please answer in details , with clear handwritten, If true give a reason and an

Question

please answer in details , with clear handwritten,

If true give a reason and an example (whenever possible) and if false give a reason and/or a counter example I. Suppose that dim (V) = 2 and dim (W) = 4. a) T V ? W (b) There exists an onto linear map : V ? W (c) There exists a one-to-one linear map T : W ? V (d) There exists an onto linear map T : W ? V (e) If T : W ? V is onto, then there exists S : V ? W such that here exists a one-to-one linear map T: (f) If T : W -» V is onto, then there exists S : V -» W such that (g) If T : V ? W is one-to-one, then there exists S : W ? V such that (h) If T : V ? W is one-to-one, then there exists S : W ? V such that TS=lv ST Iw ST = 1w

Explanation / Answer

By rank nulity theorem

Kernel T + Rank T= dimension of V=2

a) If T is injective then kernel =0 so rank must be 2 . It will be subspace of W.. Which has dimension 4...no problem.. There can be a injective

b) Not possible. Because . Onto map means rank must be 4...but then kernel =-2 which is impossible.

Now Kernel +Rank =4=dimension of W

c) Injective map means... Kernel =0 so rank =4 but impossible as Rank is a subspace of V of dimension 2,in this case.

d) Onto means rank =2 no problem there can be a such map.

For St.. . Rank +Kernel =dimension of V=2..and also use the fact rank ST less than equal to min {rank T, rank S}

And conclude like above ...

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