Assume that \( \overrightarrow{u}\in\mathbb{P}^{*}\) is a non zero vector.

Then, as \( \overrightarrow{u}=\overrightarrow{0}\) implies \( a=0\) , the ordered set \( S=(\overrightarrow{u})\) is linearly independent.

Assume that \( (\overrightarrow{u}_1,\overrightarrow{u}_2)\in(\mathbb{P}^*)^2\) are non aligned non zero vectors.

Then, assume that \( a_1\overrightarrow{u}_1+a_2\overrightarrow{u}_2=\overrightarrow{0}\) for some \( (a_1,a_2)\) that are not zero together.

Let’s say that \( a_{2}\ne 0\) , so that \( \overrightarrow{u}_2=\frac{a_{1}}{a_{2}}\overrightarrow{u}_1\) .

That would imply that \( \overrightarrow{u}_1\) and \( \overrightarrow{u}_2\) are aligned, on the contrary of the hypothesis.

Consequently, the ordered set \( S=(\overrightarrow{u}_1,\overrightarrow{u}_2)\) is linearly independent.

Assume that \( (\overrightarrow{u}_1,\overrightarrow{u}_2)\in(\mathbb{P}^*)^2\) are aligned non zero vectors.

Consider the coefficient \( k\in\mathbb{R}^*\) such that \( \overrightarrow{u}_2=k\overrightarrow{u}_1\) .

Then \( a_1\overrightarrow{u}_1+a_2\overrightarrow{u}_2=\overrightarrow{0}\) for \( a_{1}=k\) and \( a_{2}=-1\ne 0\) .

Consequently, the ordered set \( S=(\overrightarrow{u}_1,\overrightarrow{u}_2)\) is linearly dependent.

Assume that \( (\overrightarrow{u}_1,\overrightarrow{u}_2,\overrightarrow{u}_3)\in(\mathbb{P}^*)^3\) are non zero vectors such that, say, \( \overrightarrow{u}_1\) and \( \overrightarrow{u}_2\) are not aligned.

Consider the cartesian coordinates of \( \overrightarrow{u}_1\) , \( \overrightarrow{u}_2\) and \( \overrightarrow{u}_3\) , \( \overrightarrow{u}_1=\begin{bmatrix} x_{1}\\y_{1} \end{bmatrix}\) , \( \overrightarrow{u}_2=\begin{bmatrix} x_{2}\\y_{2} \end{bmatrix}\) and \( \overrightarrow{u}_3=\begin{bmatrix} x_{3}\\y_{3} \end{bmatrix}\) .

Then the sytem of linear equations in \( (a_{2},a_{3})\) :

has exactly one solution \( (a_{2},a_{3})\) .

This is because \( (\overrightarrow{u}_1,\overrightarrow{u}_2)\) is a linearly independant ordered set, so that the system of linear equations in \( (a_{2},a_{3})\) :

has exactly one solution \( (0,0)\), and thus its determinant \( x_2y_3-x_3y_2\) is non zero.

Consequently, the linear combination of the vectors in \( S=(\overrightarrow{u}_1,\overrightarrow{u}_2,\overrightarrow{u}_3)\) with a non zero coefficient \( 1\) is equal to

\( 1\times\overrightarrow{u}_1+(-a_{2})\overrightarrow{u}_2+(-a_{3})\overrightarrow{u}_3 =\overrightarrow{0}\) .

Hence \( S=(\overrightarrow{u}_1,\overrightarrow{u}_2,\overrightarrow{u}_3)\) is linearly dependent.

Assume that \( \overrightarrow{u}\in\mathbb{P}^{*}\) is a non zero vector.

Then, as none of the non zero vectors that are not aligned with \( \overrightarrow{u}\) is a scalar multiple of \( \overrightarrow{u}\) , the ordered set \( S=(\overrightarrow{u})\) is a non spanning ordered set.

Assume that \( (\overrightarrow{u}_1,\overrightarrow{u}_2)\in(\mathbb{P}^*)^2\) are non aligned non zero vectors.

Assume that \( \overrightarrow{v}\in\mathbb{P}\) is any vector in the plane.

Consider the cartesian coordinates of \( \overrightarrow{u}_1\) , \( \overrightarrow{u}_2\) and \( \overrightarrow{v}\) , \( \overrightarrow{u}_1=\begin{bmatrix} x_{1}\\y_{1} \end{bmatrix}\) , \( \overrightarrow{u}_2=\begin{bmatrix} x_{2}\\y_{2} \end{bmatrix}\) and \( \overrightarrow{v}=\begin{bmatrix} z\\t \end{bmatrix}\) .

