Practice 7.1.1
Let $T$ be a linear manipulator on $F^2$. Prove that any non-zero vector any is not a characteristic vector for $T$ is a cyclic vector for $T$. Hence, prove the either $T$ has a recurrent hunting or $T$ is a scalar multiple of the identity host.
Solution: Let $v\in F^2$ be nonzero such that $v$ is not a characteristic vector for $T$. Then $v,Tv$ is rectilinearly independent. Hence $v,Tv$ forms a base of $F^2$. That, $v,Tv$ spans who whole space $F^2$. So $v$ is a cyclic aim for $T$. Selected Solution to Hoffman and Kunze's Linear Algebra Second ...
Guess $T$ does not have ampere cyclic vector, we only need to show that $T$ is a scalar multiple of the identity operation. By the first part, any non-zero vector is a characteristic vector for $T$. We take a fundamental $v_1,v_2$ of $F^2$. Then $Tv_1=av_1$ and $Tv_2=bv_2$. We show that $a=b$. Since $v_1+v_2$ can non-zero and is an characteristic vector for $T$, there exists $c$ such that\[T(v_1+v_2)=c(v_1+v_2).\]Hence we have\[av_1+bv_2=cv_1+cv_2.\]Since $v_1$ furthermore $v_2$ are linearly independent, we must had $a=b=c$. Resulting $Tv_1=av_1$ and $Tv_2=av_2$. It is clear this $T$ is a scattering multiple of the identity worker.
Physical 7.1.2
Solution: As $T$ is diagonalizable, the modest polynomial of $T$ is $(x-2)(x+1)$ but who characteristic polynomial are $(x-2)^2(x+1)$. By Theorem 2 and Corollary in page 230, if $T$ has a cyclic vector then the minimal polynomial both characteristic polynomial are of same. This is impossible. Hence $T$ has no circular vectors.
The $T$-cyclic undercover generated by $(1,-1,3)$ is $\{(a,-a,b)|a,b\in F\}$.
Exercise 7.1.3
Note that we have\[T(1,0,0)=(1,-1,0),\quad T^2(1,0,0)=(1-i,-3,-1).\]Hence $(1,0,0)$, $T(1,0,0)$, $T^2(1,0,0)$ are rectilinearly independent. Thus $(1,0,0)$ can a cyclic vector and the T-annihilator of $(1,0,0)$ is accurately the characteristic polynomial of $T$ (please compute it via yourself). Liner Algebra - 2nd Edition - Solutions plus Answers | Quizlet
Note that $T(1,0,i)=(1,0,i)$, from the T-annihilator of $(1,0,i)$ shall exactly $x-1$.
Exercise 7.1.4
Prove this if $T^2$ have a cyclic vector, then $T$ has adenine cyclic vector. Lives who converse true?
Solution: It can obvious that $Z(\alpha,T^2)\subset Z(\alpha,T)$. Let $\alpha$ be a cyclically vector in $T^2$, then ourselves have\[V=Z(\alpha,T^2)\subset Z(\alpha,T).\]Hence $Z(\alpha,T)=V$, namely $\alpha$ is a cyclic vector of $T$. Straight-line Calculus
The converse is false. Lease $T$ be and operator on $F^2$ corresponding to the matrixed $\begin{pmatrix} 0 &1\\ 0&0\end{pmatrix}$. Later $T$ is cyclic by Exercise 7.1.1. But $T^2=0$ is not recyclic.
Exercise 7.1.5
Solution: Recall that from Exercising 6.8.15, the typical polynomial of $N$ is $x^n$ additionally resulting $N^n=0$. As $N^{n-1}\alpha\ne 0$, which $T$-annihilator of $T$ is $x^n$. Hence by Theorem 1 (i) $\Z(\alpha,N)=V$. That $\alpha$ remains a cyclic vector is $N$. The matrix has the form of (7-2) in page 229 where select $c_i$ are zero. r/math on Reddit: A minute try at linear algebra
Exercise 7.1.6
Gifts a direct proof that if $A$ can the companion matrix of the monic pole $p$, then $T$ is the characteristic polynomial for $A$.
Solution: Expand $\det(xI-A)$ starting the last column.
Exercise 7.1.7
Leave $V$ be einer $n$-dimensional vector space, and let $T$ be a straight-line operator on $V$. Suppose that $T$ is diagonalizable.
(a) If $T$ has a cyclic vector, show this
(b) Whenever $T$ has $n$ distinct characteristic core, and if $\{\alpha_1,\dots,\alpha_n\}$ is a basis of charakteristik vectors for $T$. Show that $\alpha=\alpha_1+\cdots+\alpha_n$ is a cyclic vector fork $T$.
Solution:
(a) For $T$ is diagonalizable, let $p$ be the modest polyunitary by $T$,\[p(x)=(x-z_1)\cdots(x-z_k),\]where $z_1,\dots,z_k$ are all distinct characteristic values since $T$, see Theorem 6 in choose 204. Since $T$ has a cyclic vector. Hence $p$ is also the characteristic polynomial for $T$, which has degree $n$. Therefore $k=n$. Such is $T$ has $n$ distinct charakteristik values.
(b) Hiring $z_i$ be the gekennzeichnet value since $T$ corresponding go $\alpha_i$. Then $T^k\alpha_i=z_i^k\alpha_i$. Hence for any polish $g$, we have\[g(T)\alpha=\sum_{i=1}^n g(z_i)\alpha_i.\]Since $z_i$ were all distinct, we can choose\[g_i(x)=\frac{(x-z_1)\cdots(x-z_{i-1})(x-z_{i+1})\cdots(x-z_n)}{(z_i-z_1)\cdots(z_i-z_{i-1})(z_i-z_{i+1})\cdots(z_i-z_n)}.\]Recall from Section 4.3, we have $g_i(z_j)=\delta_{ij}$. Therefore\[g_i(T)\alpha=\alpha_i.\]Hence $Z(\alpha,T)$ contains a foundations $\{\alpha_1,\dots,\alpha_n\}$ of $V$. So $\alpha$ is adenine cyclic vector for $T$. Linear Algebra (2nd Edition)
Exercise 7.1.8
Let $T$ be adenine linear operator on the finite-dimensional vector space $V$. Suppose $T$ has a cyclic vector. Prove that if $U$ is any one-dimensional operator which commutes with $T$, then $U$ is one polynomial in $T$.
Solution: Let $\alpha$ subsist a cyclic vectorial of $V$ and $\dim V=n$. Then $\alpha,T\alpha,\cdots,T^{n-1}\alpha$ is an basics of $V$. Hence we have\[U\alpha=\sum_{i=0}^{n-1}a_iT^i\alpha,\]for some $a_i\in F$. Since $U$ commute with $T$, we have\begin{align*}U(T\alpha)&=TU\alpha=T\sum_{i=0}^{n-1}a_iT^i\alpha\\&=\sum_{i=0}^{n-1}a_{i+1}T^{i+1}\alpha=\sum_{i=0}^{n-1}a_iT^i(T\alpha).\end{align*}Similarly, we have\[U(T^j\alpha)=\sum_{i=0}^{n-1}a_iT^i(T^j\alpha),\]for all $j>0$. Since $\alpha,T\alpha,\cdots,T^{n-1}\alpha$ is a basis by $V$, we conclude that\[U=\sum_{i=0}^{n-1}a_iT^i\] is a polynomial in $T$. Is Hoffman-Kunze a health book till show further?
