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Apr
22
awarded  Good Answer
Mar
29
comment finite field arithemtic
Notice that you can easily unclutter the problem by observing that your equation can be replaced by two equations: $M_1 B_1 = M(A_1 \cdot B_1)$ for any $B_1$, and $M_2 B_2 = M(A_2 \cdot B_2)$ for any $B_2$. Your equation follows from these two, and these two follow from your equation if you take one of $B_1$, $B_2$ to be equal to $0$.
Mar
24
revised Linear operator problem: $T^n=0$ and $T^{n-1}\neq0$
Make step 1 more succinct, it was more convoluted than necessary.
Mar
24
awarded  Enlightened
Mar
24
awarded  Nice Answer
Mar
24
answered Linear operator problem: $T^n=0$ and $T^{n-1}\neq0$
Mar
24
comment Linear operator problem: $T^n=0$ and $T^{n-1}\neq0$
@BigbearZzz Could be... But really, the OP should clarify this.
Mar
24
comment Linear operator problem: $T^n=0$ and $T^{n-1}\neq0$
I would assume $r$ means rank and $d$ means determinant, but then $d(T)$ should be 0 instead of 1...
Mar
23
awarded  Enlightened
Mar
23
awarded  Nice Answer
Mar
21
comment Does the ring in this Fast Fourier Transform image of a hexagonally close packed structure have significance?
I don't actually have a clue, but here's a question: why is one of the dots on the original picture blue instead of black? Is this on purpose? Does the FFT change much when this dot is black like the others?
Mar
19
comment Does $arg \left(\dfrac {z-1}{z+1}\right)=\dfrac{\pi}{3}$ represent a circle or arc of a circle?
Shouldn't it be $6 + 4\sqrt{3}\sin t$ instead of $7 + 4 \sqrt{3} \sin t$?
Feb
29
comment Find all differentiable functions $f$ such that $f\circ f=f$
@marmistrz Because if $y \in X$, then $y = f(x)$ for some $x \in \mathbb{R}$, and therefore $f(y) = f(f(x)) = (f \circ f)(x) = f(x) = y$.
Feb
26
answered Is the interior of a closed subset of the closure of an open set included into the open set?
Feb
15
comment Can a graph be non 3-colourable without having k4 as a sub graph?
@LeBron This example is correct, but it doesn't prove the theorem wrong. The graph on the picture does contain a subdivision of $K_4$, you just need to look for it carefully. To see the subdivision, just erase any two edges incident with the central vertex.
Feb
14
comment $x$ is within 3 units of $c$
"Within" doesn't seem to be standard terminology. If you want to use it, you can just define it yourself as either (i) or (ii), whichever is more comfortable. If you've seen it in another text, then you need to guess the meaning from the context.
Feb
6
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.
Feb
6
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.
Feb
6
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 $D = \operatorname{diag}(2, 2, 3)$.
Feb
6
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?