Proof

It is the definition of the coordinates of a vector, combined with the fact that both its abscissa and ordinate are positive.

Proof

It is the definition of the coordinates of a vector, combined with the fact that its abscissa is negative and its ordinate is positive.

Proof

It is the definition of the coordinates of a vector, combined with the fact that both its abscissa and ordinate are negative.

Proof

It is the definition of the coordinates of a vector, combined with the fact that its abscissa is positive and its ordinate is negative.

Proof

It is the definition of the coordinates of a vector aligned with the \( x\) axis, combined with the fact that its abscissa is positive.

Proof

It is the definition of the coordinates of a vector aligned with the \( y\) axis, combined with the fact that its ordinate is positive.

Proof

It is the definition of the coordinates of a vector aligned with the \( x\) axis, combined with the fact that its abscissa is negative.

Proof

It is the definition of the coordinates of a vector aligned with the \( y\) axis, combined with the fact that its ordinate is negative.

Proof

It is the definition of the coordinates of the null vector.

Proof

Assume \( (x_1,y_1,x_2,y_2,x_3,y_3)\in\mathbb{R}^6\) are real numbers, and consider the three column vectors with 2 elements, \( \overrightarrow{u}_1=\begin{bmatrix}x_1\\ y_1\end{bmatrix}\) , \( \overrightarrow{u}_2=\begin{bmatrix}x_2\\ y_2\end{bmatrix}\) and \( \overrightarrow{u}_3=\begin{bmatrix}x_3\\ y_3\end{bmatrix}\) .

Commutativity The addition of vectors is commutative because it is so for the addition of real numbers.

Indeed:

Associativity The addition of vectors is associative because it is so for the addition of real numbers.

Indeed:

Neutral Element The null vector \( \overrightarrow{0}=\begin{bmatrix}0\\ 0\end{bmatrix}\) is neutral for the addition of vectors because \( 0\) is neutral for the addition of real numbers.

Indeed:

and

Opposite of a Vector The opposite \( -\overrightarrow{u_{1}}=\begin{bmatrix}-x_{1}\\ {-y_{1}}\end{bmatrix}\) of \( \overrightarrow{u_{1}}\) is its reciprocal for the addition of vectors because the opposite of a real number is its reciprocal for the addition of real numbers.

Indeed:

and

Proof

Assume \( (x,y)\in\mathbb{R}^2\) are real numbers, and consider the column vector \( \overrightarrow{u}=\begin{bmatrix}x\\ y\end{bmatrix}\) and its opposite \( -\overrightarrow{u}=\begin{bmatrix}-x\\ {-y}\end{bmatrix}\) .

Then \( -\overrightarrow{u}=\begin{bmatrix}-x\\ -y\end{bmatrix} =\begin{bmatrix}(-1)x\\ (-1)y\end{bmatrix}\) , so that the abscissa and ordinate of \( -\overrightarrow{u}\) are proportionnal to the abscissa and ordinate of \( \overrightarrow{u}\) .

Consequently, \( -\overrightarrow{u}\) is aligned with \( \overrightarrow{u}\) .

Moreover, the abscissa and ordinate of \( -\overrightarrow{u}\) have the oppposite signs than the abscissa and ordinate of \( \overrightarrow{u}\) respectively.

Consequently, \( -\overrightarrow{u}\) is in the direction opposite to the direction of \( \overrightarrow{u}\) .

Endly, \( \left\|-\overrightarrow{u}\right\| =\sqrt{(-x)^{2}+(-y)^{2}}=\sqrt{x^{2}+y^{2}} =\left\|\overrightarrow{u}\right\|\) .

Proof (of the last assertion)

Assume \( (x,y,z,t)\in\mathbb{R}^4\) are real numbers, and consider the two column vectors with 2 elements, \( \overrightarrow{u}=\begin{bmatrix}x\\ y\end{bmatrix}\) and \( \overrightarrow{v}=\begin{bmatrix}z\\ t\end{bmatrix}\) .

