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Jan
4
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.
Jan
4
reviewed Approve Why is this $0 = 1$ proof wrong?
Jan
4
reviewed Edit Maximizing volume in Calculus
Jan
4
revised Maximizing volume in Calculus
improved formatting
Jan
4
reviewed Edit How is this a counter example to “ $A$ and $B$ are isomorphic but $G/A \ncong G/B”$?
Jan
4
revised How is this a counter example to “ $A$ and $B$ are isomorphic but $G/A \ncong G/B”$?
improved version
Jan
4
reviewed Looks OK How to compute sum with non consecutive indices in Maple?
Jan
3
revised $0$ and $p^n \mathbb{Z}_p$ are the only ideals of $\mathbb{Z}_p$
edited body
Jan
3
revised $0$ and $p^n \mathbb{Z}_p$ are the only ideals of $\mathbb{Z}_p$
added 8 characters in body
Jan
3
answered $0$ and $p^n \mathbb{Z}_p$ are the only ideals of $\mathbb{Z}_p$
Jan
3
comment $0$ and $p^n \mathbb{Z}_p$ are the only ideals of $\mathbb{Z}_p$
@fretty: $n$ can be $0$.
Jan
3
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!
Jan
3
revised Inversible Antisymmetric Matrix
added 21 characters in body; edited tags
Jan
3
answered Inversible Antisymmetric Matrix
Jan
3
answered Why $\|Ax\|^2=\frac{1}{4}\left(2\left<A(\beta x), v\right>+2\left<Av,\beta x\right>\right)$
Jan
3
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
Jan
3
reviewed Approve Near-rings: why are ideals defined like that?
Jan
3
reviewed Approve Regression vs Classification
Jan
3
reviewed No Action Needed Equation of a cubic function with inflection point on (0.5,0.5) and contains (0,0), (1,1)
Jan
2
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.