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Homework 2.pdf

Algebra A - Winter 2024

Homework 2

90

(40 pts) Check whether the following sets $V$ with the defined operations are vector spaces over $F$ or not.
If they are vector spaces, prove the properties. If not, what property fails?

(a) The set $V = \{f : \mathbb{R} \rightarrow \mathbb{R} : f \text{ is an even function} \}$ and $F = \mathbb{R}$ with the addition $(f + g)(x) = f(x) + g(x)$ and the scalar multiplication $(af)(x) = af(x)$.

(Remember that an even function is a function $f : \mathbb{R} \rightarrow \mathbb{R}$ which satisfies $f(t) = f(-t)$ for all $t \in \mathbb{R}$).

Proof

Let's check the property

First, we need to check the sum and scalar is still in the vector space

$(f+g)(x)=f(x)+g(x)=f(-x)+g(-x)=(f+g)(-x)$ $\checkmark$

$(af)(x)=af(x)=af(-x)=(af)(-x)$ $\checkmark$
1. The sum of associativity and commutativity $\checkmark$
  
  $\left(f+\left(g+h\right)\right)(x)=f(x)+\left(g+h\right)(x)=f\left(x\right)+g\left(x\right)+h\left(x\right)$
  
  $\left(\left(f+g)+h\right)\right)(x)=\left(f+g)(x)+h(x\right)=f\left(x\right)+g\left(x\right)+h\left(x\right)$
  
  Thus $\left(f+\left(g+h\right)\right)(x)=\left(\left(f+g\right)+h\right)(x)$
  
  $(f + g)(x) = f(x) + g(x)$, $(g + f)(x) = g(x) + f(x)=f(x)+g(x)$
  
  Thus $(f+g)(x)=(g+f)(x)$ 2. Neutral element $\checkmark$
  
  Let $g(x)=0,\forall x\in R$
  
  $(f+g)(x)=f(x)+g(x)=f(x)$ 3. For every $v\in V,\exists w\in V$ such that $v+w=0$ $\checkmark$
  
  $\forall f(x)\in V,\exists g(x)\in V,\left(f+g\right)\left(x\right)=f(x)+g(x)=0$. Then $g(x)=-f(x)$ 4. $1\cdot v=v,\forall v\in V$ $\checkmark$
  
  Since $1\in \R$, then $\forall f(x)\in V,\left(1\cdot f\right)(x)=1\cdot f(x)=f\left(x\right)$ 5. $\lambda\cdot(v+w)=\lambda\cdot v+\lambda\cdot w,\forall\lambda\in\mathbb{F},\forall v,w\in V$ $\checkmark$
  
  $\left(\lambda\left(f+g)\right)(x\right)=\lambda(f+g)(x)=\lambda\left(f(x)+g(x)\right)=\lambda f(x)+\lambda g(x)=\left(\lambda f\right)(x)+\left(\lambda g\right)(x)$ 6. $\left(\lambda+\mu\right)\cdot v=\lambda\cdot v+\mu\cdot v,\forall\lambda,\mu\in\mathbb{F},\forall v\in V$ $\checkmark$
  
  $\left(\left(\lambda+\mu\right)\cdot f\right)\left(x\right)=\left(\lambda+\mu\right)\cdot f\left(x\right)=\lambda\cdot f\left(x\right)+\mu\cdot f\left(x\right)$ 7. $\lambda\cdot\left(\mu\cdot v\right)=\left(\lambda\cdot\mu\right)\cdot v,\forall\lambda,\mu\in\mathbb{F},\forall v\in V$ $\checkmark$
  
  $\left(\lambda\cdot\left(\mu\cdot f)\left(x\right)\right)=\lambda\cdot\left(\mu\cdot f\right)\left(x\right)=\lambda\cdot\left(\mu\cdot f\left(x\right)\right)=\left(\lambda\cdot\mu\right)\right.\cdot f\left(x\right)=\left(\left(\lambda\cdot\mu\right)\cdot f\right)\left(x\right)$
Thus it is a vector space

(b) The set $ V={A\in M_n(\R):A\text{ is invertible}}$ and $F = \mathbb{R}$ with the natural addition $A+B$ of matrices and the natural scalar multiplication $c \cdot A$ (i.e., coordinate by coordinate).