Consider the vector \( \overrightarrow{w}=\begin{bmatrix}r\\ s\end{bmatrix}\) built as on the figure 12.

Let’s prove geometrically that \( r=x+z\) , considering different cases of signs of \( x\) and \( z\) .

The proof that \( s=y+t\) is made similarly, with the \( y\) axis instead of the \( x\) axis.

The different cases are the followings:

1. \( x\) or \( z\) equal to \( 0\) .

2. \( x\) and \( z\) positive.

3. \( x\) and \( z\) of different signs.

4. \( x\) and \( z\) negative.

Let’s make the proof.

1. Assume that \( x\) or \( z\) is equal to \( 0\) .

   As the construction of the figure 12 is commutative, we may assume that \( x=0\) .

1. Assume that \( z=0\) .

   

   If, starting from the origin, we draw end-to-beginning the vertical vectors \( \overrightarrow{u}=\begin{bmatrix}0\\ y\end{bmatrix}\) and \( \overrightarrow{v}=\begin{bmatrix}0\\ t\end{bmatrix}\) , then the resulting vector \( \overrightarrow{w}=\begin{bmatrix}r\\ s\end{bmatrix}\) is vertical as well.

   Consequently, \( s=0\) , that is equal to \( x+z\) .

2. Assume that \( z>0\) .

   

   Starting from the origin, we draw end-to-beginning the vertical vector

   \( \overrightarrow{u}=\begin{bmatrix}0\\ y\end{bmatrix}\) and the vector \( \overrightarrow{v}=\begin{bmatrix}z\\ t\end{bmatrix}\) , with \( z<0\) .

   Then the distances to the origin of the projections on the \( x\) axis of the ends of the vector \( \overrightarrow{v}\) and the resulting vector \( \overrightarrow{w}\) are both equal to \( z\) .

   Moreover, the projection on the \( x\) axis of the end of the vector \( \overrightarrow{w}\) is to the right of the origin, so that \( s=|s|=z\) .

   Consequently, \( s=z\) , that is equal to \( x+z\) because \( x=0\) .

3. Assume that \( z<0\) .

   

   Starting from the origin, we draw end-to-beginning the vertical vector

   \( \overrightarrow{u}=\begin{bmatrix}0\\ y\end{bmatrix}\) and the vector \( \overrightarrow{v}=\begin{bmatrix}z\\ t\end{bmatrix}\) , with \( z<0\) .

   Then the distances to the origin of the projections on the \( x\) axis of the ends of the vector \( \overrightarrow{v}\) and the resulting vector \( \overrightarrow{w}\) are both equal to \( |z|\) , so that \( |s|=|z|=-z\) .

   Moreover, the projection on the \( x\) axis of the end of the vector \( \overrightarrow{w}\) is to the left of the origin, so that \( s=-|s|=-(-z)=z\) .

   Consequently, \( s=z\) , that is equal to \( x+z\) because \( x=0\) .

2. Assume that \( x\) and \( z\) are both positive.

   

   Starting from the origin, we draw end-to-beginning the vector \( \overrightarrow{u}=\begin{bmatrix}x\\ y\end{bmatrix}\) with \( x>0\) and the vector \( \overrightarrow{v}=\begin{bmatrix}z\\ t\end{bmatrix}\) , with \( z>0\) .

   Then the distances between the projections on the \( x\) axis of the ends of the different vectors are the following.

• For the vector \( \overrightarrow{u}\) , it is \( x\) .

• For the vector \( \overrightarrow{v}\) , it is \( z\) .

• And for the vector \( \overrightarrow{w}\) , it is \( x+z\) , so that \( |s|=x+z\) .

• Moreover, the projection of the end of the vector \( \overrightarrow{w}\) on the \( x\) axis is to the right of the origin, so that \( s=|s|=x+z\) .

3. Assume that \( x\) and \( z\) are of different signs.

   As the construction of the figure 12 is commutative, we may assume that \( x>0\) and thus \( z<0\) .

