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[2015-5-02. : 5:23 pm]  Devourer -- Dem0nDem0n shouted: Mini Moose 2707 That's the exact definition my textbook gives me. differently put: n-Tuple(x1, x2, ..., xn) is linear independent, when SUM[i=0;i<n] ( ai * xi) = 0 => a = 0[2015-5-02. : 5:16 pm]  Dem0n -- For some reason, it took me forever to understand what linear independence was. I fucking suck at math.[2015-5-02. : 5:16 pm] Dem0n -- Mini Moose 2707Mini Moose 2707 shouted: A set of vectors is said to be linearly dependent if one of the vectors in the set can be defined as a linear combination of the other vectors. That's the exact definition my textbook gives me.[2015-5-02. : 5:07 pm] jjf28 -- SacrieurSacrieur shouted: Yo jjf how's the weather up in Twin Cities? I'm not in the cities right now, I'm in the great port of Minnesota[2015-5-02. : 4:56 pm] Devourer -- Mini Moose 2707Mini Moose 2707 shouted: So yeah, I guess 0 tuple in a 0-space is linearly independent Mini Moose 2707Mini Moose 2707 shouted: A set of vectors is said to be linearly dependent if one of the vectors in the set can be defined as a linear combination of the other vectors. No other vectors to make linear combinations with anyway. ok thanks <3[2015-5-02. : 4:54 pm]  Moose -- A set of vectors is said to be linearly dependent if one of the vectors in the set can be defined as a linear combination of the other vectors. No other vectors to make linear combinations with anyway.[2015-5-02. : 4:47 pm] O)FaRTy1billion[MM] -- Mini Moose 2707Mini Moose 2707 shouted: You do gradually lose mass over time (maybe depending on size), which I was happy to find ou I read a post where someone suggested that ;o[2015-5-02. : 4:46 pm] Devourer -- Mini Moose 2707Mini Moose 2707 shouted: But a vector space of dimension zero doesn't have a zero vector. well apparently it does have the zero vector.[2015-5-02. : 4:44 pm] Devourer -- http://de.wikipedia.org/wiki/Nullvektorraum well the german wikipedia entry exists, apparently no equal english one[2015-5-02. : 4:41 pm] Devourer -- DevourerDevourer shouted: The nullvectorsubspace is a subspace of any subspace or vectorroom (since every vectorspace/vectorsubspace, per definition, has to contain the 0, this is true)[2015-5-02. : 4:39 pm] Devourer -- The nullvectorsubspace is a subspace of any subspace or vectorroom[2015-5-02. : 4:39 pm] Moose -- DevourerDevourer shouted: The Null-Vector-Subspace only containing (0, ..., 0) as the only element obviously is a vectorial subspace (null is element, addition and scales 'completed' (idk english word for this)). subspace of WHAT[2015-5-02. : 4:38 pm] Devourer -- the empty tuple however is a 0-tuple and the linear combination, which basically is a sum of some coefficient ai over the elements of this tuple is a sum of 0, defined as 0 and therefore (0,...,0) is defined in only that way. It makes sense that the empty tuple is linear independent / a base :<[2015-5-02. : 4:36 pm] Devourer -- But that does not apply to (0,...,0) since 10 * (0,...,0) = (0,...,0) but also 20 * (0,...,0) = (0,...,0).[2015-5-02. : 4:36 pm] Devourer -- (one better definition of linear indepedence is: given any vector in that space and it can only be created in exactly one way (linear combination) of the given tuple, then this tuple is linear independent[2015-5-02. : 4:34 pm] Devourer -- The Null-Vector-Subspace only containing (0, ..., 0) as the only element obviously is a vectorial subspace (null is element, addition and scales 'completed' (idk english word for this)).[2015-5-02. : 4:33 pm] Devourer -- Mini Moose 2707Mini Moose 2707 shouted: A set of vectors is said to be linearly dependent if one of the vectors in the set can be defined as a linear combination of the other vectors. correct as far as I know[2015-5-02. : 4:31 pm] Moose -- There are no vectors or other vectors to do this in a space of no vectors[2015-5-02. : 4:30 pm] Moose -- A set of vectors is said to be linearly dependent if one of the vectors in the set can be defined as a linear combination of the other vectors.[2015-5-02. : 4:27 pm] Devourer -- Isn't a "base" of a vector(sub)space always linear independent? If that is true, considering that the empty tuple is a base for the null-sub-vectorspace, the empty tuple must be independent D:[2015-5-02. : 4:23 pm] Moose -- I'm just not sure if it makes sense or is meaningful to even talk about empty vectors[2015-5-02. : 4:23 pm] Devourer -- but that should not matter too much, should it? Regarding the quesiton, it is.[2015-5-02. : 4:20 pm] Devourer -- The (only) base of the Null-Room is the empty tuple p=() and has dimension 0, but is that tuple linear independent?[2015-5-02. : 4:19 pm] Devourer -- Not (0, 0, ...., 0); I mean the latter. The tuple p = () with zero "content", esily put[2015-5-02. : 4:16 pm] Devourer -- DevourerDevourer shouted: The empty tuple is linear independent, right? ?[2015-5-02. : 4:12 pm] Moose -- But I am math major, so there are probably more physics details I don't know[2015-5-02. : 4:12 pm] Devourer -- (On a sidenote, doesn't the "mass increase" the closer something is to the speed of light?[2015-5-02. : 4:10 pm] Moose -- (Though if force is constant and mass approaches infinity, then acceleration gets arbitrarily small)[2015-5-02. : 4:09 pm] Moose -- force = mass*acceleration <=> acceleration = force/mass. If force and mass both > 0, then acceleration > 0[2015-5-02. : 3:32 pm] Devourer -- Because the slightest projecticle, e.g. a sun-ray, does move any object, but just very very little. Or am I wrong?[2015-5-02. : 3:31 pm] Devourer -- There is this concept of unstopable force and inmoveable object, which can not exist at the same time. K. But isn't an unmoveable object impossible unless there is no (kinetic) energy impacting that object at all?[2015-5-02. : 3:13 pm] Moose -- You do gradually lose mass over time (maybe depending on size), which I was happy to find ou |
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