Proof

Let's check the property

First, we need to check the sum and scalar is still in the vector space

Let $A,B\in M_{n}(\mathbb{R})\text{ and are invertible}$

But $A+B$ may equal to $0$ when $A=-B$ No!

Also, let $a\in\R$, then $a\cdot A$ may equal to $0$ when $a=0$ No!

Thus it is not vector space since $+:V\times V\rightarrow V$ and $\cdot :$ $\mathbb{F}\times V\rightarrow V$ is not satisfied

(c) Let $V = \mathbb{R} \cup \{-\infty\} \cup \{\infty\}$ and $F = \mathbb{R}$. We consider the addition and multiplication of two real numbers as usual and for $x \in \mathbb{R}$ we define:

$x + \infty = \infty + x = \infty, x + (-\infty) = (-\infty) + x = -\infty,$

$\infty + \infty = \infty, -\infty + (-\infty) = -\infty, \infty + (-\infty) = 0,$

and $x \cdot \infty = \begin{cases} -\infty & \text{if } x < 0, \\ 0 & \text{if } x = 0, \\ \infty & \text{if } x > 0, \end{cases}$ $x \cdot (-\infty) = \begin{cases} \infty & \text{if } x < 0, \\ 0 & \text{if } x = 0, \\ -\infty & \text{if } x > 0. \end{cases}$

Proof

Let's check the property

First, we need to check the sum and scalar is still in the vector space

Since we know $\R$ is a vector space over $\R$, then we only need to check when we choose $-\infty~and~\infty$ in vector space

For sum, for $x\in \R$,

$x + \infty = \infty + x = \infty, x + (-\infty) = (-\infty) + x = -\infty,$

$\infty + \infty = \infty, -\infty + (-\infty) = -\infty, \infty + (-\infty) = 0,$

Thus it is still in vector space

Then since $x \cdot \infty = \begin{cases} -\infty & \text{if } x < 0, \\ 0 & \text{if } x = 0, \\ \infty & \text{if } x > 0, \end{cases}$ $x \cdot (-\infty) = \begin{cases} \infty & \text{if } x < 0, \\ 0 & \text{if } x = 0, \\ -\infty & \text{if } x > 0. \end{cases}$ is also in the vector space

Thus the scalar is satisfied
1. The sum of associativity and commutativity
  
  $x + \infty = \infty + x = \infty, x + (-\infty) = (-\infty) + x = -\infty,$ $\infty+\infty=\infty+\infty,-\infty+(-\infty)=-\infty+(-\infty),\infty+(-\infty)=(-\infty)+\infty,$
  
  Then $\left(\infty+\infty\right)+\infty=\infty+\infty+\infty=\infty+\left(\infty+\infty\right)=\infty$ and $\left(-\infty+\left(-\infty\right)\right)+\left(-\infty\right)=-\infty+\left(-\infty\right)+\left(-\infty\right)=-\infty+\left(-\infty+\left(-\infty\right)\right)=-\infty$
  
  $\left(x+\infty)+\infty=x+\infty+\infty=x+\left(\infty\right.+\infty\right)=\infty$ and $\left(x+(-\infty))+(-\infty)=x+(-\infty)+(-\infty)=x+\left(-\infty\right.+(-\infty)\right)=-\infty$
  
  $\left(x+\infty)+\left(-\infty\right)=x+\infty+\left(-\infty\right)=x+\left(\infty\right.+\left(-\infty\right)\right)$ but the left side is 0 and the right side is $x$
  
  Not equal! Thus assocaitivity is not satisfied. 2. Neutral element $\checkmark$
  
  $x+0=x$ and $\infty+0=\infty$ and $-\infty+0=-\infty$ 3. For every $v\in V,\exists w\in V$ such that $v+w=0$ $\checkmark$
  