1. Assume that \( x>|z|\) .

   

   Starting from the origin, we draw end-to-beginnig the vector

   \( \overrightarrow{u}=\begin{bmatrix}x\\ y\end{bmatrix}\) with \( x>0\) and the vector \( \overrightarrow{v}=\begin{bmatrix}z\\ t\end{bmatrix}\) , with \( z<0\) and \( x>|z|\) .

   Then the distances tbetween the projections on the \( x\) axis of the beginnigs and the ends of the different vectors are the following.

• For the vector \( \overrightarrow{u}\) , it is \( x\) .

• For the vector \( \overrightarrow{v}\) , it is \( |z|=-z\) .

• And for the vector \( \overrightarrow{w}\) , it is \( x-|z|=x+z\) , so that \( |s|=x+z\) .

• Moreover, the projection on the \( x\) axis of the end of the vector \( \overrightarrow{w}\) is to the right of the origin, so that \( s=|s|=x+z\) .

2. Assume that \( x=|z|\) , so that \( z=-x\) .

   

   Starting from the origin, we draw end-to-beginnig the vector

   \( \overrightarrow{u}=\begin{bmatrix}x\\ y\end{bmatrix}\) with \( x>0\) and the vector \( \overrightarrow{v}=\begin{bmatrix}z\\ t\end{bmatrix}\) , with \( z<0\) and \( z=-x\) .

   Then the distances between the projections on the \( x\) axis of the beginnigs and the ends of the vectors \( \overrightarrow{u}\) and \( \overrightarrow{v}\) , is \( x=|z|=-z\) .

   Moreover, the vector \( \overrightarrow{w}\) is along the \( y\) axis, so that \( s=0=x+z\) .

   Consequently, \( s=x+z\) .

3. Assume that \( x<|z|\) .

   

   Starting from the origin, we draw end-to-beginnig the vector

   \( \overrightarrow{u}=\begin{bmatrix}x\\ y\end{bmatrix}\) with \( x>0\) and the vector \( \overrightarrow{v}=\begin{bmatrix}z\\ t\end{bmatrix}\) , with \( z<0\) and \( x>|z|\) .

   Then the distances tbetween the projections on the \( x\) axis of the beginnigs and the ends of the different vectors are the following.

• For the vector \( \overrightarrow{u}\) , it is \( x\) .

• For the vector \( \overrightarrow{v}\) , it is \( |z|=-z\) .

• And for the vector \( \overrightarrow{w}\) , it is \( |z|-x=-(x+z)\) , so that \( |s|=-(x+z)\) .

• Moreover, the projection on the \( x\) axis of the end of the vector \( \overrightarrow{w}\) is to the left of the origin, so that \( s=-|s|=x+z\) .

4. Assume that \( x\) and \( z\) are both negative.

   

   Starting from the origin, we draw end-to-beginnig the vector \( \overrightarrow{u}=\begin{bmatrix}x\\ y\end{bmatrix}\) with \( x<0\) and the vector \( \overrightarrow{v}=\begin{bmatrix}z\\ t\end{bmatrix}\) , with \( z<0\) .

   Then the distances tbetween the projections on the \( x\) axis of the beginnigs and the ends of the different vectors are the following.

• For the vector \( \overrightarrow{u}\) , it is \( |x|=-x\) .

• For the vector \( \overrightarrow{v}\) , it is \( |z|=-z\) .

• And for the vector \( \overrightarrow{w}\) , it is \( |x|+|z|=-(x+z)\) , so that \( |s|=-(x+z)\) .

• Moreover, the projection on the \( x\) axis of the end of the vector \( \overrightarrow{w}\) is to the left of the origin, so that \( s=-|s|=x+z\) .

Proof

Subtract a vector is add its opposite because subtract a real number is add its opposite.

Indeed, if we assume \( (x,y,z,t)\in\mathbb{R}^4\) are real numbers, and if we consider the two column vectors with 2 elements, \( \overrightarrow{u}=\begin{bmatrix}x\\ y\end{bmatrix}\) and \( \overrightarrow{v}=\begin{bmatrix}z\\ t\end{bmatrix}\) , then:

Proof (of theorem 13)

Assume \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^2\) are column vectors with two real elements.

