Problem 1. 1. Let $V$ and $W$ be vector spaces over a field $F$. Let $0_{\mathrm{V}}$ and $0_{\mathrm{W}}$ be the zero vectors of $V$ and $W$ respectively. Let $T: V \rightarrow W$ be a function. Prove the following. (a) If $T$ is a linear transformation, then $T\left(0_{V}\right)=0_{\mathrm{W}}$. (b) $T$ is a linear transformation if and only if $$T(\alpha x+\beta y)=\alpha T(x)+\beta T(y)$$ for all $x, y \in V$ and $\alpha, \beta \in F$. (c) $T$ is a linear transformation if and only if $$T\left(\sum_{i=1}^{n} \alpha_{i} x_{i}\right)=\sum_{i=1}^{n} \alpha_{i} T\left(x_{i}\right)$$ for all $x_{1}, \ldots, x_{n} \in V$ and $\alpha_{1}, \ldots, \alpha_{n} \in F$
Proof . (a) We have that $T\left(O_{v}\right)=T\left(O_{v}+U_{v}\right)=T\left(O_{v}\right)+T\left(0_{v}\right)$. Adding $-T\left(0_{v}\right)$ to both sider we get $-\underbrace{T\left(0_{v}\right)+T\left(0_{v}\right)}_{O_{w}}=-\underbrace{T\left(0_{v}\right)+T\left(0_{v}\right)}_{O_{w}}+T\left(0_{v}\right)$ So, $\quad O_{W}=T\left(U_{v}\right)$ (b) This follows from (c) with $n=2$, (c) $\Rightarrow$ Suppose $T$ is linear. We prove the resvit by induction, If $n=1$, then $T\left(\alpha_{1} x_{1}\right)=\alpha_{1} T\left(x_{1}\right)$ since $T$ is linear. If $n=2$, Then $T\left(x_{1} x_{1}+\alpha_{2} x_{2}\right)=T\left(\alpha_{1} x_{1}\right)+T\left(\alpha_{2} x_{2}\right)$ $=\alpha_{1} T\left(x_{1}\right)+\alpha_{2} T\left(x_{2}\right)$, Let $k \geqslant 2$ in formula. Suppose the formula is true for $n=k$, Then $$T\left(\alpha_{1} x_{1}+\alpha_{2} x_{2}+\ldots+\alpha_{k} x_{k}+\alpha_{k+1} x_{k+1}\right)=T\left(\alpha_{1} x_{1}+\ldots+\alpha_{k} x_{k}\right)+T\left(\alpha_{k+1} x_{k+1}\right)=…$$ $$=\alpha_{1} T\left(x_{1}\right)+\alpha_{2} T\left(x_{2}\right)+\alpha_{k} T\left(x_{k}\right)+\alpha_{k+1} T\left(x_{k+1}\right)$$ $(\Leftarrow)$ Suppose $T\left(\sum_{i=1}^{n} \alpha_{i} x_{i}\right)=\sum_{i=1}^{n} \alpha_{i} T\left(x_{i}\right)$ for all $\alpha_{i} \in F$ and $x_{i} \in V_{1}$ Setting $n=2$ and $\alpha_{1}=\alpha_{2}=1$ we get $T\left(x_{1}+x_{2}\right)=T\left(x_{1}\right)+T\left(x_{2}\right)$ for all $x_{1}, x_{2} \in V$. Setting $n=1$, we get $T\left(\alpha_{1} x_{1}\right)=\alpha_{1} T\left(x_{1}\right)$ for all $\alpha_{1} \in F$ and $x_{1} \in V$, Thus, $T$ is linear,
