Category: Chapter 3

Chapter 3 Exercise F 0

Chapter 3 Exercise F

1. Solution: For any $\vp\in\ca L(V,\mb F)$, if $\dim \m{range} \vp=0$, then $\vp$ is the zero map. If $\dim \m{range} \vp=1$, then $\vp$ is surjective since $\dim\mb F=1$. Moreover, $\dim \m{range} \vp\leqslant \dim \mb...

Chapter 3 Exercise E 0

Chapter 3 Exercise E

1. Solution: Suppose $T$ is linear. Let $(v_1, Tv_1), (v_2, Tv_2) \in \operatorname{graph of} T$. We have $$ \begin{aligned} (v_1, Tv_1) + (v_2, Tv_2) &= (v_1 + v_2, Tv_1 + Tv_2)\\ &= (v_1 +...

Chapter 3 Exercise D 0

Chapter 3 Exercise D

1. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 22. It is almost the same. 2. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 25. 3. Solution: If...

Chapter 3 Exercise C 0

Chapter 3 Exercise C

1. Solution: Suppose for some basis $v_1$, $\cdots$, $v_n$ of $V$ and some basis $w_1$, $\cdots$, $w_m$ of $W$, the matrix of $T$ has at most $\dim \m{range} T-1$ nonzero entries. Then there are at...

Chapter 3 Exercise B 0

Chapter 3 Exercise B

1. Solution: Assume $V$ is 5-dimensional vector space with a basis $e_1$, $\cdots$, $e_5$. Define $T\in\ca L(V,V)$ by \[Te_1=e_1,Te_2=e_2,Te_3=Te_4=Te_5=0.\]Then $\mathrm{null} T=\mathrm{span}(e_3,e_4,e_5)$, hence $\dim \mathrm{null} T=3$. Similarly, $\mathrm{range} T=\mathrm{span}(e_1,e_2)$, hence $\dim \mathrm{range} T=2$. 2. Solution:...

Chapter 3 Exercise A 0

Chapter 3 Exercise A

1. Solution: If $T$ is linear, then \[(0,0)=T(0,0,0)=(b,0)\]by 3.11, hence $b=0$. We also have \[T(1,1,1)=T(1,1,0)+T(0,0,1),\]it is equivalent to \[(1+b,6+c)=(b-2,6)+(3+b,0)=(1+2b,6).\]Thus $6+c=6$ implies $c=0$. Conversely, if $b=c=0$, $T$ is obviously linear. See 3.4 or Problem 3....