Location
LADR4eSolutions经典最全（By Axler？）.pdf, Exercise 3.F.22(b)
(The README already notes this answer is wrong. Below is an error analysis along with two alternative proofs. Discussion is welcome!)
Error Analysis
The original proof takes subspaces $X, Y$ such that $V = U \oplus X$ and $V = W \oplus Y$, then defines:
$$\psi(u + x) = \tfrac{1}{2}\varphi(x), \quad \beta(w + y) = \tfrac{1}{2}\varphi(y)$$
When verifying $\varphi = \psi + \beta$, for $v = u + x = w + y$, the proof claims:
$$\varphi(v) = \tfrac{1}{2}\varphi(v) + \tfrac{1}{2}\varphi(v) = \tfrac{1}{2}\varphi(x) + \tfrac{1}{2}\varphi(y) = \psi(v) + \beta(v)$$
The step $\tfrac{1}{2}\varphi(v) = \tfrac{1}{2}\varphi(x)$ requires $\varphi(u) = 0$. However, $u \in U$ does not necessarily belong to $U \cap W$, and $\varphi \in (U \cap W)^0$ only vanishes on $U \cap W$, not on all of $U$. The same issue applies to $\tfrac{1}{2}\varphi(v) = \tfrac{1}{2}\varphi(y)$.
The Easy Direction: $U^0 + W^0 \subseteq (U \cap W)^0$
If $\varphi \in U^0$ and $\psi \in W^0$, then for any $v \in U \cap W$, we have $v \in U$ and $v \in W$, so $(\varphi + \psi)(v) = \varphi(v) + \psi(v) = 0 + 0 = 0$. Hence $\varphi + \psi \in (U \cap W)^0$.
The Hard Direction: $(U \cap W)^0 \subseteq U^0 + W^0$
Proof 1: Direct Construction via Direct Sum Decomposition
Take $\varphi \in (U \cap W)^0$. Choose subspaces $U_1, W_1, X$ such that
$$U = (U \cap W) \oplus U_1, \quad W = (U \cap W) \oplus W_1$$
$$V = (U \cap W) \oplus U_1 \oplus W_1 \oplus X$$
Define $\alpha, \beta \in V'$ by: for $v = a + b + c + d$ with $a \in U \cap W$, $b \in U_1$, $c \in W_1$, $d \in X$, let
$$\alpha(v) = \varphi(c) + \tfrac{1}{2}\varphi(d), \quad \beta(v) = \varphi(b) + \tfrac{1}{2}\varphi(d)$$
$\alpha \in U^0$: For $u \in U = (U \cap W) \oplus U_1$, we have $u = a + b$ (with $c = d = 0$), so $\alpha(u) = 0$.
$\beta \in W^0$: For $w \in W = (U \cap W) \oplus W_1$, we have $w = a + c$ (with $b = d = 0$), so $\beta(w) = 0$.
$\alpha + \beta = \varphi$: For any $v = a + b + c + d$,
$$\alpha(v) + \beta(v) = \varphi(b) + \varphi(c) + \varphi(d) = \varphi(a) + \varphi(b) + \varphi(c) + \varphi(d) = \varphi(v)$$
where $\varphi(a) = 0$ since $a \in U \cap W$ and $\varphi \in (U \cap W)^0$.
Therefore $\varphi = \alpha + \beta \in U^0 + W^0$.
Proof 2: Dimension Argument
Alternatively, the hard direction can be avoided entirely by combining the easy inclusion with a dimension count.
Using $\dim S^0 = \dim V - \dim S$:
$$\dim(U \cap W)^0 = \dim V - \dim(U \cap W)$$
For the right-hand side, by the dimension formula for sums of subspaces:
$$\dim(U^0 + W^0) = \dim U^0 + \dim W^0 - \dim(U^0 \cap W^0)$$
By part (a), $U^0 \cap W^0 = (U + W)^0$, so $\dim(U^0 \cap W^0) = \dim V - \dim(U + W)$. Substituting:
$$\dim(U^0 + W^0) = (\dim V - \dim U) + (\dim V - \dim W) - (\dim V - \dim(U + W))$$
$$= \dim V - \dim U - \dim W + \dim(U + W)$$
Applying $\dim(U + W) = \dim U + \dim W - \dim(U \cap W)$:
$$= \dim V - \dim(U \cap W)$$
Since $U^0 + W^0 \subseteq (U \cap W)^0$ and both sides have the same dimension, equality holds.
Location
LADR4eSolutions经典最全（By Axler？）.pdf, Exercise 3.F.22(b)
(The README already notes this answer is wrong. Below is an error analysis along with two alternative proofs. Discussion is welcome!)
Error Analysis
The original proof takes subspaces$X, Y$ such that $V = U \oplus X$ and $V = W \oplus Y$ , then defines:
When verifying$\varphi = \psi + \beta$ , for $v = u + x = w + y$ , the proof claims:
The step$\tfrac{1}{2}\varphi(v) = \tfrac{1}{2}\varphi(x)$ requires $\varphi(u) = 0$ . However, $u \in U$ does not necessarily belong to $U \cap W$ , and $\varphi \in (U \cap W)^0$ only vanishes on $U \cap W$ , not on all of $U$ . The same issue applies to $\tfrac{1}{2}\varphi(v) = \tfrac{1}{2}\varphi(y)$ .
The Easy Direction:$U^0 + W^0 \subseteq (U \cap W)^0$
If$\varphi \in U^0$ and $\psi \in W^0$ , then for any $v \in U \cap W$ , we have $v \in U$ and $v \in W$ , so $(\varphi + \psi)(v) = \varphi(v) + \psi(v) = 0 + 0 = 0$ . Hence $\varphi + \psi \in (U \cap W)^0$ .
The Hard Direction:$(U \cap W)^0 \subseteq U^0 + W^0$
Proof 1: Direct Construction via Direct Sum Decomposition
Take$\varphi \in (U \cap W)^0$ . Choose subspaces $U_1, W_1, X$ such that
Define$\alpha, \beta \in V'$ by: for $v = a + b + c + d$ with $a \in U \cap W$ , $b \in U_1$ , $c \in W_1$ , $d \in X$ , let
where$\varphi(a) = 0$ since $a \in U \cap W$ and $\varphi \in (U \cap W)^0$ .
Therefore$\varphi = \alpha + \beta \in U^0 + W^0$ .
Proof 2: Dimension Argument
Alternatively, the hard direction can be avoided entirely by combining the easy inclusion with a dimension count.
Using$\dim S^0 = \dim V - \dim S$ :
For the right-hand side, by the dimension formula for sums of subspaces:
By part (a),$U^0 \cap W^0 = (U + W)^0$ , so $\dim(U^0 \cap W^0) = \dim V - \dim(U + W)$ . Substituting:
Applying$\dim(U + W) = \dim U + \dim W - \dim(U \cap W)$ :
Since$U^0 + W^0 \subseteq (U \cap W)^0$ and both sides have the same dimension, equality holds.