  $x+(-x)=0$ and $\infty+(-\infty)=0$ 4. $1\cdot v=v,\forall v\in V$ $\checkmark$
  
  $1\cdot x=x$ and $1\cdot \infty=\infty$ and $1\cdot (-\infty)=-\infty$ 5. $\lambda\cdot(v+w)=\lambda\cdot v+\lambda\cdot w,\forall\lambda\in\mathbb{F},\forall v,w\in V$ $\checkmark$
  
  $\lambda\cdot(x+\infty)=\lambda\cdot x+\lambda\cdot\infty=\infty$ and $\lambda\cdot(x+(-\infty))=\lambda\cdot x+\lambda\cdot(-\infty)=-\infty$ and $\lambda\cdot(\infty+\infty)=\lambda\cdot\infty+\lambda\cdot\infty=\infty$ and $\lambda\cdot(-\infty+(-\infty))=\lambda\cdot(-\infty)+\lambda\cdot(-\infty)=-\infty$ and $\lambda\cdot(\infty+(-\infty))=\lambda\cdot\infty+\lambda\cdot(-\infty)=0$ 6. $\left(\lambda+\mu\right)\cdot v=\lambda\cdot v+\mu\cdot v,\forall\lambda,\mu\in\mathbb{F},\forall v\in V$
  
  $(\lambda+\mu)\cdot\infty=\lambda\cdot\infty+\mu\cdot\infty=\infty$
  
  But left side can be $0$ when $\lambda+\mu=0$, then the right side is still $\infty$, then they are not equal 7. $\lambda\cdot\left(\mu\cdot v\right)=\left(\lambda\cdot\mu\right)\cdot v,\forall\lambda,\mu\in\mathbb{F},\forall v\in V$ $\checkmark$
  
  $\lambda\cdot\left(\mu\cdot\infty)=\lambda\cdot\mu\cdot\infty=\left(\lambda\cdot\mu\right)\right.\cdot\infty=\infty$ and $\lambda\cdot\left(\mu\cdot\left(-\infty\right))=\lambda\cdot\mu\cdot\left(-\infty\right)=\left(\lambda\cdot\mu\right)\right.\cdot\left(-\infty\right)=-\infty$
Thus it is not a vector space since the associativity (a) and (f) distributivity are not satisfied
(10 pts) Is $\mathbb{R}$ a vector space over $\mathbb{C}$? Is $\mathbb{C}$ a vector space over $\mathbb{R}$? Explain.

First, Field is $\mathbb{C}$ and vector space is $\R$. We need to check $+:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ and $\cdot :$ $\mathbb{C}\times\mathbb{R}\rightarrow\mathbb{R}$ first.

$+:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ is true

But $\cdot :$ $\mathbb{C}\times\mathbb{R}\rightarrow\mathbb{R}$ can not hold since complex number $\times$ real number may be complex number, such as $-i\times 1=-i$

Thus $\mathbb{R}$ is not a vector space over $\mathbb{C}$

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Then we check another one (Field is $\mathbb{R}$ and vector space is $\mathbb{C}$.) We need to check $+:\mathbb{C}\times\mathbb{C}\rightarrow\mathbb{C}$ and $\cdot :$ $\mathbb{R}\times\mathbb{C}\rightarrow\mathbb{C}$ first.

$+:\mathbb{C}\times\mathbb{C}\rightarrow\mathbb{C}$ is true and $\cdot :$ $\mathbb{R}\times\mathbb{C}\rightarrow\mathbb{C}$ is also true $\checkmark$

Then we check the property of vector space
1. The sum of associativity and commutativity $\checkmark$
  
  This is true because the property of complex number 2. Neutral element $\checkmark$
  
  Yes, it is $0$ and $0\in \mathbb{C}$ 3. For every $v\in V,\exists w\in V$ such that $v+w=0$ $\checkmark$
  
  Yes, there exists additive inverse in $\mathbb{C}$ and just let $v=-w$ 4. $1\cdot v=v,\forall v\in V$ $\checkmark$
  
  Yes, $1$ is in the complex number and satisfies $1\cdot v=v$ 5. $\lambda\cdot(v+w)=\lambda\cdot v+\lambda\cdot w,\forall\lambda\in\mathbb{F},\forall v,w\in V$ $\checkmark$
  