Then, because of the followint facts:

1. subtract a vector is add its opposite,

2. the addition of vectors is associative,

3. the opposite of a vector is its reciprocal for the addition of vectors,

4. and the null vector is neutral for the addition of vectors,

the following calculations may be performed:

• \( (\overrightarrow{u}+\overrightarrow{v})-\overrightarrow{v} =(\overrightarrow{u}+\overrightarrow{v})+(-\overrightarrow{v}) =\overrightarrow{u}+(\overrightarrow{v}+(-\overrightarrow{v})) =\overrightarrow{u}+\overrightarrow{0} =\overrightarrow{u}\)

• \( (\overrightarrow{u}-\overrightarrow{v})+\overrightarrow{v} =(\overrightarrow{u}+(-\overrightarrow{v}))+\overrightarrow{v} =\overrightarrow{u}+((-\overrightarrow{v})+\overrightarrow{v}) =\overrightarrow{u}+\overrightarrow{0} =\overrightarrow{u}\)

Proof (of corrollary 1)

Assume \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^2\) are column vectors with two real elements.

1. If we subtract \( \overrightarrow{v}\) to both members of the equality \( \overrightarrow{w}=\overrightarrow{u}+\overrightarrow{v}\) , we see that it is equivalent to: \( \overrightarrow{w}-\overrightarrow{v}=(\overrightarrow{u}+\overrightarrow{v})-\overrightarrow{v}=\overrightarrow{u}\) , because of theorem 13.

2. If we add \( \overrightarrow{v}\) to both members of the equality \( \overrightarrow{w}=\overrightarrow{u}-\overrightarrow{v}\) , we see that it is equivalent to: \( \overrightarrow{w}+\overrightarrow{v}=(\overrightarrow{u}-\overrightarrow{v})+\overrightarrow{v}=\overrightarrow{u}\) , because of theorem 13.

3. If we add \( \overrightarrow{u}\) to both members of the equality \( \overrightarrow{w}=\overrightarrow{v}-\overrightarrow{u}\) , we see that it is equivalent to: \( \overrightarrow{w}+\overrightarrow{u}=(\overrightarrow{v}-\overrightarrow{u})+\overrightarrow{u}=\overrightarrow{v}\) , because of theorem 13.

   Because the addition of vectors is commutative, the last equality is equivalent to: \( \overrightarrow{v}=\overrightarrow{u}+\overrightarrow{w}\) .

   And if we exchange the roles of \( \overrightarrow{v}\) and \( \overrightarrow{w}\) in item (I), we see that the last equality is equivalent to: \( \overrightarrow{u}=\overrightarrow{v}-\overrightarrow{w}\)

Proof

Assume \( (x,y)\in\mathbb{R}^2\) are real numbers, and consider the column vector \( \overrightarrow{u}=\begin{bmatrix}x\\ y\end{bmatrix}\) with 2 elements.

Consider the canonical base of the vector plane \( \mathbb{P}\) : \( \overrightarrow{i}=\begin{bmatrix}1\\ 0\end{bmatrix}\) and \( \overrightarrow{j}=\begin{bmatrix}0\\ 1\end{bmatrix}\) .

Then, as we have seen in the paragraph 3, the abscissa of the vector \( \overrightarrow{u}\) in the canonical base is its first element \( x\) and its ordinate is its second element \( y\) .

Moreover, the following calculations may be performed:

\( x\overrightarrow{i}+y\overrightarrow{j} =x\begin{bmatrix}1\\ 0\end{bmatrix}+y\begin{bmatrix}0\\ 1\end{bmatrix} =\begin{bmatrix}x\times 1+y\times 0\\ x\times 0+y\times 1\end{bmatrix} =\begin{bmatrix}x\\ y\end{bmatrix} =\overrightarrow{u}\)

Proof (of the lemma 1)

Assume \( (x,y)\in\mathbb{R}^2\) are real numbers, and consider the column vector \( \overrightarrow{u}=\begin{bmatrix}x\\ y\end{bmatrix}\) with 2 elements.

