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Let V be a vector space and T is an element of L(V), show that the kernel of T^n is a subspace of the kernel of T^(n+1).
L(V) is the set of all linear maps from V to V and T is a linear isomorphism and the kernel of T={for all v that are elements of V, V=Tv=0} (???)

2007-01-01 12:38:09 · 9 answers · asked by kate 1 in Science & Mathematics Mathematics

9 answers

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 · answer #1 · answered by Anonymous · 2 0

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 · answer #2 · answered by rozinante 3 · 0 0

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 · answer #3 · answered by koolciaran 2 · 0 1

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 · answer #4 · answered by Anonymous · 0 1

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2016-11-25 21:16:06 · answer #5 · answered by Anonymous · 0 0

I thought kernels only dealt with popcorn! :P

2007-01-01 12:49:28 · answer #6 · answered by teekshi33 4 · 1 3

baked beans on toast

2007-01-01 19:43:52 · answer #7 · answered by Anonymous · 0 1

Jesus wept.

2007-01-01 13:02:51 · answer #8 · answered by Gerbil 4 · 0 1

you lost me at "let v" !

2007-01-01 12:45:20 · answer #9 · answered by Anonymous · 0 3

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