Then the sytem of linear equations in \( (a_{1},a_{2})\) :

has at exactly one solution \( (a_{1},a_{2})\) .

This is because \( (\overrightarrow{u}_1,\overrightarrow{u}_2)\) is a linearly independant ordered set, so that the system of linear equations in \( (a_{2},a_{3})\) :

has exactly one solution \( (0,0)\), and thus its determinant \( x_2y_3-x_3y_2\) is non zero.

Consequently, \( \overrightarrow{v}=a_1\overrightarrow{u}_1+a_2\overrightarrow{u}_2\) , so that \( S=(\overrightarrow{u}_1,\overrightarrow{u}_2)\) is a spanning ordered set.

Assume that \( (\overrightarrow{u}_1,\overrightarrow{u}_2)\in(\mathbb{P}^*)^2\) are aligned non zero vectors.

Then, as any linear combination of \( \overrightarrow{u}_1\) and \( \overrightarrow{u}_2\) is aligned with \( \overrightarrow{u}_1\) , no vector that is not aligned with \( \overrightarrow{u}_1\) may be written as a linear combination of \( \overrightarrow{u}_1\) and \( \overrightarrow{u}_2\) .

Consequently, \( S=(\overrightarrow{u}_1,\overrightarrow{u}_2)\) is a non spanning ordered set.

Assume that \( (\overrightarrow{u}_1,\overrightarrow{u}_2,\overrightarrow{u}_3)\in(\mathbb{P}^*)^3\) are non zero vectors such that, say, \( \overrightarrow{u}_1\) and \( \overrightarrow{u}_2\) are not aligned.

Then, as \( S'=(\overrightarrow{u}_1,\overrightarrow{u}_2)\) is a spanning ordered set, any vector in the plane is a linear combination of \( \overrightarrow{u}_1\) and \( \overrightarrow{u}_2\) , and consequently of \( \overrightarrow{u}_1\) , \( \overrightarrow{u}_2\) and \( \overrightarrow{u}_3\) , with a zero coefficient for \( \overrightarrow{u}_3\) .

Consequently, \( S=(\overrightarrow{u}_1,\overrightarrow{u}_2,…,\overrightarrow{u}_n)\) is a spanning ordered set.

Assume that \( (\overrightarrow{I},\overrightarrow{J})\in(\mathbb{P}^*)^2\) are non aligned non zero vectors.

Assume that \( (a,b)\in(\mathbb{R}^*_+)^2\) are scalars such that \( a>0\) and \( b>0\) .

Then the following assertions hold.

• The vector \( a\overrightarrow{I}\) is positively aligned with \( \overrightarrow{I}\) and its norm is \( \left\|a\overrightarrow{I}\right\|=a\left\|\overrightarrow{I}\right\|\) .

• The vector \( b\overrightarrow{J}\) is positively aligned with \( \overrightarrow{J}\) and its norm is \( \left\|b\overrightarrow{J}\right\|=b\left\|\overrightarrow{J}\right\|\) .

• Because of the geometrical view of the sum of two non zero vectors studied in lecture 15 of section 4, the vector \( a\overrightarrow{I}+b\overrightarrow{J}\) is built by drawing the vector \( b\overrightarrow{J}\) at the end of the vector \( a\overrightarrow{I}\) .

\( (\overrightarrow{I},\overrightarrow{J})\in(\mathbb{P}^*)^2\) are non aligned non zero vectors.

Assume that \( (a,b)\in\mathbb{R}^*_+\times\mathbb{R}^*_-\) are scalars such that \( a>0\) and \( b<0\) .

Then the following assertions hold.

• The vector \( a\overrightarrow{I}\) is positively aligned with \( \overrightarrow{I}\) and its norm is \( \left\|a\overrightarrow{I}\right\|=a\left\|\overrightarrow{I}\right\|\) .

• The vector \( b\overrightarrow{J}\) is negatively aligned with \( \overrightarrow{J}\) and its norm is \( \left\|b\overrightarrow{J}\right\|=|b|\left\|\overrightarrow{J}\right\|\) .

• The vector \( a\overrightarrow{I}+b\overrightarrow{J}\) is built by drawing the vector \( b\overrightarrow{J}\) at the end of the vector \( a\overrightarrow{I}\) .

Assume that \( (\overrightarrow{I},\overrightarrow{J})\in(\mathbb{P}^*)^2\) are non aligned non zero vectors.

