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 Jan4 comment Eigenvalue of the substraction of 2 matrices If you have to prove or disprove the statement "If $A$ and $B$ have the same eigenvalues, then $A-B$ has eigenvalue $0$", then one counterexample is sufficient to disprove the statement. If you want to prove that the statement is true, you'll need to show it for general matrices $A$ and $B$ and not just using one example. Jan4 reviewed Approve Why is this $0 = 1$ proof wrong? Jan4 reviewed Edit Maximizing volume in Calculus Jan4 revised Maximizing volume in Calculus improved formatting Jan4 reviewed Edit How is this a counter example to “ $A$ and $B$ are isomorphic but $G/A \ncong G/B”$? Jan4 revised How is this a counter example to “ $A$ and $B$ are isomorphic but $G/A \ncong G/B”$? improved version Jan4 reviewed Looks OK How to compute sum with non consecutive indices in Maple? Jan3 revised $0$ and $p^n \mathbb{Z}_p$ are the only ideals of $\mathbb{Z}_p$ edited body Jan3 revised $0$ and $p^n \mathbb{Z}_p$ are the only ideals of $\mathbb{Z}_p$ added 8 characters in body Jan3 answered $0$ and $p^n \mathbb{Z}_p$ are the only ideals of $\mathbb{Z}_p$ Jan3 comment $0$ and $p^n \mathbb{Z}_p$ are the only ideals of $\mathbb{Z}_p$ @fretty: $n$ can be $0$. Jan3 comment Diagonalization versus s.d. product for non-commuting Hermitian matrices @nomadreid: Yes, they are equivalent, which is what I have shown in my post. First, I have shown that $[A, B] \neq 0$ is equivalent to (1) and then, I have given you the theorem that it is also equivalent to (2). Because equivalence of statements is an equivalence relation, (1) and (2) are equivalent to each other. This also applies to your third point: The point of my post was to show that (1) and (2) are not just consequences of $A$ and $B$ being non-commutative, but all three statements are equivalent! Jan3 revised Inversible Antisymmetric Matrix added 21 characters in body; edited tags Jan3 answered Inversible Antisymmetric Matrix Jan3 answered Why $\|Ax\|^2=\frac{1}{4}\left(2\left+2\left\right)$ Jan3 revised System of vectors $\{f_1, f_2, \ldots, f_n\} \in V^*$ is a basis of $V^*$ if and only if $\ker f_1 \cap \ker f_2 …\cap \ker f_n = \{0\}$ formatting Jan3 reviewed Approve Near-rings: why are ideals defined like that? Jan3 reviewed Approve Regression vs Classification Jan3 reviewed No Action Needed Equation of a cubic function with inflection point on (0.5,0.5) and contains (0,0), (1,1) Jan2 comment Diagonalization versus s.d. product for non-commuting Hermitian matrices @nomadreid: Which bases do you mean? My point was that both statement (1) and (2) are equivalent to $[A,B] \neq 0$, in case it wasn't clear. I thought you were trying to understand how one comes up with these two results.