About exercise 3.F 23 on p.115 in "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
There is the following exercise in this book:
23 Suppose $V$ is finite-dimensional and $U$ and $W$ are subspaces of $V$. Prove that $(U \cap W)^0 = U^0 + W^0$.
$U^0$ is the set of linear functionals $\phi$ on $V$ such that $\phi(x) = 0$ for all $U$.
$W^0$ is the set of linear functionals $\phi$ on $V$ such that $\phi(x) = 0$ for all $W$.
$(U \cap W)^0$ is the set of linear functionals $\phi$ on $V$ such that $\phi(x) = 0$ for all $U \cap W$.
I think the assumption $\dim V < +\infty$ is not necessary.
But the author assumed that $\dim V < +\infty$.
Why?
My solution is here:
If $\phi \in (U \cap W)^0$, then $\phi(x) = 0$ for all $x \in U + W$.
Since $U \subset U + W$ and $W \subset U + W$, $\phi(x) = 0$ for all $x \in U \cap W$.
Conversely, if $\phi \in U^0 + W^0$, $\phi = \phi_1 + \phi_2$ for some $\phi_1 \in U^0$ and $\phi_2 \in W^0$.
For all $x \in U \cap W$, $\phi(x) = \phi_1(x) + \phi_2(x) = 0 + 0 = 0$.
So, $\phi \in (U \cap W)^0$.
2 Answers
$\begingroup$Regarding your solution: it looks like you have not proven that $(U \cap W)^0 \subset U^0 + W^0$.
If $\phi \in (U \cap W)^0$, then $\phi(x) = 0$ for all $x \in U + W$.
This is incorrect.
Since $U \subset U + W$ and $W \subset U + W$, $\phi(x) = 0$ for all $x \in U \cap W$.
I'm not sure what you're trying to prove here, since we already have $\phi \in (U \cap W)^0$, which by definition means that $\phi(x) = 0$ for all $x \in U \cap W$.
Conversely, if $\phi \in U^0 + W^0$, $\phi = \phi_1 + \phi_2$ for some $\phi_1 \in U^0$ and $\phi_2 \in W^0$.
For all $x \in U \cap W$, $\phi(x) = \phi_1(x) + \phi_2(x) = 0 + 0 = 0$. So, $\phi \in (U \cap W)^0$.
This is correct. Indeed, we have $U^0 + W^0 \subset (U \cap W)^0$, even in the infinite-dimensional setting.
$\endgroup$ 1 $\begingroup$I wonder I need the assumption $\dim V < +\infty$.
At least, if we assume that $\dim V < +\infty$, we can solve the exercise much more easily.
Let $t_1, \cdots, t_k$ be a basis of $U \cap W$.
Let $t_1, \cdots, t_k, u_1, \cdots, u_l$ be a basis of $U$.
Let $t_1, \cdots, t_k, w_1, \cdots, w_m$ be a basis of $W$.
Then, $t_1, \cdots, t_k, u_1, \cdots, u_l, w_1, \cdots, w_m$ is a basis of $U + W$.
Let $t_1, \cdots, t_k, u_1, \cdots, u_l, w_1, \cdots, w_m, v_1, \cdots, v_n$ be a basis of $V$.
Let $\phi \in (U \cap W)^0$.
Then $\phi(t_i) = 0$ for all $i \in \{1, \cdots, k\}$.
Let $\phi(u_i) = \beta_i$ for all $i \in \{1, \cdots, l\}$ and $\phi(w_i) = \gamma_i$ for all $i \in \{1, \cdots, m\}$ and $\phi(v_i) = \delta_i$ for all $i \in \{1, \cdots, n\}$.
Let $\phi_1$ be a linear functional such that
$\phi_1(t_i) = 0$ for all $i \in \{1, \cdots, k\}$ and
$\phi_1(u_i) = 0$ for all $i \in \{1, \cdots, l\}$ and
$\phi_1(w_i) = \gamma_i$ for all $i \in \{1, \cdots, m\}$ and
$\phi_1(v_i) = \delta_i$ for all $i \in \{1, \cdots, n\}$.
Let $\phi_2$ be a linear functional such that
$\phi_2(t_i) = 0$ for all $i \in \{1, \cdots, k\}$ and
$\phi_2(u_i) = \beta_i$ for all $i \in \{1, \cdots, l\}$ and
$\phi_2(w_i) = 0$ for all $i \in \{1, \cdots, m\}$ and
$\phi_2(v_i) = 0$ for all $i \in \{1, \cdots, n\}$.
Then, $\phi_1 \in U^0$ and $\phi_2 \in W^0$ and $\phi = \phi_1 + \phi_2$.
So, $\phi \in U^0 + W^0$.
So, $(U \cap W)^0 \subset U^0 + W^0$.
The converse is easy(We don't need the assumption $\dim V < +\infty$).
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