Exercise 1, Section 7.2 of Hoffman’s Linear Algebra

Question

Let $T$ be the linear operator on $F^2$ which is represented in the standard ordered basis by the matrix $$\begin{bmatrix}0&0\\1&0\end{bmatrix}.$$ Let $\alpha_1=(0,1)$. Show that $F^2\neq Z(\alpha_1;T)$, and that there is no non-zero vector $\alpha_2$ in $F^2$ with $Z(\alpha_2;T)$ disjoint from $Z(\alpha_1;T)$.

My attempt: Since $T(\alpha_1)=(0,0)=0\cdot \alpha_1$, we have $\alpha_1$ is eigenvector of $T$ corresponding eigenvalue $0$. By example 1 section 7.1, $\dim Z(\alpha_1;T)=1$. Thus $Z(e_2;T)\neq F^2$. Assume towards contradiction, $\exists \alpha_2\in F^2\setminus \{0\}$ such that $Z(\alpha_1;T)\cap Z(\alpha_2;T)=\{0\}$. Since $\alpha_2\neq 0$, we have $\dim Z(\alpha_2;T)\neq 0$. Clearly $\dim Z(\alpha_2;T)\neq 2$. So $\dim Z(\alpha_2;T)=1$. By example 1 section 7.1, $\alpha_2$ is eigenvector of $T$. Let $\alpha_2=(a,b)=a\cdot e_1+b\cdot e_2$, where $a\neq 0$ or $b\neq 0$. Then $T(\alpha_2)=c\cdot \alpha_2$. So $$T(\alpha_2)=a\cdot T(e_1)+b\cdot T(e_2)=(0,a)=(c\cdot a,c\cdot b).$$ Which implies $c\cdot a=0$ and $c\cdot b=a$. If $c\neq 0$, then $a=0$ and $b=0$. Thus we reach contradiction. If $c=0$, then $a=0$. So $\alpha_2=(0,b)=b\cdot \alpha_1$. That is $\alpha_2\in \text{span}(\alpha_1)\subseteq Z(\alpha_1;T)$. So $0\neq \alpha_2\in Z(\alpha_1;T)\cap Z(\alpha_2;T)$. Thus we reach contradiction. Hence $\nexists \alpha_2\in F^2\setminus \{0\}$ such that $Z(\alpha_1;T)\cap Z(\alpha_2;T)=\{0\}$. Is my proof correct?

Answer 1

I concur with your comment. A little shorter still, discarding the unnecessary index: If $\alpha=(a,b)$, then $$ T\alpha=(0,a)=a\alpha_1\in Z(\alpha_1,T)\Rightarrow a=0\Rightarrow\alpha=b\alpha_1\in Z(\alpha_1,T)\Rightarrow b=0\Rightarrow \alpha=0. $$ Note that $Z(\alpha,T)=\langle\alpha,T\alpha,T^2\alpha,\ldots\rangle$.