Linear Algebra Hello. I need help with my practice final. I am not trying to copy. I want to see If I am doing it right So please help. Thank you very much! Here\'s the problem Let V and W be (finite-dimensional) vector spaces. In fact, unless otherwise stated in one of the problems or definitions below you may assume that dim V n and dimW m. Also, unless otherwise stated, let T V W be a linear transformation. Definitions and Basic Results 1. The kernel of T, written ker T, is the set of vectors UE v such that T(i) 30w. 2. The range of T, written range T is the set of vectors i e W such that T(i) if for some i E V. 3. T is injective means that whenever i E V then T(ii) T(i2) if and only if ii J v2 4. T is surjective means range T W 5. T is bijective if and only if it is both injective and surjective. 6. T is called a linear operator if V W 7. Let T be a linear operator in V and let u be subspace of V. u is said to be invariant under T if T(ii) Eu for every i Eu 8. Let u and u be aces of V (a) The sum of these subspaces, written uitu2, is the set of vectors i EV that can be written as a sum of vectors ii ii2 where ii E uh and ii2 Eu2. (b) V is said to be a direct sum of u and up, written V Juh Du2, if and only if for every if E V there exists a unique pair of vectors iii E ui and ii2 Eur such that if ii ii2. Examples R2 z-axis) (y-axis). Ra (ary-plane) axis A) span span Fact V u if and only if v un +uo and un nu Solution if T is injective then there should be a map of v in w for every v belonging to V and w belonging to W. Consider the basis of W as all elements are maped then the dim of W should be at least equal to the dimension of V as linear transformation map each basis element of V to some basis element in W If T is surjective then same argument above applies to just opposite because now mapping covers all the elements of W. if T is bijective then it is both injective and surjective .So now only equality can hold hence dim(V) = dim(W).

1. Linear Algebra
Hello. I need help with my practice final. I am not trying to copy. I want to see If I am doing it
right So please help. Thank you very much!
Here's the problem Let V and W be (finite-dimensional) vector spaces. In fact, unless otherwise
stated in one of the problems or definitions below you may assume that dim V n and dimW m.
Also, unless otherwise stated, let T V W be a linear transformation. Definitions and Basic
Results 1. The kernel of T, written ker T, is the set of vectors UE v such that T(i) 30w. 2. The
range of T, written range T is the set of vectors i e W such that T(i) if for some i E V. 3. T is
injective means that whenever i E V then T(ii) T(i2) if and only if ii J v2 4. T is surjective means
range T W 5. T is bijective if and only if it is both injective and surjective. 6. T is called a linear
operator if V W 7. Let T be a linear operator in V and let u be subspace of V. u is said to be
invariant under T if T(ii) Eu for every i Eu 8. Let u and u be aces of V (a) The sum of these
subspaces, written uitu2, is the set of vectors i EV that can be written as a sum of vectors ii ii2
where ii E uh and ii2 Eu2. (b) V is said to be a direct sum of u and up, written V Juh Du2, if and
only if for every if E V there exists a unique pair of vectors iii E ui and ii2 Eur such that if ii ii2.
Examples R2 z-axis) (y-axis). Ra (ary-plane) axis A) span span Fact V u if and only if v un +uo
and un nu
Solution
if T is injective then there should be a map of v in w for every v belonging to V and w belonging
to W. Consider the basis of W as all elements are maped then the dim of W should be at least
equal to the dimension of V as linear transformation map each basis element of V to some basis
element in W
If T is surjective then same argument above applies to just opposite because now mapping covers
all the elements of W.
if T is bijective then it is both injective and surjective .So now only equality can hold hence
dim(V) = dim(W)