# Tag Archives: Exercise B

## Chapter 10 Exercise B

## Chapter 9 Exercise B

## Chapter 8 Exercise B

## Chapter 7 Exercise B

1. Solution: It is true. Consider the standard orthonormal basis $e_1,e_2,e_3$ of $\mb R^3$. Define $T\in \ca L(\mb R^3)$ by the rule:\[Te_1=e_1,\quad Te_2=2e_2+e_1,\quad Te_3=3e_3.\]Since we have\[\langle Te_1,e_2\rangle =\langle e_1,e_2\rangle =0,\]\[\langle e_1,Te_2\rangle =\langle e_1,e_1+2e_2\rangle =1,\]it follows that $T$ is not self-adjoint. … Continue reading

## Chapter 6 Exercise B

2. Solution: If $v\in \m{span}(e_1,\cdots,e_m)$, then $e_1$, $\cdots$, $e_m$ is an orthonormal basis of $\m{span}(e_1,\cdots,e_m)$ by 6.26. By 6.30, it follows that\[\|v\|^2=|\langle v,e_1\rangle|^2+\cdots+|\langle v,e_m\rangle|^2.\] If $\|v\|^2=|\langle v,e_1\rangle|^2+\cdots+|\langle v,e_m\rangle|^2$, we denote \[ \xi=v-(\langle v,e_1\rangle e_1 +\cdots+\langle v,e_m\rangle e_m). \]It is easily … Continue reading

## Chapter 5 Exercise B

1. Suppose $T\in\ca L(V)$ and there exists a positive integer $n$ such that $T^n=0$. (a) Prove that $I-T$ is invertible and that \[(I-T)^{-1}=I+T+\cdots+T^{n-1}.\] (b) Explain how you would guess the formula above. Solution: (a) Note that \[ (I-T)(I+T+\cdots+T^{n-1})=I-T^n=I \]and \[ … Continue reading

## Chapter 3 Exercise B

1. Give an example of a linear map $T$ such that $\dim \mathrm{null} T=3$ and $\dim \mathrm{range} T = 2$. 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 $\mathrm{null} T=\mathrm{span}(e_3,e_4,e_5)$, … Continue reading

## Chapter 2 Exercise B

1. Find all vector spaces that have exactly one basis. 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 $2v$. … Continue reading

## 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 $a\ne 0$, then $a$ has inverse $a^{-1}$ such that $a^{-1}a=1$. … Continue reading