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…
Chapter 3 Exercise E
Exercises 1,2 and 4. For Problem 2, please also see Carson Rogers’s comment. 4. Solution: For any $f\in \ca L(V_1\times \cdots\times V_m,W)$ and given $i\in \{1,\cdots,m\}$, define $f_i:V_i\to W$ by…
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…
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…
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…
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…