Answer to Question #254788 in Linear Algebra for Sabelo Xulu

Question

5.a) Given a vector space V, when is a subset T ⊆ V of V said to be a vector subspace of V? 

b) Given a linear transformation V $\to$ W show that the kernel Ker f of f is a subspace of V

c) Given any linear transformation

V $\to$ W show that the subsetI m[f] ={y∈W y=f(x), for some x∈V} is a vector subspace of W

d) Given any linear transformation V $\to$ W, show that f is one-to-one, if and only if, its kernel is trivial i.e, Ker f={0}

e) Given a linear transformation V $\to$ W between finite dimensional vector spaces, show that dim Ker f + dim Im[f] =dim V

Hint: Show Ker f is finite dimensional and then starting with a basis B_o for Ker f extend to a basis of V

Answer 1

(a) A subset T of a vector space V is a subspace of v if:

- T is non-empty

-For every v,t $\in$ T and scalars

a,b $\in$ $R$

av+bt $\in$ T

Implying that a non- empty subset of v is a subspace if it is closed under linear combination

(b) Suppose u,v are in ker (f)

(1) f(u+v) = f(u)+f(v) = 0+0=0

(11) For a scalar, c

f(cv) =cf(v)=c.0=0

(iii)f(0)=f{(v)+(-v)}=f(v)+f(-v) =0+0=0

Therefore ker(f) is closed under zero, addition and multiplication

(c) Let y₁ and y₂ $\in$ im (f) and let r₁

,r₂ be scalars

Consider r₁y₁+r₂y₂

There exist x₁ and x₂ $\in$ v such

y₁=f(x₁) and y₂=f(x₂) and

that r₁f(x₁)+r₂f(x₂)= f(r₁x₁)+f(r₂x₂)

=f(r₁x₁+r₂x₂)

But f(r₁x₁+r₂x₂)$\in$ Im (f)

Hence linear combination of y₁ and y₂ is in Im (f)

Therefore Im (f) is a subspace of w

(d)

Let ker (f)={0}

Let f(x)=f(y) for x,y $\in$ V

f(x)-f(y)=0

f(x-y)=0

$\therefore (x-y)\in$ ker f

x-y=0

x=y

(e)

Let dim(Ker(f)) =k , dim(v)=n and n>k

Let B_o= {v₁,...,v_k} be a basis for ker (f)

Extending this basis to a full basis of V,

(n-k) vectors are added to the basis of the ker (f) to create a basis for V

Basis of V, B_v= {v₁,...,v_k,v_k+1,...,v_n}

The Image of f

Im [f]={f(v₁),...,f(v_k),f(v)_k+1,...,f(v_n)}

As v₁,...,v_k are in the kernel, f(v₁),...,f(v_k)=0

Therefore the basis for the Image of f is

{f(v_k+1),...,f(v_n)}

And it spans the range of f

Suppose there exists a_k+1,…,a_n∈R ,not all zero such that

a_k+1f(v_k+1)+⋯+a_nf(v_n)=0.

Using linearity of f, this is equivalent to f(a_k+1v_k+1+1+⋯+a_nv_n)=0,

a_k+1v_k+1+⋯+a_nv_n∈kerf

thus there exists b₁,…,b_k∈R

not all zero such that

a_k+1v_k+1+⋯+a_nv_n=b₁v₁+⋯+b_kv_k

$\implies$ All the constants are zero, which proves linear independence of

{f(v_k+1),. . . . .f(v_n)

$\therefore$ dim ker f + dim Im [f] = dim v