Category: Chapter 5

Chapter 5 Exercise C 0

Chapter 5 Exercise C

1. Solution: It is not said $V$ is finite-dimensional, but I will do it by assuming $\dim V<\infty$. If $T$ is invertible, then $\m{null}{T}=0$ and $\m{range} T=V$ since $T$ is bijective and surjective. Hence...

Chapter 5 Exercise B 0

Chapter 5 Exercise B

1. Solution: (a) Note that \[ (I-T)(I+T+\cdots+T^{n-1})=I-T^n=I \]and \[ (I+T+\cdots+T^{n-1})(I-T)=I-T^{n}=I ,\](in fact we just need to check only one) it follows that $I-T$ is invertible and \[(I-T)^{-1}=I+T+\cdots+T^{n-1}.\] (b) From the familiar formula \[1-x^n=(1-x)(1+x+\cdots+x^{n-1}).\] 2....

Chapter 5 Exercise A 0

Chapter 5 Exercise A

1. Solution: (a) For any $u\in U$, then $Tu=0\in U$ since $U\subset \m{null} T$, hence $U$ is invariant under $T$. (b) For any $u\in U$, then $Tu\in\m{range} T \subset U$, hence $U$ is invariant...