Answer to Question #298455 in Linear Algebra for Silvester

Question

5. 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.

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

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

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

Answer 1

1. First of all, we remark that $0\in \ker \varphi$ by linearity of $\varphi$. Now let $v,w \in \ker \varphi, \; \lambda \in F$, we have $\varphi(v+w)=\varphi(v)+\varphi(w)=0+0=0$ and $\varphi(\lambda v)=\lambda \cdot \varphi(v)=0$, so the kernel is a subspace.
2. As $\varphi(0)=0$ we have $0\in \text{Im }\varphi$. Now let $v,w \in \text{Im } \varphi, \;\lambda \in F$, by definition of $\text{Im}$ there exist vectors $v',w'\in V$ with $\varphi(v')=v, \; \varphi(w')=w$. Then we have $v+w = \varphi(v')+\varphi(w')=\varphi(v'+w')\in \text{Im }\varphi$ and $\lambda v = \lambda \varphi(v')=\varphi(\lambda v')\in \text{Im }\varphi$, so the image is a subspace.
3. Suppose $\varphi$ is injective. By definition $\varphi(x)=\varphi(0)=0$ iff $x=0$, so $\ker \varphi = \{0 \}$. Now suppose that $\ker \varphi=\{ 0 \}$ and let $x,y \in V$ such that $\varphi(x)=\varphi(y)$. Therefore, we have $\varphi(x)-\varphi(y)=\varphi(x-y)=0$, $x-y\in \ker \varphi \Rightarrow x-y=0$ so $\varphi$ is injective.
4. $\varphi$ is surjective iff $(\forall w\in W)(\exists v\in V)\:|\: \varphi(v)=w$ and this means exactly $(\forall w\in W)\: w\in \text{Im } \varphi$ which is true iff $\text{Im }\varphi=W$.