Prove the statements : Vectors/Matrices

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In summary, the conversation discusses properties of multiplication over the set of $2\times 2$ matrices with real entries, $M_2(\mathbb{R})$. The properties of associativity, commutativity, and the existence of a neutral element are explored. It is proven that multiplication is associative, not commutative, and that the identity matrix serves as the neutral element. Additionally, the definitions of addition and scalar multiplication in $\mathbb{R}^n$ are provided. The conversation ends by discussing the different expressions $a^{3}+b^{3}+c^{3}-3abc$ and $(a+b+c)(a^{2} +b^{2}+c^{2}-ab
  • #1

mathmari

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Hey! :eek:

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? (Wondering)
    $$$$
  • Could you give me also a hint for this one? (Wondering)
    $$$$
    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? (Wondering)
 
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  • #2
Hey mathmari!

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

The rest looks all good to me. (Smile)
 
  • #3
Klaas van Aarsen said:
What is the definition of addition in $\mathbb R^n$? (Wondering)
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? (Wondering)
 
  • #4
Yep. All correct. (Nod)
 
  • #5
Klaas van Aarsen said:
Yep. All correct. (Nod)

Great! Thank you! (Sun)
 
  • #6
mathmari said:
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.
 

What is a vector?

A vector is a mathematical object that has both magnitude and direction. It is represented by an arrow, where the length of the arrow represents its magnitude and the direction of the arrow represents its direction.

What is a matrix?

A matrix is a rectangular array of numbers or variables arranged in rows and columns. It is used to represent data or perform mathematical operations such as addition, subtraction, and multiplication.

How do you prove that two vectors are equal?

To prove that two vectors are equal, you need to show that they have the same magnitude and direction. This can be done by comparing the components of the vectors, or by using geometric methods such as vector addition and subtraction.

What is the dot product of two vectors?

The dot product of two vectors is a scalar quantity that represents the product of their magnitudes and the cosine of the angle between them. It is used to calculate the angle between two vectors, determine if they are perpendicular, and perform projections.

How do you prove that two matrices are equal?

To prove that two matrices are equal, you need to show that they have the same dimensions and that each corresponding element is equal. This can be done by comparing each element of the matrices or by using algebraic methods such as matrix multiplication and addition.

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