Let $\alpha \in L(U, V)$ be a linear map between finite-dimensional vector spaces. Let

$$\begin{gathered} M^{l}(\alpha)=\{\beta \in L(V, U): \beta \alpha=0\} \quad \text { and } \\ M^{r}(\alpha)=\{\beta \in L(V, U): \alpha \beta=0\} . \end{gathered}$$

(a) Prove that $M^{l}(\alpha)$ and $M^{r}(\alpha)$ are subspaces of $L(V, U)$ of dimensions

$$\begin{gathered} \operatorname{dim} M^{l}(\alpha)=(\operatorname{dim} V-\operatorname{rank} \alpha) \operatorname{dim} U \quad \text { and } \\ \operatorname{dim} M^{r}(\alpha)=\operatorname{dim} \operatorname{ker}(\alpha) \operatorname{dim} V \end{gathered}$$

[You may use the result that there exist bases in $U$ and $V$ so that $\alpha$ is represented by

$$\left(\begin{array}{cc} I_{r} & 0 \\ 0 & 0 \end{array}\right)$$

where $I_{r}$ is the $r \times r$ identity matrix and $r$ is the rank of $\left.\alpha .\right]$

(b) Let $\Phi: L(U, V) \rightarrow L\left(V^{*}, U^{*}\right)$ be given by $\Phi(\alpha)=\alpha^{*}$, where $\alpha^{*}$ is the dual map induced by $\alpha$. Prove that $\Phi$ is an isomorphism. [You may assume that $\Phi$ is linear, and you may use the result that a finite-dimensional vector space and its dual have the same dimension.]

(c) Prove that

$$\Phi\left(M^{l}(\alpha)\right)=M^{r}\left(\alpha^{*}\right) \quad \text { and } \quad \Phi\left(M^{r}(\alpha)\right)=M^{l}\left(\alpha^{*}\right)$$

[You may use the results that $(\beta \alpha)^{*}=\alpha^{*} \beta^{*}$ and that $\beta^{* *}$ can be identified with $\beta$ under the canonical isomorphism between a vector space and its double dual.]

(d) Conclude that $\operatorname{rank}(\alpha)=\operatorname{rank}\left(\alpha^{*}\right)$.