# Tag Archives: Exercise B

## Chapter 10 Exercise B

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## Chapter 9 Exercise B

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## Chapter 8 Exercise B

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## 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

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## 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

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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 Posted in Chapter 5 | Tagged ## 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

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## 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

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## Chapter 1 Exercise B

1. Prove that $-(-v)=v$ for every $v\in V$. 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. Suppose $a\in\mathbb F$, $v\in V$, … Continue reading

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