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分块矩阵的秩怎么求,分块矩阵的转置公式

时间:2023-05-04 09:09:39 阅读:194681 作者:102

Lesson 12: Making Block Matrices in LATEX

按列分块:

[AAb⋯An−1b]

$$xjkfdleft[xjkfdbegin{array}{c|c|c|c}A & Ab & xjkfdcdots & A^{n-1}bxjkfdend{array}xjkfdright]$$左右分块:

⎡⎣⎢⎢⎢⎢⎢a11a21⋮an1a12a22an2⋯⋯⋱⋯a1na2n⋮annb1b2⋮bn⎤⎦⎥⎥⎥⎥⎥

$$xjkfdleft[xjkfdbegin{array}{cccc|c}a_{11}&a_{12}&xjkfdcdots&a_{1n}&b_1xjkfdxjkfda_{21}&a_{22}&xjkfdcdots&a_{2n}&b_2xjkfdxjkfdxjkfdvdots&&xjkfdddots&xjkfdvdots&xjkfdvdotsxjkfdxjkfda_{n1}&a_{n2}&xjkfdcdots&a_{nn}&b_nxjkfdxjkfdxjkfdend{array}xjkfdright]$$四个对角:

C=[ACBD]

$$C=xjkfdleft[xjkfdbegin{array}{c|c}A& B xjkfdxjkfd xjkfdhline C& Dxjkfdend{array}xjkfdright]$$

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