From dccac079c7820e1071f9a73c5739edb5c44bc58f Mon Sep 17 00:00:00 2001 From: Jet Hughes Date: Fri, 19 Aug 2022 00:22:10 +1200 Subject: [PATCH] vault backup: 2022-08-19 00:22:10 --- content/notes/HW2.md | 14 ++++++++++++-- 1 file changed, 12 insertions(+), 2 deletions(-) diff --git a/content/notes/HW2.md b/content/notes/HW2.md index bb3e54afd..27d2e8d33 100644 --- a/content/notes/HW2.md +++ b/content/notes/HW2.md @@ -38,7 +38,7 @@ $dim(U\cap W) \leq 4$ 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 $dim(U+W) \leq 10$. 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. +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. ==need to reword== (d). Since $dim(U+W) \leq 10$ and $dim(U+W) = 11 - dim(U\cap W)$, we have $dim(U\cap W) \geq 11-10= 1$ @@ -46,6 +46,16 @@ We have $dim(U+W) \leq dim(U) + dim(W) = 11$. But since $U+W$ is a subspace of $ # 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. -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 $\{(x-3), x(x-3)\}$ \ No newline at end of file +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 $\{(x-3), x(x-3)\}$ + +To prove this is a basis we need to show that it is linearly independent and that is spans $U$ . To show linear independece if $a(x-3)+b(x^2+3x) = (0x^2 + 0x + 0)$, we have $-3a=0, a-3b=0$ and $b =0$. So $a=b=0$ and the only linear combination equal to the zero vector is the one where $a=b=0$, hence this set is linearly independent. ==need to add proof of span== + +(b). Find another subspace $W$ of $P_2(\mathbb{R})$ such that $U+W$ is a direct sum. + +Need to find $W$ where $U\cap W = \{0_v\}$ and $U+W = P_2(\mathbb{R})$. We have a basis of $U = \{(x-3), (x^2-3x)\}$. We can adjoin a basis $\{1, x, x^2\}$ of $P_2(\mathbb{R})$ to get a spanning set $\{x-3, x^2-3x,1, x, x^2\}$ of $P_2(\mathbb{R})$. + +The 3rd vector is not a linear combination of the first two. The 4th vector is a linar combination of the 1st and 3rd vectors: $(x)=(x-3)+3(1)$ +