Linear Mathematics | Part IB, 2003

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

Ml(α)={βL(V,U):βα=0} and Mr(α)={βL(V,U):αβ=0}.\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 Ml(α)M^{l}(\alpha) and Mr(α)M^{r}(\alpha) are subspaces of L(V,U)L(V, U) of dimensions

dimMl(α)=(dimVrankα)dimU and dimMr(α)=dimker(α)dimV\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 UU and VV so that α\alpha is represented by

(Ir000)\left(\begin{array}{cc} I_{r} & 0 \\ 0 & 0 \end{array}\right)

where IrI_{r} is the r×rr \times r identity matrix and rr is the rank of α.]\left.\alpha .\right]

(b) Let Φ:L(U,V)L(V,U)\Phi: L(U, V) \rightarrow L\left(V^{*}, U^{*}\right) be given by Φ(α)=α\Phi(\alpha)=\alpha^{*}, where α\alpha^{*} is the dual map induced by α\alpha. Prove that Φ\Phi is an isomorphism. [You may assume that Φ\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

Φ(Ml(α))=Mr(α) and Φ(Mr(α))=Ml(α)\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 (βα)=αβ(\beta \alpha)^{*}=\alpha^{*} \beta^{*} and that β\beta^{* *} can be identified with β\beta under the canonical isomorphism between a vector space and its double dual.]

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

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