1.8

Revision

Decide if the following statements are true or false.
1. Let $\mathbb{F}$ be a field and $a \in \mathbb{F}$. If $\exists n \in \mathbb{N}$ such that $a^n = 0$, $\big( \underbrace{a \cdot a \cdot \dots \cdot a}_{n \text{ times}} = 0 \big) \implies a = 0$
  
  True.
  - If $n = 1$, $a = 0$ trivially.
  - If $n = 2$, $a \cdot a = 0 \implies a = 0$ or $a = 0$. So $a = 0$. (Exercise from the first week.)
  - If $n = 3$, $a^2 \cdot a = 0 \implies a^2 = 0$ or $a = 0 \implies a = 0$.
  - In this way, recursively, if $a^n = 0 \implies a^{n-1} \cdot a = 0 \implies a^{n-1} = 0$ or $a = 0 \implies \dots \implies a = 0$.
    1. Let $\mathbb{F}$ be a field and $a \in \mathbb{F}$. If $n \cdot a = 0$, $\big( n \cdot a = \underbrace{a + a + \dots + a}_{n \text{ times}} \big) \implies a = 0$.
  FALSE: Let's give a counterexample.
  
  Think of $\mathbb{F} = \mathbb{Z}_p$ for $p$ prime and $1 \in \mathbb{Z}_p$$\implies p \cdot 1 = 1 + \dots + 1 = p = 0$ in $\mathbb{Z}_p$, but $1 \neq 0$ in $\mathbb{Z}_p$. 3. Let $\mathbb{F}$ be a field. If $a \in \mathbb{F} \setminus \{0\}$, $n \in \mathbb{N}$ such that $n \cdot a = 0 \implies n \cdot x = 0 \ \forall x \in \mathbb{F}$.
  
  TRUE: We have that $n \cdot a = 0$.
  
  Now, let $x \in \mathbb{F}$, we have that $n \cdot x = x + \dots + x = \big( (a a^{-1}) x + \dots + (a a^{-1}) x \big) = a \big( a^{-1} x + \dots + a^{-1} x \big)$ (Since $a \neq 0$, $a^{-1} \in \mathbb{F}$).
  
  Using associativity and distributivity: $a \big( (a^{-1}) x + \dots + (a^{-1}) x \big) = a \cdot \big( n \cdot (a^{-1} x) \big) = (n \cdot a) \cdot (a^{-1} x) = 0$.
Let $V = \mathbb{R}_{>0}$ and $\mathbb{F} = \mathbb{Q}$. Prove that $V$ is a vector space over $\mathbb{F}$ with the operations: $a \oplus b = ab, \quad \frac{n}{m} \odot a = a^{n/m} = \sqrt[m]{a^n}$.
1. Commutativity: $a \oplus b = ab = ba = b \oplus a \ \checkmark$ (The product in $\mathbb{R}_{>0}$ commutes.)
2. Associativity:
  - $(a \oplus b) \oplus c = (ab) \cdot c = a \cdot (bc) = a \oplus (b \oplus c) \ \checkmark$ (The product is associative in $\mathbb{R}_{>0}$.)
  - $\frac{n}{m} \odot \left(\frac{p}{q} \odot a\right) ?= \left(\frac{n}{m} \cdot \frac{p}{q}\right) \odot a$. Let's check:$\left(\frac{n}{m}\cdot\frac{p}{q}\right)\odot a=\frac{np}{mq}\odot a=a^{\frac{np}{mq}}=\left(a^{\frac{p}{q}}\right)^{\frac{m}{n}}=\left(\frac{p}{q}\odot a\right)^{\frac{n}{m}}=\frac{n}{m}\odot\left(\frac{p}{q}\odot a\right)$
    1. Additive Identity: We need $\tilde{0} \in V$ such that $a \oplus \tilde{0} = a \ \forall a \in V$. In this case, we have $a \oplus 1 = a \cdot 1 = a \ \forall a \in V$$\implies 1$ is the additive identity.
    2. Multiplicative Identity: Since $1$ is the multiplicative identity in $\mathbb{R}_{>0}$, we have to check that $\frac{1}{1} \odot a = a^{1/1} = a \ \forall a \in V$. This is true since $1 \odot a = a^{1/1} = a \ \checkmark$.
    3. Additive Inverses: For all $a \in V$, there exists $-a \in V$ such that $a \oplus (-a) = 1$ (where $1$ is the additive identity in $V$). Since $a \in V = \mathbb{R}_{>0}$, there exists $\frac{1}{a} \in V$ and we have that $a \oplus \frac{1}{a} = a \cdot \frac{1}{a} = 1$.
    4. Distributivity: $\frac{n}{m} \odot (a \oplus b) = \frac{n}{m} \odot (ab) = (ab)^{n/m} = a^{n/m} \cdot b^{n/m} = \frac{n}{m} \odot a \oplus \frac{n}{m} \odot b$. Hence, distributivity holds.
    5. Scalar Addition and Multiplication: $\left(\frac{n}{m} + \frac{p}{q}\right) \odot a = \left(\frac{nq + pm}{mq}\right) \odot a = a^{\frac{nq + pm}{mq}}$$=a^{\frac{n}{m}+\frac{p}{q}}=\left(a^{\frac{n}{m}}\right)\cdot\left(a^{\frac{p}{q}}\right)=\left(\frac{n}{m}\odot a\right)\cdot\left(\frac{p}{q}\odot a\right)=\left(\frac{n}{m}\odot a\right)\oplus\left(\frac{p}{q}\odot a\right)$.
  Therefore, $V$ is a vector space over $\mathbb{F}$.
Suppose $U = \{(x, x, y, y) \in \mathbb{F}^4 : x, y \in \mathbb{F}\}$. Find a subspace $W$ of $\mathbb{F}^4$ such that $\mathbb{F}^4 = U \oplus W$.

