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Dec
13
comment Why are every structures I study based on Real number?
@TaxxiDriver: Actually I had as possibility in mind the cardinal numbers, for distances between sets. Unlike natural numbers, you cannot extend the cardinal numbers to a field (because of equations like $\aleph_0 + 1 = \aleph_0$).
Dec
13
comment Why are every structures I study based on Real number?
@mbork: Interesting. That one has not even a total order. Is the "addition" here join or meet, or a different operation?
Dec
13
comment Why are every structures I study based on Real number?
@rschwieb: Except that a metric space need not be a vector space.
Dec
13
comment Why are every structures I study based on Real number?
@rschwieb: I wasn't asking about inner products/bilinear forms (there it is obvious that you need a multiplication) but about metrics (a function $d(x,y)$ that is is zero iff $x=y$, is positive otherwise, is symmetric, and fulfils the triangle inequality).
Dec
13
comment Why are every structures I study based on Real number?
Why does it have to be a field for generalized metrics? As far as I can see, the metric axioms only use addition, not multiplication, so why additionally require a multiplication?
Dec
8
awarded  Popular Question
Dec
6
answered Water displacement in volume
Dec
6
comment How can I check these equations if they have a solution?
Ah, so your actual question is: For which values of $k$ and $m$ are there no solutions?
Dec
6
answered Find the last non-zero digit of $30^{2345}$
Dec
6
answered Prove that $\binom{p}{k}/p $ is integral for $k\in \{1,..,p-1\}$ with $p$ a prime number
Dec
6
comment How can I check these equations if they have a solution?
@HTehlike: $2^3+6=14\equiv 0\,(\mod 7)$ and $(3\cdot 2^2 + 3\cdot 2\cdot 2 + 2^2)\cdot 2 = 28 \equiv 0\,(\mod 7)$ where $6,7,2 > 0$ and $2\ge 0$ and $7\nmid 2$.
Nov
17
awarded  Good Question
Nov
4
revised Trace norm identity (in bra-ket notation)
added 29 characters in body
Nov
4
answered Trace norm identity (in bra-ket notation)
Oct
19
comment What are the possible dimensions of all $n \times n$ matrices such that $A\vec{x} = \vec{0}$?
That matrix would not do a basis change. But then, you don't need to explicitly find a matrix that does it. You just have to answer the question: "Does there exist a basis for $\mathbb R^n$ that has $\vec x$ as basis vector? About the dimension: Is the dimension a property of the space, or just a property of the individual basis?
Oct
19
comment What are the possible dimensions of all $n \times n$ matrices such that $A\vec{x} = \vec{0}$?
No. $(1,1,0,\ldots,0)^T$ will also be sent to $\vec 0$ e.g. if the first column is all $1$ and the second column is all $-1$. To elaborate on my previous hint: Is it possible to change the basis to one having $\vec x$ as basis vector? And what happens to dimensions when you change the basis?
Oct
19
comment What are the possible dimensions of all $n \times n$ matrices such that $A\vec{x} = \vec{0}$?
The $n^2-n$ is right. The $n$ is not. Hint: What happens for a basis change?
Oct
19
comment What are the possible dimensions of all $n \times n$ matrices such that $A\vec{x} = \vec{0}$?
@JennParker: You're right about the form of the matrix, but not the dimension. Hint: How many matrix elements can you still choose freely?
Oct
19
comment What are the possible dimensions of all $n \times n$ matrices such that $A\vec{x} = \vec{0}$?
@JennParker: The dimension for $\vec x=0$ you've got right, but in the other cases you're wrong. Hint: What matrices will map the vector $(1,0,\dots,0)^T$ to $\vec 0$?
Oct
19
comment Showing that $e^{-2} < \ln 2$
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