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…
Chapter 2 Exercise C
1. Solution: Let $u_1,u_2,\cdots,u_n$ be a basis of $U$. Thus $n=\dim U=\dim V$. Hence $u_1,u_2,\cdots,u_n$ is a linearly independent list of vectors in V with length $\dim V$. By 2.39,…
Chapter 2 Exercise B
1. Solution: The only vector spaces is $\{0\}$. For if there is a nonzero vector $v$ in a basis, then we can get a new basis by changing $v$ to…
Chapter 2 Exercise A
1. Solution: We just need to show that $v_1$, $v_2$, $v_3$, $v_4$ can be expressed as linear combination of $v_1-v_2$, $v_2-v_3$, $v_3-v_4$, $v_4$. Note that \[v_1=(v_1-v_2)+(v_2-v_3)+(v_3-v_4)+v_4,\]\[v_2=(v_2-v_3)+(v_3-v_4)+v_4,\]\[v_3=(v_3-v_4)+v_4,\quad v_4=v_4.\] 2. Solution: (a)…
Chapter 1 Exercise C
1. Solution: (a) $\{(x_1,x_2,x_3)\in\mathbb F^3:x_1+2x_2+3x_3=0\}$ is a subspace of $\mathbb F^3$. By 1.34, to show a subset is a subspace, we just need to check Additive identity, Closed under addition…
Chapter 1 Exercise B
1. Solution: By definition, we have\[(-v)+(-(-v))=0\quad\text{and}\quad v+(-v)=0.\]This implies both $v$ and $-(-v)$ are additive inverses of $-v$, by the uniqueness of additive inverse, it follows that $-(-v)=v$. 2. Solution: If…
Chapter 1 Exercise A
1.Solution: Because $(a+bi)(a-bi)=a^2+b^2$, one has\[\frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}.\]Hence\[c=\frac{a}{a^2+b^2},d=-\frac{b}{a^2+b^2}.\] 2. Solution1:From direct computation, we have\[\left(\frac{-1+\sqrt{3}i}{2}\right)^2=\frac{-1-\sqrt{3}i}{2},\]hence \[\left(\frac{-1+\sqrt{3}i}{2}\right)^3=\frac{-1-\sqrt{3}i}{2}\cdot\frac{-1+\sqrt{3}i}{2}=1.\]This means $\dfrac{-1+\sqrt{3}i}{2}$ is a cube root of 1. Solution2: Note that \[(a+bi)+(a-bi)=2a\] and \[(a+bi)(a-bi)=a^2+b^2,\] it follows that $\dfrac{-1+\sqrt{3}i}{2}$…