We need to be able to write any $(x, y, z, w) \in \mathbb{F}^4$ as $u + w$ for $u \in U$ and $w \in W$.

Therefore, we consider $W = \{(0, x, y, 0) \in \mathbb{F}^4 : x, y \in \mathbb{F}\}$.

Let's see that $\mathbb{F}^4 = U \oplus W$:
1. $\mathbb{F}^4 = U + W$: We have that $(x, y, z, w) = (x, x, z, w) + (0, y - x, z - w, 0)$ where $(x, x, z, w) \in U$ and $(0, y - x, z - w, 0) \in W$.
  So every $v \in \mathbb{F}^4$ can be written as $u + w$ with $u \in U$ and $w \in W$.
2. $U \cap W = \{0\}$: Let $(x, y, z, w) \in U \cap W$.
  
  This tells us Since $(x, y, z, w) \in U \implies x = y$ and $z = w$.
  Since $(x, y, z, w) \in W \implies x = 0$ and $w = 0$.
  Therefore $x = y = z = w = 0$. So $U \cap W = \{0\}$.
Find the values of $k$ for which the system $\begin{cases}2kx_1+4x_2+2kx_3=2\\ kx_1+(k+4)x_2+3x_3=-2\\ -kx_1-2x_2+x_3=1\\ (k+2)x_2+(3k+1)x_3=-1\end{cases}$

Write in matrix and row reduced it, we have$\begin{pmatrix}\begin{array}{ccc|c}0 & 0 & 2k+2 & 4\\ 0 & k+2 & 4 & -1\\ k & 2 & -1 & -1\\ 0 & 0 & 3k-3 & 0\end{array}\end{pmatrix}$

Then $k=1$ if it has solution. Then $\begin{pmatrix}\begin{array}{ccc|c} 0 & 0 & 4 & 4 \ 0 & 3 & 4 & -1 \ 1 & 2 & -1 & -1 \ 0 & 0 & 0 & 0 \end{array}\end{pmatrix} $ where the solution must be unique

The solutions is $\begin{cases}x_1=\frac{10}{3}\\ x_2=-\frac53\\ x_3=1\end{cases}$
1. No solution.
  
  $k \neq 1$ 2. An infinite amount of solutions.
  
  No way! 3. A unique solution.
  
  Yes
Suppose $\{v_1, \ldots, v_n\}$ is L.I. in $V$ and $w \in V$. Prove that $\dim(\langle v_1 + w, \ldots, v_n + w \rangle) \geq n - 1$.

We need to find $n - 1$ L.I. vectors $\{w_1, \ldots, w_{n-1}\}$ contained in $\langle v_1 + w, \ldots, v_n + w \rangle$.

$\implies \langle w_1, \ldots, w_{n-1} \rangle \subseteq \langle v_1 + w, \ldots, v_n + w \rangle$

$\implies n - 1 \leq \dim(\langle w_1, \ldots, w_{n-1} \rangle) \leq \dim(\langle v_1 + w, \ldots, v_n + w \rangle)$

Note that $v_2 - v_1 = (v_2 + w) - (v_1 + w) \in \langle v_1 + w, \ldots, v_n + w \rangle$.

