題目

Let $V$ be a finite-dimensional vector space over a field $F$, and let $W$ be a subspace of $V$. Show that there exists a linear transformation $T: V \to V$ whose null space and range are both $W$ if and only if $\dim V = 2 \dim W$.

答題

Denote the nullity of $f$ by $\operatorname{nullity}(f)$, and the rank of a linear function $f$ as $\operatorname{rank}(f)$.

Suppose first that there exists a linear transformation $T: V \to V$ whose null space and range are both $W$. Then $$ \operatorname{nullity}(T)=\dim W $$ and $$ \operatorname{rank}(T)=\dim W. $$ By the dimension theorem (Friedberg et al. (2002), Theorem 2.3), $$ \dim V =\operatorname{nullity}(T)+\operatorname{rank}(T) =2\dim W. $$ This proves the necessity.

Conversely, suppose that $\dim V=2\dim W$. We first construct $T$ and verify that it is linear, and then show that both its range and its null space are equal to $W$. Extend a basis of $W$ to a basis of $V$, and let $U$ be the subspace spanned by the additional basis vectors. Then $$ \dim U=\dim V-\dim W=\dim W. $$ After the substitution of $U$ for $V$ in Friedberg et al.'s (2002) Theorem 2.19, $U$ is isomorphism to $W$. By the definition of the isomorphism (Friedberg et al. (2002), p. 102), there exists an isomorphism linear transformation $S: U \to W$.

Define $T:V\to V$ by $T(v)=S(u)$ whenever $v=w+u$, with $w\in W$ and $u\in U$. Let $v_1,v_2\in V$ and $c \in F$. After the substitution of $2$ for $k$, $W$ for $W_1$, and $U$ for $W_2$ in Friedberg et al.'s (2002) Theorem 5.10, every $v \in V$ has a unique representation $v=w+u$, where $w\in W$ and $u\in U$. Thus for $i=1,2$, there exist unique vectors $w_i\in W$ and $u_i\in U$ such that $v_i=w_i+u_i$. The linearity of $S$ gives $$ T(v_1+v_2)=S(u_1+u_2)=S(u_1)+S(u_2)=T(v_1)+T(v_2), $$ and $$ T(cv_1)=T(cw_1+cu_1)=S(cu_1)=cS(u_1)=cT(v_1). $$ Therefore $T$ is a linear transformation by its definition (Friedberg et al. (2002), p. 65).

We now show that the range of $T$ is equal to $W$. As $S(U)=W$, the range of $T$ is $W$.

We next show that the null space of $T$ is equal to $W$. Let $v\in V$ and suppose that $T(v)=0$. There exist unique vectors $w\in W$ and $u\in U$ such that $v=w+u$. Then $S(u)=0$. By the definition of the isomorphism (Friedberg et al. (2002), p. 102), $S$ is invertible and hence one-to-one (Friedberg et al. (2002), p. 100). Therefore $S(u)=S(0)$ implies that $u=0$. As $v=w+u$, it follows that $v=w\in W$. Thus every vector in the null space of $T$ belongs to $W$. Conversely, let $x\in W$. As $W$ is a subspace of $V$, $x\in V$. As $x=x+0$ and $0\in U$, $T(x)=S(0)=0$. Thus every vector in $W$ belongs to the null space of $T$. Therefore the null space of $T$ is $W$.

Therefore there exists a linear transformation $T:V\to V$ whose null space and range are both $W$. This proves the sufficiency and completes the proof.

Reference

Friedberg, S. H., Insel, A. J., and Spence, L. E. (2002). Linear Algebra. Pearson, Upper Saddle River, 4th edition.