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content/notes/HW2.md

@@ -38,8 +38,14 @@ $dim(U\cap W) \leq 4$ so from the formula $dim(U+W) = dim(U) + dim(W) - dim(U \c
 
 (c). Max value of $dim(U+W)$
 
-We have $dim(U+W) \leq dim(U) + dim(W) = 11$. But since $U+W$ is a subspace of $V$ which has dimension $10$ , we must have $dim(U+W) \leq 10$. 
+We have $dim(U+W) \leq dim(U) + dim(W) = 11$. But since $U+W$ is a subspace of $V$ which has dimension $10$ , we must have $dim(U+W) \leq 10$. To show that this can be achieved, we could let $\{v_1,...,v_{10}\}$ be a basis for V, and choose $U = span\{v_1, v_2, v_3, v_4, v_5, v_5, v_7)\}$ and $W=span\{v_7,v_8,v_9,v_{10}\}$. Then $U+W$ will contain the basis for $V$ and hence $U+W=V$. So the maximum possible value of dim(U+W) is 10.
+
+(d). Since $dim(U+W) \leq 10$ and $dim(U+W) = 11 - dim(U\cap W)$, we have $dim(U\cap W) \geq 11-10= 1$ 
+
+(c). No, because if $U+W$ is direct then $dim(U+W) = 0$, but we showed that $dim(U\cap W) \geq 1$
 
 # 3.
 Let $U$ = $\{p\in P_2(\mathbb{R}): p(x)$ is divisible by $x-3\}$. Then U is a subspace of P2
-(a) Find a basis of U.
+(a) Find a basis of U.
+
+Suppose a polynomial of degree 2 is divisible by $x-3$ . It can be written as $(x-3)(ax+b)$ where $a,b \in \mathbb{R}$. Then we have a basis $\{(x-3), x(x-3)\}$