In the same way, for all $i \in \{2, \ldots, n\}$, we have that $v_i - v_1 = (v_i + w) - (v_1 + w) \in \langle v_1 + w, \ldots, v_n + w \rangle$.

So $\{v_2 - v_1, v_3 - v_1, \ldots, v_n - v_1\}$ ($n - 1$ vectors) $\subseteq \langle v_1 + w, \ldots, v_n + w \rangle$.

Are they L.I.?

Suppose $a_1(v_2 - v_1) + a_2(v_3 - v_1) + \ldots + a_{n-1}(v_n - v_1) = 0$.

We want to show that $a_1 = a_2 = \ldots = a_{n-1} = 0$.

We have that $(-a_1 - a_2 - \ldots - a_{n-1})v_1 + a_1v_2 + a_2v_3 + \ldots + a_{n-1}v_n = 0$.

And since $\{v_1, \ldots, v_n\}$ is L.I., it follows that $-a_1 - a_2 - \ldots - a_{n-1} = 0$, $a_1 = 0$, $a_2 = 0$, $\ldots$, $a_{n-1} = 0$.

Therefore, $\dim(\langle v_1 + w, \ldots, v_n + w \rangle) \geq \dim(\langle v_2 - v_1, \ldots, v_n - v_1 \rangle) = n - 1$.
Consider the real vector space $M_2(\mathbb{R})$. Let $W_1=\left\{\begin{pmatrix}x & -x\\ y & z\end{pmatrix}\in M_2(\mathbb{R})\right\},W_2=\left\{\begin{pmatrix}x & y\\ -x & z\end{pmatrix}\in M_2(\mathbb{R})\right\}.$
1. Show that $W_1$ and $W_2$ are subspaces of $M_2(\mathbb{R})$.
  
  We just need to check that $A + \alpha B \in W_i$ for every $A, B \in W_i$ and $\alpha \in \mathbb{R}$.
  1. $W_1$: $\begin{pmatrix} x & -x \\ y & z \end{pmatrix} + \alpha \begin{pmatrix} a & -a \\ b & c \end{pmatrix} = \begin{pmatrix} x + \alpha a & -(x + \alpha a) \\ y + \alpha b & z + \alpha c \end{pmatrix} \in W_1$$\implies W_1$ is a subspace.
  2. $W_2$: Exercise (it’s the same as $W_1$).
    1. Find the dimension of $W_1, W_2, W_1 + W_2$, and $W_1 \cap W_2$. Give an explicit basis for each of these spaces.
  3. $W_1$: $W_1=\left\{\begin{pmatrix}x & -x\\ y & z\end{pmatrix}\in M_2(\mathbb{R})\right\}=\left\lbrace x\begin{pmatrix}1 & -1\\ 0 & 0\end{pmatrix}+y\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}+z\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix},\,x,y,z\in\mathbb{R}\right\rbrace=\langle\begin{pmatrix}1 & -1\\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}\rangle$
    
    Since these matrices are L.I., a basis for $W_1$ is $\left\{ \begin{pmatrix} 1 & -1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right\}$ and $\dim(W_1) = 3$.
    
    $W_2$: In the same way as for $W_1$, we have that $\left\{ \begin{pmatrix} 1 & 0 \\ -1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right\}$ is a basis for $W_2$ and $\dim(W_2) = 3$.
    
    $W_1 + W_2$: Note that $W_1 \subseteq W_1 + W_2 \subseteq M_2(\mathbb{R})$$\implies \dim(W_1) \leq \dim(W_1 + W_2) \leq \dim(M_2(\mathbb{R}))$. $\implies 3 \leq \dim(W_1 + W_2) \leq 4$.
    
    But note that $\begin{pmatrix} 1 & 0 \\ -1 & 0 \end{pmatrix} \in W_2 \implies \begin{pmatrix} 1 & 0 \\ -1 & 0 \end{pmatrix} \in W_1 + W_2$ but $\begin{pmatrix}1 & 0\\ -1 & 0\end{pmatrix}\notin W_1\implies\dim(W_1+W_2)\neq3$. ($W_1 \subsetneq W_1+W_2$)
    
    $\implies \dim(W_1 + W_2) = 4$ and therefore $W_1 + W_2 = M_2(\mathbb{R})$.
    
    So a basis for $W_1 + W_2$ is $\left\{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right\}$. 2. $W_1 \cap W_2$: We have that $\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2)$ $\implies \dim(W_1 \cap W_2) = 3 + 3 - 4 = 2$.
    
    Find a basis for $W_1 \cap W_2$ (let $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in W_1 \cap W_2$).

‍