Exercise 11, Section 3.5 of Hoffman’s Linear Algebra

Question

Let $W_1$ and $W_2$ be subspaces of a finite-dimensional vector space $V$.

(a) Prove that $(W_1+W_2)^0=W_1^0\cap W_2^0$.

(b) Prove that $(W_1\cap W_2)^0=W_1^0+W_2^0$.

My attempt: (a) Let $f\in (W_1+W_2)^0$. Then $f(x)=0_F$, $\forall x\in W_1+W_2$. Since $W_1,W_2\subseteq W_1+W_2$, we have $f(W_1)$, $f(W_2)=\{0_F\}$. Thus $f\in W_1^0 \cap W_2^0$. Hence $(W_1+W_2)^0\subseteq W_1^0\cap W_2^0$. Conversely, let $f\in W_1^0\cap W_2^0$. Then $f(W_1)$, $f(W_2)=\{0_F\}$. Let $x\in W_1+W_2$. Approach(1): Since $W_1+W_2$ $=\mathrm{span}(W_1\cup W_2)$, we have $x=\sum_{i=1}^nc_i \cdot_V u_i$, where $n\in \Bbb{N}$, $c_i\in F$, $u_i\in W_1\cup W_2$. So $f(x)$ $=f(\sum_{i=1}^nc_i \cdot_V u_i)$ $=\sum_{i=1}^nc_i \cdot_F f(u_i)$ $= \sum_{i=1}^nc_i \cdot_F 0_F$ $=0_F$. Thus $f(x)=0_F$, $\forall x\in W_1+W_2$. So $f\in (W_1+W_2)^0$. Hence $W_1^0\cap W_2^0\subseteq (W_1+W_2)^0$. Approach(2): Then $x=w_1+w_2$, for some $w_1,w_2\in W_1,W_2$. So $f(x)$ $=f(w_1+w_2)$ $=f(w_1)+f(w_2)$ $=0_F+0_F$ $=0_F$. Thus $f(x)=0_F$, $\forall x\in W_1+W_2$. So $f\in (W_1+W_2)^0$.

(b) Let $f\in W_1^0+W_2^0$. Then $f=g+h$, for some $g\in W_1^0$, $h\in W_2^0$. So $g(W_1)=\{0_F\}$ and $h(W_2)=\{0_F\}$. Let $x\in W_1\cap W_2$. Then $g(x)=0_F$ and $h(x)=0_F$. So $f(x)$ $=g+h(x)$ $=g(x)+h(x)$ $=0_F$. Thus $f(x)=0_F$, $\forall x\in W_1\cap W_2$. So $f\in (W_1\cap W_2)^0$. Hence $W_1^0+W_2^0\subseteq (W_1\cap W_2)^0$. Conversely, let $f\in (W_1 \cap W_2)^0$. Approach(1): Then $f(x)=0_F$, $\forall x\in W_1\cap W_2$. How to show $f=g+h$, where $g\in W_1^0$, $h\in W_2^0$? Approach(2): $(W_1\cap W_2)^0$ $=W_1^0+W_2^0$ $=\mathrm{span}(W_1^0\cup W_2^0)$. So we need to show $(W_1\cap W_2)^0$ is the smallest subspace containing $W_1^0\cup W_2^0$. It’s easy to check $(W_1\cap W_2)^0$ is subspace of $V^*$. Let $W$ be a subspace of $V$ such that $W_1^0\cup W_2^0\subseteq W$. We need to show $(W_1\cap W_2)^0\subseteq W$. I think, which uses essentially same proof idea as of $(W_1\cap W_2)^0\subseteq W_1^0+W_2^0$.

Answer 1

Since $W_1\cap W_2$ is a subspace of $W_2$, there is a complementary subspace $U_2$ such that $W_2=(W_1\cap W_2)\oplus U_2$. Let $U$ be a complement of $W_1+W_2$ in $V$, So $U\oplus U_2$ is a complement of $W_1$.

Now define $g$ to be zero on $W_1$ and $f$ on $U\oplus U_2$. By linearity $g$ is well-defined on all of $V$ and clearly it is in $W_1^0$. Now let $h=f-g$. It is enough to show that $h\in W_2^0$: on $W_1\cap W_2$ both $f$ and $g$ are zero, and on $U_2$, $g$ is equal to $f$ by definition, so $h$ is zero on both of these subspaces and hence on all of $W_2$. ($h$ is zero on $U$ too but this is irrelevant).

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Addenda:

1. Suppose $V=V_1\oplus V_2$, and if we have some linear function $f_1$ on $V_1$ and some linear functions $f_2$ o $V_2$. Each $v\in V$ has unique representation $v=v_1+v_2$, so we can define $f(v)=f_1(v_1)+f_2(v_2)$ and this definition is well-defined thanks to the uniqueness.
   In our case, we defined $g$ on $W_1$ as zero and as $f$ on $U\oplus U_2$. Since $V=W_1\oplus(U\oplus U_2)$, $g$ is well-defined on all of $V$.
2. I've defined $g$ to be equal to $f$ on the subspace $U\oplus U_2$, so in particular $g$ is equal to $f$ in its subspace $U_2$.
3. I didn't claim that $f$ is zero on $U\oplus U_2$, but $h$ is: Since $g$ equal $h$ on this subspace, $h=f-g=0$ there.
4. When I write "complement" of a subspace $U\subset V$ here I mean in the linear sense: another subspace $W\subset V$ such that $V=U\oplus W$. When we write $W^c$ we mean the set-complement, which is not a subspace.