Let $E$ and $F$ be two finite dimension vector spaces over the same field $K$, $V$ is sub-space of $E$, $L_V(E,F)$ is the set of linear maps from $E$ to $F$ which vanish on $V$. And let $W$ such that $E = V \oplus W$

a) Show that $L_V(E,F)$ is a subspace of $L(E,F)$ isomorphic to $L(E/V,F)$ and $L(W,F)$.

b) Show that $E^* = V^\perp \oplus W^\perp$ and that $V$ and $W$ are isomorphic to $W^\perp$ and $V^\perp$.

c) Show that if $f$ is a linear map from $E$ to $F$, $t_{f(F^*)}$ and $(f(E))^*$.

My work:

for a) I proved that it's a subspace. and about the isomorphism between $L_V(E,F)$ and $L(W,F)$ I consedred the following map $\Phi$ from $L_V(E,F)$ to $L(W,F)$ wich associate $\phi \in L_V(E,F)$ to to the restrection of $f \in L(W,F)$ but I don't know how to show that it's an isomorphism.

any help please for the others

  • $\begingroup$ What do you mean by the notation $V^{\perp}$? $\endgroup$ – Tobias Kildetoft Jan 16 '14 at 10:51
  • $\begingroup$ The orthogonal of $V$ $\endgroup$ – Mohamez Jan 16 '14 at 10:52
  • $\begingroup$ Orthogonal with respect to what inner product/bilinear form? $\endgroup$ – Christoph Jan 16 '14 at 10:54
  • 1
    $\begingroup$ Oh I see! I guess $V^\perp$ is the subspace of elements of $E^*$ that vanish on $V$? $\endgroup$ – Christoph Jan 16 '14 at 10:56
  • $\begingroup$ What is $f(F^*)$ and $t_{f(F^*)}$, and what do you mean by "$t_{f(F^*)}$ and $(f(E))^*$"? $\endgroup$ – Luiz Cordeiro Jan 16 '14 at 11:28

a) That $L_V(E,F)$ is a subspace of $L(E,F)$ is trivial. The problem is to show that $L_V(E,F)\cong L(E/V,F)\cong L(W,F)$. Let $\pi:E\rightarrow E/V$ be the canonical projection.

Let $\phi:L_V(E,F)\rightarrow L(E/V,F)$ be given by the following: If $T\in L_V(E,F)$, then $\phi(T):E/V\rightarrow F$ be given by $\phi(T)(\pi x)=Tx$. You should verify that $\phi$ is well-defined and is linear. Its inverse is given by $\phi^{-1}(T)=T\circ\pi$ for $T\in L(E/V,F)$. This shows that $L_V(E,F)$ is isomorphic to $L(E/V,F)$.

Let $\psi:L_V(E,F)\rightarrow L(W,F)$ be given by the following: If $T\in L(E/V,F)$, then $\psi(T)=T|_W$, that is, $\psi(T)$ is just the restriction of $T$ to $W$. Then $\psi$ is obviously linear and its inverse is given by $\psi^{-1}(T)(v+w)=T(w)$ for $T\in L(W,F)$ and $v\in V$, $w\in W$. You should first verify that $\psi^{-1}$ given by that is, indeed, well-defined. This shows that $L_V(E,F)$ is isomorphic to $L(W,F)$.

b)Given $T\in E^*$, define $T_V(v+w)=Tw$ and $T_W(v+w)=Tv$ for $v\in V$ and $w\in W$ (show that these are well-defined). Then $T_V\in V^\perp$, $T_W\in W^\perp$ and $T=T_V+T_W$. This shows that $E^*=V^\perp+W^\perp$. It is clear that $V^\perp+W^\perp=\left\{0\right\}$, so $E=V^\perp \oplus W^\perp$. To show that $V\cong W^\perp$, you should take a basis $B_V$ for $V$, a basis $B_W$ for $W$. Then, define $\varphi:V\rightarrow W^\perp$ linearly so that for every $b\in B_V$, $\varphi(b)(b)=1$ and $\varphi|_{B_V\setminus\left\{b\right\}\cup B_W}$. Then, show that $\varphi$ is an isomorphism. That $W\cong V^\perp$ is similar.


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