- #1

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Let $1\leq n\in \mathbb{N}$.

- Prove that for all $v\in \mathbb{R}^n$ it holds that $v+0_{\mathbb{R}^n}=v=0_{\mathbb{R}^n}+v$.
- Prove that for all $\lambda\in \mathbb{R}$ and $v,w\in \mathbb{R}$ it holds that $\lambda (v+w)=\lambda v+\lambda w$.
- Let $M_2(\mathbb{R}):=\left \{\begin{pmatrix}a & b \\ c & d\end{pmatrix}\mid a, b, c, d\in \mathbb{R}\right \}$ the set of all $2\times 2$-matrices over $\mathbb{R}$. We define also the multiplication on that set as \begin{equation*}\begin{pmatrix}a & b \\ c & d\end{pmatrix}\cdot \begin{pmatrix}a' & b' \\ c' & d'\end{pmatrix}=\begin{pmatrix}aa'+bc' & ab'+bd' \\ ca'+dc' & cb'+dd'\end{pmatrix}\end{equation*}
- Show that the multiplication over $M_2(\mathbb{R})$ is associative.
- Is the multiplication over $M_2(\mathbb{R})$ commutative?
- Is there a neutral element in respect of the multiplication over $M_2(\mathbb{R})$ ?

- How can we prove this property? (Wondering)

$$$$ - Could you give me also a hint for this one? (Wondering)

$$$$ -
- Let $A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}, \ B=\begin{pmatrix}e & f \\ g & h\end{pmatrix}, \ C=\begin{pmatrix}i & j \\ k & \ell\end{pmatrix}$.

Then we have the following: \begin{align*}(A\cdot B)\cdot C&=\left (\begin{pmatrix}a & b \\ c & d\end{pmatrix}\cdot \begin{pmatrix}e & f \\ g & h\end{pmatrix}\right )\cdot \begin{pmatrix}i & j \\ k & \ell\end{pmatrix}= \begin{pmatrix}ae+bg & af+bh \\ ce+dg & cf+dh\end{pmatrix}\cdot \begin{pmatrix}i & j \\ k & \ell\end{pmatrix}\\ & = \begin{pmatrix}(ae+bg)i+(af+bh)k & (ae+bg)j+(af+bh)\ell \\ (ce+dg)i+(cf+dh)k & (ce+dg)j+(cf+dh)\ell\end{pmatrix}\\ & = \begin{pmatrix}aei+bgi+afk+bhk & aej+bgj+af\ell+bh\ell \\ cei+dgi+cfk+dhk & cej+dgj+cf\ell+dh\ell\end{pmatrix}\end{align*}

\begin{align*}A\cdot (B\cdot C)&=\begin{pmatrix}a & b \\ c & d\end{pmatrix}\cdot\left ( \begin{pmatrix}e & f \\ g & h\end{pmatrix}\cdot \begin{pmatrix}i & j \\ k & \ell\end{pmatrix}\right )= \begin{pmatrix}a & b \\ c & d\end{pmatrix}\cdot\begin{pmatrix}ei+fk & ej+f\ell \\ gi+hk & gj+h\ell\end{pmatrix} \\ & = \begin{pmatrix}a(ei+fk)+b(gi+hk) & a(ej+f\ell)+b(gj+h\ell) \\ c(ei+fk)+d(gi+hk) & c(ej+f\ell)+d(gj+h\ell)\end{pmatrix} \\ & = \begin{pmatrix}aei+afk+bgi+bhk & aej+af\ell+bgj+bh\ell \\ cei+cfk+dgi+dhk & cej+cf\ell+dgj+dh\ell\end{pmatrix}\\ & = \begin{pmatrix}aei+bgi+afk+bhk & aej+bgj+af\ell+bh\ell \\ cei+dgi+cfk+dhk & cej+dgj+cf\ell+dh\ell\end{pmatrix}\end{align*}

The results are the same. Therefore it holds that $(A\cdot B)\cdot C=A\cdot (B\cdot C)$ which means the multiplication over $M_2(\mathbb{R})$ is associative.

- Let $A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}, \ B=\begin{pmatrix}e & f \\ g & h\end{pmatrix}$.

Then we have the following: \begin{equation*}A\cdot B=\begin{pmatrix}a & b \\ c & d\end{pmatrix}\cdot \begin{pmatrix}e & f \\ g & h\end{pmatrix}=\begin{pmatrix}ae+bg & af+bh \\ ce+dg & cf+dh\end{pmatrix} \end{equation*}

Then we have the following: \begin{equation*}B\cdot A=\begin{pmatrix}e & f \\ g & h\end{pmatrix}\cdot \begin{pmatrix}a & b \\ c & d\end{pmatrix}=\begin{pmatrix}ea+fc & eb+fd \\ ga+hc & gb+hd\end{pmatrix} \end{equation*}

We see that $A\cdot B\neq B\cdot A$, which means that the multiplication over $M_2(\mathbb{R})$ is not commutative.

- The neutral element in respect of the multiplication over $M_2(\mathbb{R})$ is the identity matrix \begin{equation*}I_2=\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\end{equation*}

- Let $A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}, \ B=\begin{pmatrix}e & f \\ g & h\end{pmatrix}, \ C=\begin{pmatrix}i & j \\ k & \ell\end{pmatrix}$.