Question #298652

Let φ : V → W be a linear transformation of vector spaces over the field F. The kernel of φ is by definition


the set ker(φ) ⊂ V of vectors v in V such that φ(v) = 0. The image of φ is the subset im(φ) of vectors w ∈ W


for which there exists some v ∈ V such that φ(v) = w.


(1) Show that the kernel of φ is a subspace of V .


(2) Show that the image of φ is a subspace of W.


(3) Show that φ is injective if and only if the kernel is 0.


(4) Show that φ is surjective if and only if the image is W.


1
Expert's answer
2022-02-17T13:26:36-0500
  1. First of all, we remark that 0kerφ0\in \ker \varphi by linearity of φ\varphi. Now let v,wkerφ,  λFv,w \in \ker \varphi, \; \lambda \in F, we have φ(v+w)=φ(v)+φ(w)=0+0=0\varphi(v+w)=\varphi(v)+\varphi(w)=0+0=0 and φ(λv)=λφ(v)=0\varphi(\lambda v)=\lambda \cdot \varphi(v)=0, so the kernel is a subspace.
  2. As φ(0)=0\varphi(0)=0 we have 0Im φ0\in \text{Im }\varphi. Now let v,wIm φ,  λFv,w \in \text{Im } \varphi, \;\lambda \in F, by definition of Im\text{Im} there exist vectors v,wVv',w'\in V with φ(v)=v,  φ(w)=w\varphi(v')=v, \; \varphi(w')=w. Then we have v+w=φ(v)+φ(w)=φ(v+w)Im φv+w = \varphi(v')+\varphi(w')=\varphi(v'+w')\in \text{Im }\varphi and λv=λφ(v)=φ(λv)Im φ\lambda v = \lambda \varphi(v')=\varphi(\lambda v')\in \text{Im }\varphi, so the image is a subspace.
  3. Suppose φ\varphi is injective. By definition φ(x)=φ(0)=0\varphi(x)=\varphi(0)=0 iff x=0x=0, so kerφ={0}\ker \varphi = \{0 \}. Now suppose that kerφ={0}\ker \varphi=\{ 0 \} and let x,yVx,y \in V such that φ(x)=φ(y)\varphi(x)=\varphi(y). Therefore, we have φ(x)φ(y)=φ(xy)=0\varphi(x)-\varphi(y)=\varphi(x-y)=0, xykerφxy=0x-y\in \ker \varphi \Rightarrow x-y=0 so φ\varphi is injective.
  4. φ\varphi is surjective iff (wW)(vV)φ(v)=w(\forall w\in W)(\exists v\in V)\:|\: \varphi(v)=w and this means exactly (wW)wIm φ(\forall w\in W)\: w\in \text{Im } \varphi which is true iff Im φ=W\text{Im }\varphi=W.

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