  Yes, since the fundamental property of complex number 6. $\left(\lambda+\mu\right)\cdot v=\lambda\cdot v+\mu\cdot v,\forall\lambda,\mu\in\mathbb{F},\forall v\in V$
  
  Yes, since the fundamental property of complex number 7. $\lambda\cdot\left(\mu\cdot v\right)=\left(\lambda\cdot\mu\right)\cdot v,\forall\lambda,\mu\in\mathbb{F},\forall v\in V$ $\checkmark$
  
  Yes, since the fundamental property of complex number
Thus $\mathbb{C}$ is a vector space over $\mathbb{R}$
(30 pts) Consider the real vector space $M_2(\mathbb{R})$ with the usual operations.

(a) Is it possible to write the zero matrix $0 \in M_2(\mathbb{R})$ in the form $0 = a \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} + b \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} + c \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, with $a, b, c \in \mathbb{R}$ not all zero?

Let $0=\begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}=a\begin{pmatrix}1 & 0\\ -1 & 1\end{pmatrix}+b\begin{pmatrix}1 & 1\\ 2 & 1\end{pmatrix}+c\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}$

Then we have four equations $\begin{cases}0=a+b\\ 0=b+c\\ 0=-a+2b+c\\ 0=a+b\end{cases}\Rightarrow\begin{cases}-a=b\\ -c=b\\ 0=2b\\ -a=b\end{cases}\Rightarrow\begin{cases}a=0\\ c=0\\ b=0\\ a=0\end{cases}$

Thus it is impossible.

(b) Is it possible to write any matrix $A \in M_2(\mathbb{R})$ uniquely in the form $A = a \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} + b \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} + c \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, with $a, b, c \in \mathbb{R}$?

Suppose we can write it in different ways, then we have

$A=\tilde{a}\begin{pmatrix}1 & 0\\ -1 & 1\end{pmatrix}+\tilde{b}\begin{pmatrix}1 & 1\\ 2 & 1\end{pmatrix}+\tilde{c}\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}$

Then we minus the original equation: $0=\begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}=\left(a-\tilde{a}\right)\begin{pmatrix}1 & 0\\ -1 & 1\end{pmatrix}+\left(b-\tilde{b}\right)\begin{pmatrix}1 & 1\\ 2 & 1\end{pmatrix}+\left(c-\tilde{c}\right)\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}$

Since we have known the coefficients must be all zero, thus $a=\tilde{a},b=\tilde{b},c=\tilde{c}$. Contradiction!

Thus it is unique

Now, we need to prove existence

Let $A=\begin{pmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix}$, then$\begin{cases}a_{11}=a+b\\ a_{12}=b+c\\ a_{21}=-a+2b+c\\ a_{22}=a+b\end{cases}\Rightarrow a_{11}=a_{22}$

However, if $a_{11}\neq a_{22}$, we cannot express it.

Thus it is impossible
(20 pts) Consider the real vector space $\mathbb{R}[x]$ with the usual operations. Is $q(x) = 3x^3 - 2x^2 + 7x - 8$ a linear combination of $r(x) = x^3 - 2x^2 - 5x - 3$ and $s(x) = 3x^3 - 5x^2 - 4x - 9$?

Suppose it is, then we have $q\left(x\right)=r\left(x\right)+s\left(x\right)\Rightarrow3x^3-2x^2+7x-8=\lambda\left(x^3-2x^2-5x-3\right)+\mu\left(3x^3-5x^2-4x-9\right)$

Then $\begin{cases}3=\lambda+3\mu\\ -2=-2\lambda-5\mu\\ 7=-5\lambda-4\mu\\ -8=-3\lambda-9\mu\end{cases}$, but the first equation contradicts to the fourth equation.

Because the first equation: $3=\lambda+3\mu\Rightarrow9=3\lambda+9\mu$

The fourth equation: $-8=-3\lambda-9\mu$

If we add them, we get $9=-8$ which is a contradiction!

Thus $q(x)$ is not a linear combination of $r(x)$ and $s(x)$

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