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## Solution to Linear Algebra Hoffman & Kunze Chapter 7.1

#### Exercise 7.1.1

Let $T$ be a linear operator on $F^2$. Prove that any non-zero vector which is not a characteristic vector for $T$ is a cyclic vector for $T$. Hence, prove that either $T$ has a cyclic vector or $T$ is a scalar multiple of the identity operator.

Solution: Let $v\in F^2$ be nonzero such that $v$ is not a characteristic vector for $T$. Then $v,Tv$ is linearly independent. Hence $v,Tv$ forms a basis of $F^2$. Namely, $v,Tv$ spans the whole space $F^2$. So $v$ is a cyclic vector for $T$.

Suppose $T$ does not have a cyclic vector, we only need to show that $T$ is a scalar multiple of the identity operator. By the first part, any non-zero vector is a characteristic vector for $T$. We take a basis $v_1,v_2$ of $F^2$. Then $Tv_1=av_1$ and $Tv_2=bv_2$. We show that $a=b$. Since $v_1+v_2$ is non-zero and is a 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 $v_1$ and $v_2$ are linearly independent, we must have $a=b=c$. Hence $Tv_1=av_1$ and $Tv_2=av_2$. It is clear that $T$ is a scalar multiple of the identity operator.

#### Exercise 7.1.2

Solution: Since $T$ is diagonalizable, the minimal polynomial of $T$ is $(x-2)(x+1)$ but the characteristic polynomial is $(x-2)^2(x+1)$. By Theorem 2 and Corollary in page 230, if $T$ has a cyclic vector then the minimal polynomial and characteristic polynomial are the same. This is impossible. Hence $T$ has no cyclic vectors.

The $T$-cyclic subspace 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 linearly independent. Thus $(1,0,0)$ is a cyclic vector and the T-annihilator of $(1,0,0)$ is exactly the characteristic polynomial of $T$ (please compute it by yourself).

Note that $T(1,0,i)=(1,0,i)$, hence the T-annihilator of $(1,0,i)$ is exactly $x-1$.

#### Exercise 7.1.4

Prove that if $T^2$ has a cyclic vector, then $T$ has a cyclic vector. Is the converse true?

Solution: It is clear that $Z(\alpha,T^2)\subset Z(\alpha,T)$. Let $\alpha$ be a cyclic vector of $T^2$, then we have$V=Z(\alpha,T^2)\subset Z(\alpha,T).$Hence $Z(\alpha,T)=V$, namely $\alpha$ is a cyclic vector of $T$.

The converse is false. Let $T$ be the operator on $F^2$ corresponding to the matrix $\begin{pmatrix} 0 &1\\ 0&0\end{pmatrix}$. Then $T$ is cyclic by Exercise 7.1.1. But $T^2=0$ is not cyclic.

#### Exercise 7.1.5

Solution: Recall that from Exercise 6.8.15, the characteristic polynomial of $N$ is $x^n$ and hence $N^n=0$. Since $N^{n-1}\alpha\ne 0$, the $T$-annihilator of $T$ is $x^n$. Hence by Theorem 1 (i) $\Z(\alpha,N)=V$. Thus $\alpha$ is a cyclic vector of $N$. The matrix has the form of (7-2) in page 229 where all $c_i$ are zero.

#### Exercise 7.1.6

Give a direct proof that if $A$ is the companion matrix of the monic polynomial $p$, then $T$ is the characteristic polynomial for $A$.

Solution: Expand $\det(xI-A)$ from the last column.

#### Exercise 7.1.7

Let $V$ be an $n$-dimensional vector space, and let $T$ be a linear operator on $V$. Suppose that $T$ is diagonalizable.

(a) If $T$ has a cyclic vector, show that

(b) If $T$ has $n$ distinct characteristic values, and if $\{\alpha_1,\dots,\alpha_n\}$ is a basis of characteristic vectors for $T$. Show that $\alpha=\alpha_1+\cdots+\alpha_n$ is a cyclic vector for $T$.

Solution:

(a) Since $T$ is diagonalizable, let $p$ be the minimal polynomial of $T$,$p(x)=(x-z_1)\cdots(x-z_k),$where $z_1,\dots,z_k$ are all distinct characteristic values for $T$, see Theorem 6 in page 204. Since $T$ has a cyclic vector. Hence $p$ is also the characteristic polynomial for $T$, which has degree $n$. Therefore $k=n$. That is $T$ has $n$ distinct characteristic values.

(b) Let $z_i$ be the characteristic value for $T$ corresponding  to $\alpha_i$. Then $T^k\alpha_i=z_i^k\alpha_i$. Thus for any polynomial $g$, we have$g(T)\alpha=\sum_{i=1}^n g(z_i)\alpha_i.$Since $z_i$ are 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 basis $\{\alpha_1,\dots,\alpha_n\}$ of $V$. So $\alpha$ is a cyclic vector for $T$.

#### Exercise 7.1.8

Let $T$ be a linear operator on the finite-dimensional vector space $V$. Suppose $T$ has a cyclic vector. Prove that if $U$ is any linear operator which commutes with $T$, then $U$ is a polynomial in $T$.

Solution: Let $\alpha$ be a cyclic vector of $V$ and $\dim V=n$. Then $\alpha,T\alpha,\cdots,T^{n-1}\alpha$ is a basis of $V$. Hence we have$U\alpha=\sum_{i=0}^{n-1}a_iT^i\alpha,$for some $a_i\in F$. Since $U$ commutes 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 of $V$, we conclude that$U=\sum_{i=0}^{n-1}a_iT^i$ is a polynomial in $T$.

### This Post Has 3 Comments

1. 2. Hence we have $$av_1+bv_2=cv_1+cv_2.$$The "v" is missing.
1. 