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An ageing Algebra teacher


8h
awarded  Nice Answer
8h
revised For all $x,y\in G$ we have: $f(xf(y))=f(x)y$. Prove that $f$ is an isomorphism?
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13h
revised a root of some polynomial over finite field
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14h
answered a root of some polynomial over finite field
14h
answered Understanding why the public exponent $e$ is chosen the way it is in RSA
16h
comment set of all $2\times 2$ matrcies having neither eigen value is real
Will be glad to try if I can find the time.
17h
comment set of all $2\times 2$ matrcies having neither eigen value is real
You're welcome. I am not sure about a proper reference, but the first one that comes to my mind is Lang's superb Linear Algebra.
17h
comment set of all $2\times 2$ matrcies having neither eigen value is real
I think you should consider a map like $g$ from matrices to the reals. And then whether $> 0$ or $\ge 0$ is a matter of definitions.
17h
comment set of all $2\times 2$ matrcies having neither eigen value is real
Actually $\ge 0$ (two coincident real roots). So you see, $g(A) = (a_{11} + a_{22})^{2} - 4 (a_{11}a_{22} - a_{12} a_{21})$ is a continuous function $g : M_{2}(\mathbb{R}) \to \mathbb{R}$, and your set is the preimage of the closed set $\{ x \in \mathbb{R} : x \ge 0 \}$.
17h
comment set of all $2\times 2$ matrcies having neither eigen value is real
So the eigenvalues are the roots of $f$. Now consider the discriminant conditions for $f$ to have two non-real roots, or two real roots. You will end up with your two sets being the counterimages of an open (resp. closed) set under a continuous function.
18h
answered For all $x,y\in G$ we have: $f(xf(y))=f(x)y$. Prove that $f$ is an isomorphism?
1d
reviewed Reject Formula for $\sum_{d|n} \frac {\mu(d)}d$
1d
comment Clarification about the definition of free module
For some details of @ThomasAndrews' answer, see for instance en.wikipedia.org/wiki/Free_object
1d
answered To show two matrices are conjugate to each other
1d
comment To show two matrices are conjugate to each other
@user35603, they are both diagonalizable.
1d
comment Kernel and Image of a group homomorphism
@learnmore, yes. To be clear, the image of $\phi$ is the whole $G$.
1d
answered Kernel and Image of a group homomorphism
1d
comment If $f$ is irreducible over a perfect field, then $f$ has no multiple zeros.
@implicitlee, you're welcome.
1d
answered If $f$ is irreducible over a perfect field, then $f$ has no multiple zeros.
1d
revised Proving sums of multinomial coefficients
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