Problem 2. Verify whether or not $T: V \rightarrow W$ is a linear transformation. If $T$ is a linear transformation then: (i) compute the nullspace of $T,$ (ii) compute the range of $T,$ (iii) compute the nullity of $T,$ (iv) compute the rank of $T,$ (v) determine if $T$ one-to-one, and (vi) determine if $T$ is onto. (a) $T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}$ given by $T(a, b, c)=(a-b, 2 c)$ (b) $T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ given by $T(a, b)=\left(a-b, b^{2}\right)$ (c) $T: M_{2,3}(\mathbb{R}) \rightarrow M_{2,2}(\mathbb{R})$ given by $$T\left(\begin{array}{lll} a & b & c \\ d & e & f \end{array}\right)=\left(\begin{array}{cc} 2 a-b & c+2 d \\ 0 & 0 \end{array}\right)$$ (d) $T: P_{2}(\mathbb{R}) \rightarrow P_{3}(\mathbb{R})$ given by $T\left(a+b x+c x^{2}\right)=a+b x^{3}$ (e) $T: P_{2}(\mathbb{R}) \rightarrow P_{2}(\mathbb{R})$ given by $T\left(a+b x+c x^{2}\right)=(1+a)+(1+b) x+(1+c) x^{2}$
Proof . (a)This linear, Let $x=(a, b, c)$ and $y=(d, e, f)$, Let $\alpha, \beta \in \mathbb{R}$, Then \begin{aligned} T(\alpha x+\beta y) &=T(\alpha a+\beta d, \alpha b+\beta e, \alpha c+\beta f) \\ &=(\alpha a+\beta d-\alpha b-\beta e, 2 \alpha c+2 \beta f) \\ &=(\alpha a-\alpha b, 2 \alpha c)+(\beta d-\beta e, 2 \beta f) \\ &=\alpha(a-b, 2 c)+\beta(d-e, 2 f) \\ &=\alpha T(x)+\beta T(y) . \end{aligned} (i)(ii) are easy to check, (iii) A basis for $N(T)$ is $\left\{\left(\begin{array}{l}1 \\ 0\end{array}\right)\right\} .$ So, Nullity $(T)=1$. (iv) $R(T)=\mathbb{R}^{2}$, So, $\operatorname{rank}(T)=2$, (v) $T$ is not one-to one. For example, $T(0,0,0)=T(1,1,0)=(0,0)$ (vi) T is onto since $R(t)=\mathbb{R}^{2}$. (b) T is not linear, For example, $T(1,1)+T(2,2)=\left(1-1,1^{2}\right)+\left(2-2,2^{2}\right)$ $=(0,1)+(0,4)=(0,5)$ but $T((1,1)+(2,2))=T(3,3)=\left(3-3,3^{2}\right)=(0,9)$ (c) $T$ is linear. Let $x=\left(\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ d_{1} & e_{1} & f_{1}\end{array}\right)$ and $y=\left(\begin{array}{ll}a_{2} & b_{2} c_{2} \\ d_{2} & e_{2} f_{2}\end{array}\right)$ and $\alpha, \beta \in \mathbb{R},$ Then \begin{aligned} &T(\alpha x+\beta y)=T\left(\begin{array}{cc}\alpha a_{1}+\beta a_{2} & \alpha b_{1}+\beta b_{2} & \alpha c_{1}+p_{2} z_{2} \\ \alpha d_{1}+\beta d_{2} & \alpha e_{1}+\beta e_{2} & \alpha f_{1}+\beta f_{2}\end{array}\right) \\ &=\left(\begin{array}{cc}2\left(\alpha a_{1}+\beta a_{2}\right)-\left(\alpha_{1}+\beta b_{2}\right) & \alpha c_{1}+\beta c_{2}+2\left(\alpha d_{1}+\beta d_{2}\right) \\ 0 & 0\end{array}\right) \\ &=\alpha\left(\begin{array}{cc}2 a_{1}-b_{1} & c_{1}+2 d_{1} \\ 0 & 0\end{array}\right)+\beta\left(\begin{array}{cc}2 a_{2}-b_{2} & c_{2}+2 d_{2} \\ 0 & 0\end{array}\right) \\ &=\alpha T(x)+\beta T(y) \end{aligned}
Problem 3. Let $a$ and $b$ be real numbers where $a<b$. Let $C(\mathbb{R})$ be the vector space of continuous functions on the real line as in HW # 1. Let $T: C(\mathbb{R}) \rightarrow \mathbb{R}$ given by $$T(f)=\int_{a}^{b} f(t) d t$$ Verify whether or not $T$ is linear.
