Problem Give an example of a function $\phi : \mathbb{C} \to \mathbb{C}$ such that $$\phi(w + z) = \phi(w) + \phi(z)$$ for all $w,z \in \mathbb{C}$ but $\phi$ is not linear. (Here $\mathbb{C}$ is thought of as a complex vector space) Solution Let $\phi(z) = \textrm{Re}(z)$ for all $z \in \mathbb{C}$, let now $a,b,c,d \in \mathbb{R}$ such that $z = a + bi$ and $w = c + di$. Then we must have: $$\phi(z + w) = \phi((a + bi) + (c + di)) \stackrel{D1....

Problem Give an example of a function $\phi : \mathbb{R}^2 \to \mathbb{R}$ such that $$\phi(av) = a\phi(v)$$ for all $a \in \mathbb{R}$ and all $v \in \mathbb{R}^2$ but $\phi$ is not linear. Solution Let $\phi((x,y)) = \textrm{sgn}(x)\sqrt{x^2 + y^2}$, then: $$\phi(a(x,y)) = \phi((ax,ay)) = \textrm{sgn}(ax)|a|\sqrt{x^2 + y^2}$$ $$= a \cdot \textrm{sgn}(x)\sqrt{x^2 + y^2} = a \phi((x,y))$$ But consider $(0,1)$ and $(1,0)$, then: $$\sqrt{2} = \phi((1,1)) = \phi((1,0) + (0,1)) = \phi((1,0)) + \phi((0,1)) = 1 + 1 = 2$$ A contradiction....

Problem Show that every linear map from a 1-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if $\dim V = 1$ and $T \in \mathcal{L}(V,V)$, then there exists $\lambda \in \mathbb{F}$ such that $Tv = \lambda v$ for all $v \in V$ Solution Let $u \in V$ be non-zero, then the list $u$ is a basis for $V$ as per Example 2.18 and theorem 2....

Problem Prove the assertions in 3.9. Solution Associativity Multiplication of linear maps is defined as function composition, therefore: $$((T_1 T_2) T_3)(v) = (T_1T_2)(T_3 v) = T_1(T_2(T_3v))$$ $$= T_1((T_2T_3)(v)) = (T_1(T_2T_3))(v)$$ Implying that $(T_1T_2)T_3 = T_1(T_2T_3)$. Identity Let $T \in \mathcal{L}(V,W)$ and $I$ be the identitymap on $V$ and $W$ respectively: $$(TI)(v) = T(Iv) = Tv$$ $$(IT)(v) = I(Tv) = Tv$$ Distributive properties Let $T, T_1, T_2 \in \mathcal{L}(U,V)$ and $S, S_1, S_2 \in \mathcal{L}(V,W)$, then:...

Problem Prove the assertion in 3.7. Solution Let $T, S, S^\prime \in \mathcal{L}(V,W)$, then: Commutativity $$(T + S)(v) \stackrel{D3.6}{=} Tv + Sv \stackrel{D1.19}{=} Sv + Tv \stackrel{D3.6}{=} (S + T)(v)$$ Associativity $$((T + S) + S^\prime)(v) \stackrel{D3.6}{=} (T + S)(v) + S^\prime(v) $$ $$\stackrel{D3.6}{=} (T(v) + S(v)) + S^\prime(v) \stackrel{D1.19}{=} T(v) + (S(v) + S^\prime(v))$$ $$\stackrel{D3.6}{=} T(v) + (S + S^\prime)(v) \stackrel{D3.6}{=} (T + (S + S^\prime))(v)$$...