Wikipedista:Matrix Computations

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Motto

• Я знаю, как управлять Вселенной. И скажите - зачем же мне бежать за миллионом?!
• I know how to control the universe. So tell me, why should I run for a million?!
• Vím, jak ovládat vesmír. Tak mi řekněte, proč bych se měl hnát za milionem?!

Grigorij Perelman, Komsomolskaja Pravda [1]

Matice, značení

Značení matic, které je rozumné používat (v článcích, které jsou primárně o maticích)

• $A,\,B,\,C\,\ldots$ matice (ne tučné);
• $A^{T}$ transponovaná matice;
• $A^{H}={\bar {A}}^{T}$ herminovsky sdružená matice ($A^{*}$ se plete s pseudoinverzí);
• $A^{-1}$ inverzní matice;
• $A^{\dagger }$ Moore-Penroseova pseudoinverze (možno též $A^{+}$, první mi příjde hezčí).

Pahýly, které jsem vyplodil

Maticové

• Polární rozklad
• Jacobiho matice (třídiagonální)
• Stochastická matice

Ostatní matika

• Tupperův vzorec
• Antonín Sochor (matematik)
• Česká matematická společnost

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Aux

Veselá limita

$\lim _{x\longrightarrow s}{\frac {x^{x^{x}}-x^{x^{s}}}{s^{x^{s}}-s^{s^{x}}}}$

Veselá rovnost

${\sqrt {2}}^{\sqrt {2}}=(2^{\frac {1}{2}})^{\sqrt {2}}=2^{{\frac {1}{2}}\cdot {\sqrt {2}}}=2^{\frac {\sqrt {2}}{2}}=2^{\frac {1}{\sqrt {2}}}={\sqrt[{\sqrt {2}}]{2}}$

$-1=(-1)^{1}=(-1)^{\frac {2}{2}}={\sqrt {(-1)^{2}}}={\sqrt {1}}=1$

Logaritmy

${\begin{array}{rcl}\log _{a}(x)+\log _{a}(y)\!\!&\!\!=\!\!&\!\!\log _{a}(x\cdot y)\\[2mm]\log _{a}(x)-\log _{a}(y)\!\!&\!\!=\!\!&\!\!\log _{a}\!{\Big (}{\frac {\displaystyle x}{\displaystyle y}}{\Big )}\\[2mm]s\cdot \log _{a}(x)\!\!&\!\!=\!\!&\!\!\log _{a}(x^{s})\\[2mm]-\log _{a}(x)\!\!&\!\!=\!\!&\!\!\log _{a}\!{\Big (}{\frac {\displaystyle 1}{\displaystyle x}}{\Big )}\\[2mm]\end{array}}\qquad \qquad \qquad {\begin{array}{rcl}\log _{a}(s)\cdot \log _{s}(x)\!\!&\!\!=\!\!&\!\!\log _{a}(x)\\[2mm]\log _{a}(x)\cdot \log _{x}(a)\!\!&\!\!=\!\!&\!\!1\\[2mm]x\!\!&\!\!=\!\!&\!\!a^{\displaystyle \,\log _{a}(x)}\\[2mm]x^{\displaystyle \,\log _{a}(y)}\!\!&\!\!=\!\!&\!\!y^{\displaystyle \,\log _{a}(x)}\\[2mm]\end{array}}\qquad \qquad \qquad {\begin{array}{rlc}\log _{a}(x)+\log _{b}(x)=\log _{c}(x),\;\;{\text{kde}}\;\;{\frac {\displaystyle 1}{\displaystyle \log _{s}(c)}}={\frac {\displaystyle 1}{\displaystyle \log _{s}(a)}}+{\frac {\displaystyle 1}{\displaystyle \log _{s}(b)}}\end{array}}$