Proof . Let $\alpha, \beta \in \mathbb{R}$ and $f, g \in C(\mathbb{R})$. Then \begin{aligned} T(\alpha f+\beta g)=\int_{a}^{b} \alpha f(t)+\beta g(t) d t &=\alpha \int_{a}^{b} f(t) d t+\beta \int_{a}^{b} g(x) d t \\ &=\alpha T(f)+\beta T(g) \end{aligned}
Problem 4. Let $F$ be a field. Recall that if $A \in M_{m, n}(F)$ then we can make a linear transformation $L_{A}: F^{n} \rightarrow F^{m}$ where $L_{A}(x)=A x$ is left-sided matrix multiplication. In each problem, calculate $L_{A}(x)$ for the given $A$ and $x$ (a) $F=\mathbb{R}, L_{A}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}, A=\left(\begin{array}{cc}1 & \pi \\ \frac{1}{2} & -10\end{array}\right), x=\left(\begin{array}{c}17 \\ -5\end{array}\right)$ (b) $F=\mathbb{C}, L_{A}: \mathbb{C}^{3} \rightarrow \mathbb{C}^{2}, A=\left(\begin{array}{ccc}-i & 1 & 0 \\ 1+i & 0 & -1\end{array}\right), x=\left(\begin{array}{c}-2 i \\ 4 \\ 1.57\end{array}\right)$
Proof . (a) $L_{A}(x)=\left(\begin{array}{cc}1 & \pi \\ y / 2 & -10\end{array}\right)\left(\begin{array}{c}17 \\ -5\end{array}\right)=\left(\begin{array}{cc}17 & -5 \pi \\ \frac{17}{2} & +50\end{array}\right)=\left(\begin{array}{c}17-5 \pi \\ 117 / 2\end{array}\right)$ (b) $L_{A}(x)=\left(\begin{array}{ccc}-i & 1 & 0 \\ 1+i & 0 & -1\end{array}\right)\left(\begin{array}{c}-2 i \\ 4 \\ 1,57\end{array}\right)=\left(\begin{array}{c}2 i^{2}+4+0 \\ -2 i-2 i^{2}+0-1,57\end{array}\right)$ $=\left(\begin{array}{c}2 \\ -2 i+0.43\end{array}\right)$
Problem 5. 5. Let $V$ and $W$ be vector spaces over a field $F$. Let $T: V \rightarrow W$ be a linear transformation. Let $v_{1}, \ldots, v_{n} \in V$ such that $\operatorname{span}\left(\left\{v_{1}, \ldots, v_{n}\right\}\right)=V$, then $\operatorname{span}\left(\left\{T\left(v_{1}\right), \ldots, T\left(v_{n}\right)\right\}\right)=R(T)$.
