wordpress.com
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| | | 1 post published by Carl D'Agostino on January 4, 2017 | ||
