Skip to main content

Problem 46: Prove that the set of all solutions to the DE $y\text{'}+9y=0$ is a vector space

Problem 46: Prove that the set of all solutions to the DE $y\text{'}+9y=0$ is a vector space

Solution: Given $\V = \{y:y'+9y=0\}$

I. Properties under addition
i. Closure Property
Let $u,v \in \V$ then $u'+9u=0$ and $v'+9v=0$ \begin{align*} u+v = (u+v)'+9(u+v) \\ = (u'+v')+ 9(u+v)\\ = u'+9u+v'+9v\\ = 0+0=0\\ \therefore u+v \in \V. \end{align*} $\Rightarrow \V$ is closed under addition.

ii. Associativity:
Let $u,v,w \in \V$ then $u'+9u=0$, $v'+9v=0$ and $w'+9w=0$ Now, \begin{align*} \left[u+v\right]+w = \left[(u'+9u)+(v'+9v)\right]+(w'+9w)\\ = \left[(u'+9u+v'+9v)\right]+(w'+9w)\\ = \left[(u'+9u+v'+9v+w'+9w)\right]\\ = \left[(u'+9u)+(v'+9v+w'+9w)\right]\\ = (u'+9u)+\left[(v'+9v+w'+9w)\right]\\ = (u'+9u)+\left[(v'+9v)+(w'+9w)\right]\\ = u+\left[(v+w\right]\\ \end{align*} $\Rightarrow$ Addition is associative in $\V$.

iii. Existence of additive identity:
For all $u \in \V, \exists 0 \in \V$
Now \begin{align*} u+0=(u'+9u)+0 = (u'+9u) = u\\ 0+u=0+(u'+9u) = (u'+9u) = u\\ \end{align*} $\Rightarrow 0$ is the additive identity in $\V$.

iv. Existence of additive inverse:
Let $u \in \V$, be any element $-u\in \V$
Now \begin{align*} u+-u= (u'+9u)+ (-u'-9u)=0\\ -u+u= (-u'-9u)+(u'+9u)=0\\ \end{align*} $\Rightarrow -u$ is the additive inverse of $u$ for each $u \in \V$.

v. Commutative Law:
Let $ u,v \in \V$, Now \begin{align*} u+v= (u'+9u)+(v'+9v)\\ = (v'+9v)+(u'+9u) = v+u \end{align*} $\Rightarrow$ addition is commutative in $\V$.

II. Properties under Scalar Multiplication
vi. Closure under Scalar Multiplication
Let $\alpha\in \R$ and $u \in \V$ \begin{align*} \alpha u = (\alpha u)'+9(\alpha u)\\ = \alpha u' + 9\alpha u\\ = \alpha (u'+9u) = \alpha \cdot 0 = 0\\ \therefore \alpha u \in \V \end{align*} $\Rightarrow \V$ is closed under Scalar Multiplication.

vii.Distributivity of Addition of Real Numbers
Let $\alpha, \beta\in \R$ and $u \in \V$ \begin{align*} (\alpha+\beta)u= ((\alpha+\beta) u)'+9(\alpha+\beta) u)\\ = \alpha u'+\beta u' + 9 \alpha u + 9 \beta u\\ = \alpha u' + 9 \alpha u + \beta u' + 9 \beta u\\ = \alpha (u'+9u) + \beta (u'+9u)\\ = \alpha u + \beta u \end{align*} $\Rightarrow (\alpha+\beta)u = \alpha u+\beta u$.
$\Rightarrow $ Distributivity of Addition of Real Numbers does exists.

viii. Distributivity of Addition of Vectors
Let $\alpha\in \R$ and $u, v \in \V$ \begin{align*} \alpha\left[u+v\right]= \alpha\left[(u'+9u)+(v'+9v)\right]\\ = \left[\alpha((u'+9u)+(v'+9v))\right]\\ = \left[\alpha(u'+9u)+\alpha(v'+9v)\right]\\ = \left[\alpha(u'+9u)+\alpha(v'+9v)\right]\\ =\alpha(u'+9u)+\alpha(v'+9v)\\ =\alpha u+\alpha v\\ \end{align*} $\Rightarrow $ Distributivity of Addition of Vectors exists.

ix.Associativity of multiplication
Let $\alpha, \beta\in \R$ and $u\in \V$ \begin{align*} (\alpha\beta)u= (\alpha\beta)(u'+9u)\\ = \alpha(\beta(u'+9u)\\ =\alpha(\beta u) \end{align*} $\Rightarrow$ Associativity of multiplication exists.

x. Unity element multiplication
Let 1 be unity element of field $\F$ and $u\in \V$
Now \begin{align*} 1.u= 1\cdot(u'+9u)\\ =(u'+9u)\\ = u \end{align*} $\Rightarrow 1.u = u \forall u \in \V$.

