$I⊗A-A⊗I$的秩

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Kronecker product
代数基础 | Kronecker积
Chapter 13Kronecker Products
相关帖子: 可交换的线性变换

$A$是$2×1$矩阵 A = Array[Subscript[a, ##] &, {2, 1}]; B = KroneckerProduct[IdentityMatrix[2], A] - KroneckerProduct[A, IdentityMatrix[2]] MatrixRank[B] Copy the Code 得到\begin{array}l B= \begin{pmatrix} 0 & 0 \\ a_{2,1} & -a_{1,1} \\ -a_{2,1} & a_{1,1} \\ 0 & 0 \\ \end{pmatrix}\\ \operatorname{rank}B=1 \end{array}

$A$是$2×2$矩阵 A = Array[Subscript[a, ##] &, {2, 2}]; B = KroneckerProduct[IdentityMatrix[2], A] - KroneckerProduct[A, IdentityMatrix[2]] MatrixRank[B] Copy the Code 得到\begin{array}l B=\begin{pmatrix} 0 & a_{1,2} & -a_{1,2} & 0 \\ a_{2,1} & a_{2,2}-a_{1,1} & 0 & -a_{1,2} \\ -a_{2,1} & 0 & a_{1,1}-a_{2,2} & a_{1,2} \\ 0 & -a_{2,1} & a_{2,1} & 0 \\ \end{pmatrix} \\ \operatorname{rank}B=2 \end{array}

$A$是$2×3$矩阵 A = Array[Subscript[a, ##] &, {2, 3}]; B = KroneckerProduct[IdentityMatrix[2], A] - KroneckerProduct[A, IdentityMatrix[2]] MatrixRank[B] Copy the Code 得到\begin{array}l B= \begin{pmatrix} 0 & a_{1,2} & a_{1,3}-a_{1,2} & 0 & -a_{1,3} & 0 \\ a_{2,1} & a_{2,2}-a_{1,1} & a_{2,3} & -a_{1,2} & 0 & -a_{1,3} \\ -a_{2,1} & 0 & -a_{2,2} & a_{1,1} & a_{1,2}-a_{2,3} & a_{1,3} \\ 0 & -a_{2,1} & 0 & a_{2,1}-a_{2,2} & a_{2,2} & 0 \\ \end{pmatrix}\\ \operatorname{rank}B=4 \end{array}

$A$是$3×3$矩阵 A = Array[Subscript[a, ##] &, {2, 3}]; B = KroneckerProduct[IdentityMatrix[2], A] - KroneckerProduct[A, IdentityMatrix[2]] MatrixRank[B] Copy the Code 得到\begin{array}l B= \begin{pmatrix} 0 & a_{1,2} & a_{1,3}-a_{1,2} & 0 & -a_{1,3} & 0 \\ a_{2,1} & a_{2,2}-a_{1,1} & a_{2,3} & -a_{1,2} & 0 & -a_{1,3} \\ a_{3,1}-a_{2,1} & a_{3,2} & a_{3,3}-a_{2,2} & 0 & -a_{2,3} & 0 \\ 0 & -a_{2,1} & 0 & a_{1,1}-a_{2,2} & a_{1,2} & a_{1,3}-a_{2,3} \\ -a_{3,1} & 0 & -a_{3,2} & a_{2,1} & a_{2,2}-a_{3,3} & a_{2,3} \\ 0 & -a_{3,1} & 0 & a_{3,1}-a_{3,2} & a_{3,2} & 0 \\ \end{pmatrix}\\ \operatorname{rank}B=6 \end{array}

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本层节省宽度, 从而避免横向滚动条.(当屏幕宽度>1150px时适用)

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对于所有$n×n(n≥3)$矩阵$A$,求证$\operatorname{rank}(I⊗A-A⊗I)$的最大值为$2n$

$n$	rank
3	6
4	8
5	10
6	12

Czhang271828

hbghlyj 发表于 2022-10-30 00:22
对于所有$n×n(n≥3)$矩阵$A$,求证$\operatorname{rank}(I⊗A-A⊗I)$的最大值为$2n$

考察 $A$ 的特征向量 $v_i$ ($Av_i=\lambda_i v_i$), 则 $$
(A\otimes I-I\otimes A)(v_i\otimes v_j)=(\lambda_i -\lambda_j) (v_i\otimes v_j).
$$
因此 $A$ 的零空间至少为 $n$ 维度. 例如取 $A=\mathrm{diag}(1,\ldots, n)$, 则 $A\otimes I-I\otimes A$ 还是对角矩阵, 对角线上有且仅有 $n$ 个 $0$.

其实凡事先考虑特殊情况嘛, 若 $P^{-1}A P=J$, 那么有相似变换
$$
(P^{-1}\otimes P^{-1})(A\otimes I-I\otimes A)(P\otimes P)= J\otimes I-I\otimes J.
$$
剩下慢慢玩就行了.