No genius needed: just note that if a vector v belongs to ker(T^n) then it also belongs to ker(T^(n+1)), i.e., that if (T^n)v = 0 then obviously (T^(n+1))v = T(T^n)v = T0 = 0.
This shows that ker(T^n) is a subset of ker(T^(n+1)). And actually it is a linear subspace, of ker(T^(n+1)), because the kernel of any linear operation A is a linear space (if Av = 0 = Au then also A(v+u) = 0 and A(av) = 0 for every scalar a).
BTW, T is a linear homomorphism (an isomorphism has a zero kernel).
Also, ker(T) = {v in V: Tv = 0}, i.e., it is the set of all those v in V for which Tv = 0.
2007-01-01 12:51:41
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answer #1
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answered by Anonymous
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Actually it's pretty simple. To show that the kernel of T^n is a subspace of the kernel of T^(n+1), all we really need to do is to show that every element of Ker(T^n) is also an element of Ker(T^(n+1)).
Suppose x is an element of Ker(T^n). Then by definition (T^n)x=0. But that means that T((T^n)x)=T(0), and by definition of linearity, T(0)=0 always (look up linear in mathworld). Since T((T^n)x)=(T^(n+1))x=0, we can conclude that x is also an element of Ker(T^(n+1)). Thus the kernel of (T^n) is a subspace of the kernel of (T^(n+1)).
2007-01-01 12:56:00
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answer #2
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answered by rozinante 3
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just note that if a vector v belongs to ker(T^n) then it also belongs to ker(T^(n+1)), i.e., that if (T^n)v = 0 then obviously (T^(n+1))v = T(T^n)v = T0 = 0.
This shows that ker(T^n) is a subset of ker(T^(n+1)). And actually it is a linear subspace, of ker(T^(n+1)), because the kernel of any linear operation A is a linear space (if Av = 0 = Au then also A(v+u) = 0 and A(av) = 0 for every scalar a).
BTW, T is a linear homomorphism (an isomorphism has a zero kernel).
Also, ker(T) = {v in V: Tv = 0}, i.e., it is the set of all those v in V for which Tv
2007-01-02 00:31:28
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answer #3
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answered by koolciaran 2
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Holy Moly, I don't even know what a kernel is. Anyone out there can explain any of this to me and I'll be very impressed.
2007-01-01 14:35:48
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answer #4
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answered by Anonymous
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2016-11-25 21:16:06
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answer #5
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answered by Anonymous
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I thought kernels only dealt with popcorn! :P
2007-01-01 12:49:28
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answer #6
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answered by teekshi33 4
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baked beans on toast
2007-01-01 19:43:52
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answer #7
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answered by Anonymous
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Jesus wept.
2007-01-01 13:02:51
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answer #8
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answered by Gerbil 4
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you lost me at "let v" !
2007-01-01 12:45:20
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answer #9
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answered by Anonymous
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