Matrices: Example where $AB = I \ne BA$
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If $AB = I$ then $BA = I$
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If $A,B$ have eigenvalues $\pm1$, what can I say about the eigenvalues of $AB,BA$?
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Given a matrix, is there always another matrix which commutes with it?
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If $A$ and $B$ two $n\times n$ matrix such that $AB=BA$, then $\dim(A\cap B)$?
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Do the non-zero eigenvalues of AB and BA have the same algebraic multiplicity (for AB and BA not square)?
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Show that matrices $A$ and $B$ commute if and only if $\begin{vmatrix} b & a-c \\ e & d-f\end{vmatrix}=0.$
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$A$ and $B$ are $3\times 3$ real matrices such that $\operatorname{rank}(AB)=1$, then $\operatorname{rank}(BA$) can not be which of the following?
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Are the eigenvalues of $AB$ equal to the eigenvalues of $BA$? (Citation needed!)
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Coordinate-free proof of $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$?
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Is it true that the order of $ab$ is always equal to the order of $ba$?
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If for two suitable matrices $B$ and $C$, $AA^TB=AA^TC$ can we conclude that $A^TB=A^TC$?
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A matrix is similar to its transpose
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$A^2+B^2=AB$ and $BA-AB$ is non-singular
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Matrix in a matrix
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A weak converse of $AB=BA\implies e^Ae^B=e^Be^A$ from "Topics in Matrix Analysis" for matrices of algebraic numbers.
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Similar matrices-Identity
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$AB=BA$ with same eigenvector matrix