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1d
comment Finding the inverse of A where A is of the form $A = D (I − N)$, where $D$ is diagonal with nonzero entries and $N$ is nilpotent
That intuition also used the fact that when you multiply two upper-triangular matrices (or an arbitrary matrix with a diagonal one), their diagonals are multiplied "component-wise". So if you multiply a unitriangular matrix by $\operatorname{diag}(x, y, z)$, the resulting matrix will have $x, y, z$ on its main diagonal.
1d
comment Finding the inverse of A where A is of the form $A = D (I − N)$, where $D$ is diagonal with nonzero entries and $N$ is nilpotent
@hermes Your $N$ doesn't mork, as Alex correctly pointed out. You may want to fix your answer.
1d
comment Finding the inverse of A where A is of the form $A = D (I − N)$, where $D$ is diagonal with nonzero entries and $N$ is nilpotent
From experience. I just know that zero-diagonal upper-triangular matrices are nilpotent. Given the triangular shape of $A$, I simply guessed that one could make $N$ zero-diagonal upper-triangular, which would make $I-N$ upper-triangular with ones on the diagonal, which would mean $D = \operatorname{diag}(2, 2, 3)$.
2d
comment Finding the inverse of A where A is of the form $A = D (I − N)$, where $D$ is diagonal with nonzero entries and $N$ is nilpotent
$D$ is $\operatorname{diag}(2, 2, 3)$, $N$ is an upper-triangular matrix with zeroes on the diagonal. Can you find $N$ now?
Jan
25
comment Prove $(g+h)\circ f=g\circ f+ h\circ f$
Hint: proving that $p=q$, where $p$ and $q$ are functions, means proving $p(x) = q(x)$ for every $x$.
Jan
10
comment What does it mean for an automorphism to centralize a subgroup?
Yes, it simply means that $\sigma$ fixes $H$ pointwise.
Jan
10
comment Does the set of all fields exist ?
In addition to what @AndréNicolas said, you can think about sizes. If there was a set of all fields, you could also take the union of all fields (as sets), and the cardinality of this union would be an upper bound on the cardinalities of all fields. But there are fields of arbitrarily large cardinalities, a contradiction.
Jan
10
awarded  abstract-algebra
Nov
21
comment Proving $\cos(x)^2+\sin(x)^2=1$
But before summing you should probably shift indices in the second summand so that you always have $x^{2k}$ instead of $x^{2(k+1)}$.
Nov
21
comment Proving $\cos(x)^2+\sin(x)^2=1$
The $x$ term is not in the way. You should fix a $k$ and sum over $l$. When $k$ is fixed the $x^{2k}$ can be carried out of the parentheses.
Nov
14
comment Why, intuitively, does the Maclaurin series for $e^x$ but not $\ln(1+x)$ converge globally?
I wouldn't say that flipping the curve destroys the "global information". In case of the logarithm you can still retrieve the whole function from the data in the vicinity of one point by analytic continuation.
Nov
7
awarded  Nice Answer
Oct
30
awarded  Yearling
Oct
17
awarded  Enlightened
Oct
17
awarded  Nice Answer
Sep
14
comment Why is the identity map of the circle not straight-line homotopic to a constant map?
@L.G. How is $(1-t)f(\theta)$ an element of $S^1$? It is either undefined or doesn't belong to $S^1$, depending on your precise definition of $S^1$.
Sep
14
comment Why is the identity map of the circle not straight-line homotopic to a constant map?
If you mean that $S^1$ is a subset of $\mathbb{R}^2$, then your $F$ maps $S^1 \times I$ to a disk in $\mathbb{R}^2$, not to $S^1$ as it should.
Sep
14
comment Why is the identity map of the circle not straight-line homotopic to a constant map?
What exactly is $(1-t)f(\theta)$? How do you multiply a number with an element of $S^1$?
Aug
22
awarded  Good Answer
Jul
22
comment Show that all abelian groups of order 21 and 35 are cyclic.
@Alex it relies on abelianness in step 4, in the "easy to see" part. One has to consider products like $a^m b^n$ and use the fact that $(a^m b^n)(a^{m'}b^{n'}) = a^{m + m'} b^{n + n'}$, which is true due to the abelianness. Without abelianness one can construct a semidirect product of $\mathbb{Z}_3$ and $\mathbb{Z}_7$ that is not isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_7$.