Matice

${\begin{array}{cc}\left[{\begin{array}{ccc}1&1&1\\1-{\sqrt {2\,}}&{\sqrt {2\,}}+1&1\\{\sqrt {2\,}}-1&1-{\sqrt {2\,}}&1\\\end{array}}\right]&\left[{\begin{array}{ccc}1&-3&-3\\1-{\sqrt {2\,}}&{\sqrt {2\,}}-3&-3\\{\sqrt {2\,}}-1&1-{\sqrt {2\,}}&1\\\end{array}}\right]\\\left[{\begin{array}{ccc}1&1&1\\{\sqrt {2\,}}+1&1-{\sqrt {2\,}}&1\\-{\sqrt {2\,}}-1&{\sqrt {2\,}}+1&1\\\end{array}}\right]&\left[{\begin{array}{ccc}1&-3&-3\\{\sqrt {2\,}}+1&-{\sqrt {2\,}}-3&-3\\-{\sqrt {2\,}}-1&{\sqrt {2\,}}+1&1\\\end{array}}\right]\\\left[{\begin{array}{ccc}-1&3&3\\-{\sqrt {2\,}}-1&{\sqrt {2\,}}+3&3\\{\sqrt {2\,}}+1&-{\sqrt {2\,}}-1&-1\\\end{array}}\right]&\left[{\begin{array}{ccc}-1&-1&-1\\-{\sqrt {2\,}}-1&{\sqrt {2\,}}-1&-1\\{\sqrt {2\,}}+1&-{\sqrt {2\,}}-1&-1\\\end{array}}\right]\\\left[{\begin{array}{ccc}-1&3&3\\{\sqrt {2\,}}-1&3-{\sqrt {2\,}}&3\\1-{\sqrt {2\,}}&{\sqrt {2\,}}-1&-1\\\end{array}}\right]&\left[{\begin{array}{ccc}-1&-1&-1\\{\sqrt {2\,}}-1&-{\sqrt {2\,}}-1&-1\\1-{\sqrt {2\,}}&{\sqrt {2\,}}-1&-1\\\end{array}}\right]\end{array}}$

Sinus

${\begin{array}{rll}&\sin \left({\frac {n(2k-1)\pi }{2n+1}}-{\frac {(2k-1)\pi }{2(2n+1)}}\right)\sin \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(sinus souctu uhlu)}}\\=&\left[\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]\sin \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(roznasobeni)}}\\=&\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\sin \left({\frac {(2k-1)\pi }{2n+1}}\right)-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\sin \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(rozsireni zlomkem 2/2)}}\\=&\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\sin \left({\frac {2}{2}}\,{\frac {(2k-1)\pi }{2n+1}}\right)-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\sin \left({\frac {2}{2}}\,{\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(sinus dvojnasobneho uhlu)}}\\=&\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\left[2\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\left[2\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]&{\text{(prerovnani vyrazu)}}\\=&\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\,2\left[\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]^{2}-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\,2\left[\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]^{2}&{\text{(vztahy }}2[\cos(x)]^{2}=1+\cos(2x){\text{ a }}2[\sin(x)]^{2}=1-\cos(2x){\text{)}}\\=&\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\left[1+\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)\right]-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\left[1-\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)\right]&{\text{(roznasobeni)}}\\=&\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)+\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)\\&-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)+\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(prerovnani a vytknuti)}}\\=&\left[\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]\\&+\left[\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)+\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(hratky se znamenky a prerovnani)}}\\=&-\left[\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)-\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]\\&+\left[\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left(-{\frac {(2k-1)\pi }{2(2n+1)}}\right)-\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left(-{\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(kosinus souctu uhlu)}}\\=&-\cos \left({\frac {n(2k-1)\pi }{2n+1}}+{\frac {(2k-1)\pi }{2(2n+1)}}\right)+\cos \left({\frac {n(2k-1)\pi }{2n+1}}-{\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(spolecny jmenovatel v prvnim kosinu)}}\\=&-\cos \left({\frac {(2n+1)(2k-1)\pi }{2(2n+1)}}\right)+\cos \left({\frac {n(2k-1)\pi }{2n+1}}-{\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(kraceni v prvnim kosinu)}}\\=&-\cos \left((2k-1){\frac {\pi }{2}}\right)+\cos \left({\frac {n(2k-1)\pi }{2n+1}}-{\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(kosinus pi/2 je nula)}}\\=&\cos \left({\frac {n(2k-1)\pi }{2n+1}}-{\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(hurrraaa, hotovo!)}}\end{array}}$