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矩阵题

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hbghlyj posted 2025-5-7 01:22 |Read mode
  • 设\(A,B\)是\(n\)阶矩阵,满足\(AB = BA\),证明:\(r(A + B) \leq r(A) + r(B) - r(AB)\)。
  • 设$A$是非零矩阵,证明$A$可以写成某个列满秩矩阵与某个行满秩矩阵的乘积。
  • 设\(A = \begin{pmatrix}1 & - 1 & - 1 & - 1 \\- 1 & 1 & - 1 & - 1 \\ - 1 & - 1 & 1 & - 1 \\ - 1 & - 1 & - 1 & 1 \end{pmatrix}\),求\(\sum_{i,j = 1}^{4}A_{ij}\)。
  • 设\(n\)阶矩阵\(A = \begin{pmatrix}2 & 2 & 2 & \cdots & 2 \\0 & 1 & 1 & \cdots & 1 \\0 & 0 & 1 & \cdots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\0 & 0 & 0 & \cdots & 1\end{pmatrix}\),求\(A\)的所有代数余子式之和;
  • 设$n$阶非零实矩阵\(A = (a_{ij})\),\(n \geq 3\),如果\(a_{ij} = A_{ij}\),证明:
    (1)\(r(A) = n\);(2)\(A\)是正交矩阵。
  • 设\(A = (a_{ij})_{n \times n}\),如果\(\sum_{j = 1}^{n}a_{ij} = 0,\) \(i = 1,2,\cdots,n\)。求证:
    \(A_{11} = A_{12} = \cdots = A_{1n}\),其中\(A_{ij}\)为\(a_{ij}\)的代数余子式。
  • (1)设\(A,B\)是\(n\)阶矩阵,\(A,B\),\(A + B\)均可逆,证明:\(A^{- 1} + B^{- 1}\)也可逆,并求
    其逆。
    (2)设\(A,I - A,I - A^{- 1}\)均为可逆矩阵,证明:\((I - A)^{- 1} + (I - A^{- 1})^{- 1} = I.\)
  • 设有分块矩阵\(\begin{pmatrix}
      A & B \\
      C & D
      \end{pmatrix}\),其中\(A,D\)都可逆,证明:
    (1)\(\left| \begin{matrix}
    A & B \\
    C & D
    \end{matrix} \right| = \det(A - BD^{- 1}C)\det D\);
    (2)\((A - BD^{- 1}C)^{- 1} = A^{- 1} - A^{- 1}B(CA^{- 1}B - D)^{- 1}CA^{- 1}\)。
  • 设\(A,B \in P^{n \times n}\),求证:\((AB)^{*} = B^{*}A^{*}\)。
  • 任给\(A,B,C \in P^{n \times n}\),证明:
    (1)\(r(A) + r(B) \leq r(AB) + n\);
    (2)\(r(AB) + r(BC) \leq r(ABC) + r(B).\)
  • 设\(A,B\)是$n$阶矩阵,证明:
    (1)\(r(A - ABA) = r(A) + r\left( I_{n} - BA \right) - n\);
    (2)若\(A + B = I_{n}\),且\(r(A) + r(B) = n\),则\(A^{2} = A,B^{2} = B\),且\(AB = BA = 0.\)

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