query: Relation between eigen values of $AB$ and $BA$
positive_passages.docid: 24454
positive_passages.text: The relationship between the eigenvalues of matrices $XY$ and $YX$
negative_passages.docid: 43570
negative_passages.text: Finding two non-zero matrices where $AX = BA$
negative_passages.docid: 30923
negative_passages.text: Is the matrix $A^*A$ and $AA^*$ hermitian?
negative_passages.docid: 50010
negative_passages.text: Is this how eigenvalues of some matrix $A$ are related to the inverse of $A$?
negative_passages.docid: 90576
negative_passages.text: Symmetric matrix eigenvalues
negative_passages.docid: 6096
negative_passages.text: If $A+B=AB$ then $AB=BA$
negative_passages.docid: 102166
negative_passages.text: What is the relation between rank of a matrix, its eigenvalues and eigenvectors
negative_passages.docid: 158406
negative_passages.text: $A,B$ be Hermitian.Is this true that $tr[(AB)^2]\le tr(A^2B^2)$?
negative_passages.docid: 62474
negative_passages.text: If $AB=I_n$ and $BA=I_m$ then $n=m$.
negative_passages.docid: 30866
negative_passages.text: Relation between rank and number of distinct eigenvalues of a matrix
negative_passages.docid: 19132
negative_passages.text: Similar matrices have the same eigenvalues with the same geometric multiplicity
negative_passages.docid: 128760
negative_passages.text: If $A$ and $B$ two $n\times n$ matrix such that $AB=BA$, then $\dim(A\cap B)$?
negative_passages.docid: 52260
negative_passages.text: Diagonalization and eigenvalues
negative_passages.docid: 8638
negative_passages.text: If I know $AB$, how can I calculate $BA$?
negative_passages.docid: 98323
negative_passages.text: If A is invertible, prove that $\lambda \neq 0$, and $\vec{v}$ is also an eigenvector for $A^{-1}$, what is the corresponding eigenvalue?
negative_passages.docid: 39646
negative_passages.text: Is there a relation between $\rho(A) + \rho(B)$ and $\rho(A+B)$, where $\rho$ is the spectral radius?
negative_passages.docid: 26570
negative_passages.text: Prove $\operatorname{rank}(AB) = \operatorname{rank}(B)$