# [SOLVED]Prove the statements : Vectors/Matrices

#### mathmari

##### Well-known member
MHB Site Helper
Hey!!

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*}
1. Show that the multiplication over $M_2(\mathbb{R})$ is associative.
2. Is the multiplication over $M_2(\mathbb{R})$ commutative?
3. Is there a neutral element in respect of the multiplication over $M_2(\mathbb{R})$ ?

I have done the following:

• How can we prove this property?

• Could you give me also a hint for this one?

1. 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.
2. 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.
3. 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*}
Is what I have done correct and complete?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey mathmari !!

What is the definition of addition in $\mathbb R^n$?
It should be defined in terms of addition on $\mathbb R$.
And of scalar multiplication?

The rest looks all good to me.

#### mathmari

##### Well-known member
MHB Site Helper
What is the definition of addition in $\mathbb R^n$?
It should be defined in terms of addition on $\mathbb R$.
And of scalar multiplication?
We define addition in $\mathbb{R}^n$ componentwise: $x+y=(x_1, \ldots , x_n)+(y_1, \ldots , y_n)=(x_1+y_1, \ldots , x_n+y_n)$.
We have that $0_{\mathbb{R}^n}=(0,\ldots , 0)$.

So we have that \begin{equation*} v+0_{\mathbb{R}^n}=(v_1, \ldots , v_n)+(0,\ldots , 0)=(v_1+0, \ldots , v_n+0)=(v_1, \ldots , v_n)=v\end{equation*}
Similarily, we get \begin{equation*}0_{\mathbb{R}^n}+v=(0,\ldots , 0)+(v_1, \ldots , v_n)=(0+v_1, \ldots , 0+v_n)=(v_1, \ldots , v_n)=v\end{equation*}

Therefore, it holds that $v+0_{\mathbb{R}^n}=v=0_{\mathbb{R}^n}+v$.

The scalar multiplication is defined as follows:
$\lambda x=\lambda (x_1, \ldots , x_n)=(\lambda x_1, \ldots , \lambda x_n)$.

Then we get \begin{align*}\lambda (v+w)&=\lambda \left ((v_1, \ldots , v_n)+(w_1, \ldots , w_n)\right )=\lambda \left ((v_1+w_1, \ldots , v_n+w_n)\right ) \\ & =(\lambda(v_1+w_1), \ldots , \lambda(v_n+w_n))=(\lambda v_1+\lambda w_1, \ldots , \lambda v_n+\lambda w_n) \\ & =(\lambda v_1, \ldots , \lambda v_n)+(\lambda w_1, \ldots , \lambda w_n) =\lambda( v_1, \ldots , v_n)+ \lambda (w_1, \ldots , w_n) \\ & =\lambda v+\lambda w\end{align*}

Are these proofs correct and complete?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Yep. All correct.

MHB Site Helper

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
We see that $A\cdot B\neq B\cdot A$
For which values of $a,\ldots,h$? For all? For some? Then for which exactly? Expressions $a^{3}+b^{3}+c^{3}-3abc$ and $(a+b+c)(a^{2} +b^{2}+c^{2}-ab-bc-ca)$ also look quite differently, yet they are equal.