Proof . Let $y \in R(T)$. Then there exists $x \in V$ with $T(x)=y$ by def of $R(T)$. Since $x \in V$ and $V=\operatorname{span}\left(\left\{v_{1}, \ldots, v_{n}\right\}\right)$ we have $x=c_{1} v_{1}+c_{2} v_{2} t_{14}+c_{1} v_{1}$ for some $c_{i} \in F$, So, $y=T(x)=T\left(c_{1} v_{1}+c_{2} v_{2}+u_{1}+c_{1} v_{n}\right)=c_{1} T\left(v_{1}\right)+c_{2} T\left(v_{2}\right)+m+c_{n} T\left(v_{n}\right)$ co $y \in \operatorname{span}\left(\left\{T\left(v_{1}\right), T\left(v_{2}\right), \ldots, T\left(v_{n}\right)\right\}\right)$ Hence $R(T) \subseteq \operatorname{span}\left(\left\{T\left(v_{1}\right), T\left(v_{2}\right)_{1}, \ldots, T\left(v_{n}\right)\right\}\right)$ Thus, $\left.R(T)=\operatorname{spn}\left(\left\{T\left(V_{1}\right)_{1}, \mu_{1}\right) T\left(V_{n}\right)\right\}\right)=R(T)$
Problem 6. 6. Let $V$ and $W$ be vector spaces over a field $F$. Let $T: V \rightarrow W$ be a linear transformation. Let $0_{\mathrm{V}}$ and $0_{\mathrm{W}}$ be the zero vectors of $V$ and $W$ respectively. (a) Prove that $T$ is one-to-one if and only if $N(T)=\left\{0_{\mathrm{V}}\right\}$. (b) Suppose that $V$ and $W$ are both finite-dimensional and $\operatorname{dim}(V)=$ $\operatorname{dim}(W)$. Prove that $T$ is one-to-one if and only if $T$ is onto. (c) Suppose that $V$ and $W$ are both finite-dimensional. Prove that if $T$ is one-to-one and onto then $\operatorname{dim}(V)=\operatorname{dim}(W)$
Proof . $\Rightarrow$ is easy. Suppuse that T is one-to-one. We know that (T\left(0_{v}\right)=0_{w}\) So, $O_{v} \in N(T)$. Since $T$ is one-to-one, $0_{V}$ is the only solution Thus, $N(T)=\{x \in V \mid T(x)=0 w\}=\left\{0_{v}\right\}$ Let $x, y \in V$ with $T(x)=T(y)$ Then $T(x)-T(y)=0 w$. So, $T(x-y)=O_{w}$, since $T$ ir linear. Thus, $x-y \in N(T)$. so, $x-y=0_{V}$. Thus, $x=y$. (b) Suppoce that $\operatorname{dim}(V)=\operatorname{dim}(W)$ Rocale that nullity $(T)+\operatorname{can} k(T)=\operatorname{dim}(V)$ $\Rightarrow$ Suppore that $T$ is one-tu-one. By pact $(a)$, $N(T)=\{0,\}\}$ so nullity $(T)=0$, So, mea $0+\operatorname{cank}(T)=\operatorname{dim}(V)$ Since $\operatorname{dim}(V)=\operatorname{dim}(\omega)$ we get 1. Since $R$ (T) is a subspace of w and since dim (R(T) Hence T is onto. $\Leftrightarrow$ Suppose that $T$ is onto. So, $R(T)=W$, So, $R(T)=W_{C}=\operatorname{dim}(w)$, Thus, rank $(T)=\operatorname{dim}(v)$ since $dim(w)=dim(v)$we get that $null(T)+dim(T)=dim(v)$ because $null(T)+dim(w)=dim(w)$, so $null(T)=0$, thus $N(T)=\{0_v\}$. (c) Supeose that $V$ and $W$ ane finite dimen ritinal cand T is $1-1$ and onto. basis for $V$, Let $w_{i}=T\left(v_{i}\right)$ for $i=1,2, \ldots, n$, By thm in the notes since $T$ By thm in the notes $\left.w_{1}, w_{2}, \ldots, w_{n}\right\}$ is a basis fur $w_{1}$ s a basis $(V)=n=\operatorname{dim}(\omega)$,
Problem 7. 7. Let $V$ and $W$ be finite dimensional vector spaces and let $T: V \rightarrow W$ be a linear transformation. (a) If $\operatorname{dim}(V)<\operatorname{dim}(W),$ then $T$ is not onto. (b) If $\operatorname{dim}(V)>\operatorname{dim}(W),$ then $T$ is not one-to-one.
Proof . (a) is easy. (b) We have that $nullity (T)+rank(T)=\operatorname{dim}(V)$. Since $\operatorname{din}(V)>\operatorname{dim}(\omega)$ we have that nullity $(T)+ranK(T)>\operatorname{dim}(W) if $$nullity(T)=0$, then $rank(T)>dim(W)$. But that would imply that $R(T)$, which is a subspace of $W$ but has bigger dimension than $W$$$, this couyld not happen, so $nullity (T) >0$, So, $\operatorname{dim}(N(T))>0$. so $N(T) \neq\left\{0_{v}\right\}$so, $T$ is not one to one b problem 6.

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