线性代数

线性代数

分块矩阵

A=(A1A2A3A4),B=(B1B2B3B4) A=\begin{pmatrix}A_1&A_2\\[2ex]A_3&A_4\end{pmatrix},\quad B=\begin{pmatrix}B_1&B_2\\[2ex]B_3&B_4\end{pmatrix} A+B=(a11+b11a12+b12a13+b13a14+b14a21+b21a22+b22a23+b23a24+b24a31+b31a32+b32a33+b33a34+b34a41+b41a42+b42a43+b43a44+b44)=(A1+B1A2+B2A3+B3A4+B4), A+B=\left(\begin{array}{cc|cc}a_{11}+b_{11}&a_{12}+b_{12}&a_{13}+b_{13}&a_{14}+b_{14}\\a_{21}+b_{21}&a_{22}+b_{22}&a_{23}+b_{23}&a_{24}+b_{24}\\\hline a_{31}+b_{31}&a_{32}+b_{32}&a_{33}+b_{33}&a_{34}+b_{34}\\a_{41}+b_{41}&a_{42}+b_{42}&a_{43}+b_{43}&a_{44}+b_{44}\end{array}\right)=\left(\begin{matrix}A_1+B_1&A_2+B_2\\A_3+B_3&A_4+B_4\end{matrix}\right), AB=(A1B1+A2B3A1B2+A2B4A3B1+A4B2A3B2+A4B4) AB=\begin{pmatrix}A_1B_1+A_2B_3&A_1B_2+A_2B_4\\A_3B_1+A_4B_2&A_3B_2+A_4B_4\end{pmatrix} λA=(λA1λA2λA3λA4),AT=(A1TA3TA2TA4T),A=(A1A2A3A4),tr(A)=tr(A1)+tr(A4) \lambda A=\begin{pmatrix}\lambda A_1&\lambda A_2\\\lambda A_3&\lambda A_4\end{pmatrix},A^T=\begin{pmatrix}A_1^T&A_3^T\\A_2^T&A_4^T\end{pmatrix},\overline{A}=\begin{pmatrix}\overline{A_1}&\overline{A_2}\\\overline{A_3}&\overline{A_4}\end{pmatrix},\:\mathrm{tr}(A)=\mathrm{tr}(A_1)+\mathrm{tr}(A_4) (A00B)1=(A100B1) {\begin{pmatrix}A&0\\[2ex]0&B\end{pmatrix}}^{-1}={\begin{pmatrix}A^{-1}&0\\[2ex]0&B^{-1}\end{pmatrix}}

(1)adbc,(abcd)1=1adbc(dbca)(1)\quad \left.ad\neq bc,\quad\left(\begin{array}{cc}a&b\\c&d\end{array}\right.\right)^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc}d&-b\\-c&a\end{array}\right)

(2)a,b,cF3,a×bc0(2)\quad a,b,c\in F^3,\text{且}a\times b\cdot c\neq0 hug

(abc)1=1a×bc(b×cc×aa×b)T \left.\left(\begin{array}{c}a\\b\\c\end{array}\right.\right)^{-1}=\frac{1}{a\times b\cdot c}\left(\begin{array}{c}b\times c\\c\times a\\a\times b\end{array}\right)^{T}
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