Assume \( \lambda\in\mathbb{R}^*\) is a real number such as \( \lambda\neq 0\) .

Then we may perform the following calculations:

\( \frac{\overrightarrow{u}}{\lambda} =\begin{bmatrix}\frac{x}{\lambda}\\ \frac{y}{\lambda}\end{bmatrix} =\begin{bmatrix}\frac{1}{\lambda}x\\ \frac{1}{\lambda}y\end{bmatrix} =\frac{1}{\lambda}\begin{bmatrix}x\\ y\end{bmatrix} =\frac{1}{\lambda}\overrightarrow{u}\)

Proof (of the lemma 2)

The associativity property is a consequence of the associativity of the multiplication of real numbers.

Assume \( (x,y)\in\mathbb{R}^2\) are real numbers, and consider the column vector \( \overrightarrow{u}=\begin{bmatrix}x\\ y\end{bmatrix}\) with 2 elements.

Assume \( (\alpha,\beta)\in\mathbb{R}^{2}\) is are real numbers.

Then we may perform the following calculations:

\( \alpha(\beta\overrightarrow{u}) =\alpha\begin{bmatrix}\beta x\\ \beta y\end{bmatrix} =\begin{bmatrix}\alpha(\beta x)\\ \alpha(\beta y)\end{bmatrix} =\begin{bmatrix}(\alpha\beta) x)\\ (\alpha\beta) y)\end{bmatrix} =(\alpha\beta)\begin{bmatrix} x)\\ y)\end{bmatrix} =(\alpha\beta)\overrightarrow{u}\)

Proof (of the theorem 23)

Assume \( \overrightarrow{u}\in\mathbb{P}\) is a column vector with two real elements, and \( \lambda\in\mathbb{R}^*\) is a real number such as \( \lambda\neq 0\) .

Then we may apply the lemmas 1 et 2 in the following calculations:

• \( \lambda\frac{\overrightarrow{u}}{\lambda} =\lambda\left( \frac{1}{\lambda}\overrightarrow{u} \right) =\left( \lambda\frac{1}{\lambda} \right)\overrightarrow{u} =1.\overrightarrow{u} =\overrightarrow{u}\) ,

• and \( \frac{\lambda\overrightarrow{u}}{\lambda} =\frac{1}{\lambda} (\lambda\overrightarrow{u}) =\left( \frac{1}{\lambda} \lambda \right)\overrightarrow{u} =1.\overrightarrow{u} =\overrightarrow{u}\) .

Proof (of the corollary 2)

Assume \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^2\) are column vectors with two real elements, and \( \lambda\in\mathbb{R}^*\) is a real number such as \( \lambda\neq 0\) .

Then, if we divide by \( \lambda\) the two sides of the equality \( \overrightarrow{v}=\lambda\overrightarrow{u}\) , we see the it is equivalent to: \( \frac{\overrightarrow{v}}{\lambda}=\frac{\lambda\overrightarrow{u}}{\lambda}=\overrightarrow{u}\) .

And if we multiply by \( \lambda\) the two sides of the equality \( \frac{\lambda\overrightarrow{u}}{\lambda}\) , we see the it is equivalent to: \( \lambda\overrightarrow{v}=\lambda\frac{\overrightarrow{u}}{\lambda}=\overrightarrow{u}\) .

Proof

Assume \( (x,y,z,t)\in\mathbb{R}^4\) are real numbers, and consider the two column vectors with 2 elements, \( \overrightarrow{u}=\begin{bmatrix}x\\ y\end{bmatrix}\) and \( \overrightarrow{v}=\begin{bmatrix}z\\ t\end{bmatrix}\) .

Assume \( (\alpha,\beta)\in\mathbb{R}^2\) are real numbers.

First distributivity law

The first distributivity law is a consequence of the distributivity of the multiplication on the addition in \( \mathbb{R}\) .

