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- Mar 2018
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www.seas.ucla.edu www.seas.ucla.educhol.pdf1
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positive semidefinite, but not positive definite, then it is singular
正半定矩阵是奇异的 $$ \begin{array}{rl} f(t) =& (x-tAx)^TA(x-tAx)\ =&-2t\Vert Ax\Vert^2 + t^2x^TA^3x \end{array} $$
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