Hence, $\V$ is a vector space over $\R$.

Labels:

Comments

Post a Comment

Popular posts from this blog

Problem 98: If 6 of 18 new buildings in a city violate the building code, what is the probability that a building inspector, who randomly selects 4 of the new buildings for inspection, will catch
a. none of the buildings that violate the building code?
b. 1 of the new buildings that violate the building code?
c. 2 of the new buildings that violate the building code?
d. at least 3 of the new buildings that violate the building code?

Problem 98: If 6 of 18 new buildings in a city violate the building code, what is the probability that a building inspector, who randomly selects 4 of the new buildings for inspection, will catch a. none of the buildings that violate the building code? b. 1 of the new buildings that violate the building code? c. 2 of the new buildings that violate the building code? d. at least 3 of the new buildings that violate the building code? Solution: a. none of the buildings that violate the building code? Given that $x=0,n=4,a=6, N=18$ Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$ $P(x=0) = h(0;4,6,18)=\frac{\left(\begin{array}{c}6\\ 0\end{array}\right)\left(\begin{array}{c}18-6\\ 4-0\end{array}\right)}{\left(\begin{array}{c}18\\ 4\end{array}\right)} = 0.16176.$ b. 1 of the new buildings that violate the building code? Given ...
Post a Comment
Read more

Problem 105: If a bank receives on the average α = 6 bad checks per day, what are the probabilities that it will receive (a) 4 bad checks on any given day?
(b) 10 bad checks over any 2 consecutive days?

Problem 105: If a bank receives on the average α = 6 bad checks per day, what are the probabilities that it will receive (a) 4 bad checks on any given day? (b) 10 bad checks over any 2 consecutive days? Solution: Poisson distribution, $f(x;\lambda )=\frac{{\lambda }^{x}{e}^{-\lambda }}{x!}$ a. 4 bad checks on any given day Given that $x=4, \lambda = \alphaT = 6\dot 1 = 6$ $f(4;6 )=\frac{{6 }^{4}{e}^{-6 }}{4!} = 0.134$ b. 10 bad checks over any 2 consecutive days Given that $x=10, \lambda = \alphaT = 6\dot 2 = 12$ $f(10;12 )=\frac{{12 }^{10}{e}^{-12 }}{10!} = 0.104837$
Post a Comment
Read more

Problem 56: Let $S = \{x \in \R^3 : x = (r-2s,3r+s,s), r,s \in \R \}$. Establish that $S$ is a subspace of $\R^3$.

Problem 56: Let $S = \{x \in \R^3 : x = (r-2s,3r+s,s), r,s \in \R \}$. Establish that $S$ is a subspace of $\R^3$. Solution: $S = \{x \in \R^3 : x = (r-2s,3r+s,s), r,s \in \R \}$. S is certainly nonempty. Let $x, y \in S$. Then for some $r, s, u,v \in \R$, $x = (r-2s,3r+s,s)$ and $y = (u-2v,3u+v,v)$. Hence, $x + y = (r-2s,3r+s,s) + (u-2v,3u+v,v)$ $= (r-2s+u-2v,3r+s+3u+v,s+v) = ((r+u)-2(s+v),3(r+u)+(s+v),(s+v))$ $= (k-2l,3k+l,l)$, where $k = r + u, l=s+v$. Consequently, $S$ is closed under addition. Further, if $c \in \R$, then $cx = c(r-2s,3r+s,s) = (c(r-2s),c(3r+s),cs) $ $= (cr-2cs,3cr+cs,cs) = (a-2b,3a+b,b)$, where $a = cr,b=cs$. Therefore $S$ is also closed under scalar multiplication. It follows that $S$ is a subspace of $\R^3$.
Post a Comment
Read more