Indeed, we may perform the following calculations:

\( \alpha(\overrightarrow{u}+\overrightarrow{v}) =\alpha\left(\begin{bmatrix}x\\ y\end{bmatrix}+\begin{bmatrix}z\\ t\end{bmatrix}\right) =\alpha\begin{bmatrix}x+z\\ y+t\end{bmatrix} =\begin{bmatrix}\alpha(x+z)\\ \alpha(y+t)\end{bmatrix} =\begin{bmatrix}\alpha x+\alpha z\\ \alpha y+\alpha t\end{bmatrix}\)

\( =\begin{bmatrix}\alpha x\\ \alpha y\end{bmatrix}+\begin{bmatrix}\alpha z\\ \alpha t\end{bmatrix} =\alpha\begin{bmatrix} x\\ y\end{bmatrix}+\alpha\begin{bmatrix}z\\ t\end{bmatrix} =\alpha\overrightarrow{u}+\alpha\overrightarrow{v}\)

Second distributivity law

The second distributivity law is a consequence of the distributivity of the multiplication on the addition in \( \mathbb{R}\) as well.

Indeed, we may perform the following calculations:

\( (\alpha+\beta)\overrightarrow{u} =(\alpha+\beta)\begin{bmatrix}x\\ y\end{bmatrix} =\begin{bmatrix}(\alpha+\beta)x\\ (\alpha+\beta)y\end{bmatrix} =\begin{bmatrix}\alpha x+\beta x\\ \alpha y+\beta y\end{bmatrix} =\begin{bmatrix}\alpha x\\ \alpha y\end{bmatrix} +\begin{bmatrix}\beta x\\ \beta y\end{bmatrix}\)

\( =\alpha\begin{bmatrix} x\\ y\end{bmatrix} +\beta \begin{bmatrix} x\\ y\end{bmatrix} =\alpha\overrightarrow{u}+\beta\overrightarrow{u}\)

The associativity law is already stated in the lemma 2.

Proof

Assume \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^2\) are column vectors with two real elements, and \( (\alpha,\beta)\in\mathbb{R}^2\) are real numbers.

Signs law It is a consequence of the associativity law and of the following facts:

• \( -\alpha=(-1)\times\alpha=\alpha\times(-1)\) ,

• \( -\overrightarrow{u}=(-1)\overrightarrow{u}\) ,

• and \( -\alpha\overrightarrow{u}=(-1)(\alpha\overrightarrow{u})\) .

Indeed, we may perform the following calculations:

\( (-\alpha)\overrightarrow{u} =(\alpha\times(-1))\overrightarrow{u} =\alpha((-1)\overrightarrow{u}) =\alpha(-\overrightarrow{u})\) ,

and

\( (-\alpha)\overrightarrow{u} =((-1)\times\alpha)\overrightarrow{u} =(-1)(\alpha\overrightarrow{u}) =-\alpha\overrightarrow{u}\) .

First distributivity law

The first distributivity law is a consequence of the signs law and the fact that subtract a vector is add its opposite.

Indeed, we may perform the following calculations:

\( \alpha(\overrightarrow{u}-\overrightarrow{v}) =\alpha(\overrightarrow{u}+(-\overrightarrow{v})) =\alpha\overrightarrow{u}+\alpha(-\overrightarrow{v}) =\alpha\overrightarrow{u}+(-\alpha\overrightarrow{v}) =\alpha\overrightarrow{u}-\alpha\overrightarrow{v}\)

Second distributivity law

The second distributivity law is a consequence of the signs law and the fact that subtract a real number is add its opposite.

Indeed, we may perform the following calculations:

\( (\alpha-\beta)\overrightarrow{u} =(\alpha+(-\beta))\overrightarrow{u} =\alpha\overrightarrow{u}+(-\beta)\overrightarrow{u} =\alpha\overrightarrow{u}+(-\beta\overrightarrow{u}) =\alpha\overrightarrow{u}-\beta\overrightarrow{u}\)