Assume that \( (a,b)\in\mathbb{R}^*_-\times\mathbb{R}^*_+\) are scalars such that \( a<0\) and \( b>0\) .

Then the following assertions hold.

• The vector \( a\overrightarrow{I}\) is negatively aligned with \( \overrightarrow{I}\) and its norm is \( \left\|a\overrightarrow{I}\right\|=|a|\left\|\overrightarrow{I}\right\|\) .

• The vector \( b\overrightarrow{J}\) is positively aligned with \( \overrightarrow{J}\) and its norm is \( \left\|b\overrightarrow{J}\right\|=b\left\|\overrightarrow{J}\right\|\) .

• The vector \( a\overrightarrow{I}+b\overrightarrow{J}\) is built by drawing the vector \( b\overrightarrow{J}\) at the end of the vector \( a\overrightarrow{I}\) .

Assume that \( (\overrightarrow{I},\overrightarrow{J})\in(\mathbb{P}^*)^2\) are non aligned non zero vectors.

Assume that \( (a,b)\in(\mathbb{R}^*_-)^2\) are scalars such that \( a<0\) and \( b<0\) .

Then the following assertions hold.

• The vector \( a\overrightarrow{I}\) is negatively aligned with \( \overrightarrow{I}\) and its norm is \( \left\|a\overrightarrow{I}\right\|=|a|\left\|\overrightarrow{I}\right\|\) .

• The vector \( b\overrightarrow{J}\) is negatively aligned with \( \overrightarrow{J}\) and its norm is \( \left\|b\overrightarrow{J}\right\|=|b|\left\|\overrightarrow{J}\right\|\) .

• Because of the geometrical view of the sum of two non zero vectors studied in lecture 15 of section 4, the vector \( a\overrightarrow{I}+b\overrightarrow{J}\) is built by drawing the vector \( b\overrightarrow{J}\) at the end of the vector \( a\overrightarrow{I}\) .

Assume that \( B=(\overrightarrow{I},\overrightarrow{J})\) is a basis in the plane.

Then the following geometrical facts occur.

• The \( X\) axis is leaded by the vector \( \overrightarrow{I}\) .
• The \( Y\) axis is leaded by the vector \( \overrightarrow{J}\) .
• The coordinates network is made of parallels to the \( X\) axis and to the \( Y\) axis.
• This is because we will have to draw the projections:
• on the \( X\) axis along the \( Y\) axis,
• and on the \( Y\) axis along the \( X\) axis.

Assume that \( B=(\overrightarrow{I},\overrightarrow{J})\) is a basis in the plane.

Assume that \( x\in\mathbb{R_+^*}\) is a real number such that \( x>0\) , and consider the vector \( \overrightarrow{v}\in\mathbb{P}\) such that \( [\overrightarrow{v}]_B=\begin{bmatrix}x\\0\end{bmatrix}\) .

Then the following geometrical facts occur.

• \( \overrightarrow{v}\) is positively along the \( X\) axis.
• Its length is \( x\) times the length of \( \overrightarrow{I}\) .

Assume that \( B=(\overrightarrow{I},\overrightarrow{J})\) is a basis in the plane.

Assume that \( (x,y)\in(\mathbb{R_+^*)^{2}}\) are real numbers such that \( x>0\) and \( y>0\) , and consider the vector \( \overrightarrow{v}\in\mathbb{P}\) such that \( [\overrightarrow{v}]_B=\begin{bmatrix}x\\y\end{bmatrix}\) .

Then the following geometrical facts occur.

• The projection of the end of \( \overrightarrow{v}\) on the \( X\) axis along the \( Y\) axis is at a distance \( x\left\|\overrightarrow{I}\right\|\) of the origin positively along the \( X\) axis.
• The projection of the end of \( \overrightarrow{v}\) on the \( Y\) axis along the \( X\) axis is at a distance \( y\left\|\overrightarrow{J}\right\|\) of the origin positively along the \( Y\) axis.

Assume that \( B=(\overrightarrow{I},\overrightarrow{J})\) is a basis in the plane.

Assume that \( y\in\mathbb{R_+^*}\) is a real number such that \( y>0\) , and consider the vector \( \overrightarrow{v}\in\mathbb{P}\) such that \( [\overrightarrow{v}]_B=\begin{bmatrix}0\\y\end{bmatrix}\) .

Then the following geometrical facts occur.

• \( \overrightarrow{v}\) is positively along the \( Y\) axis.
• Its length is \( y\) times the length of \( \overrightarrow{J}\) .

Assume that \( B=(\overrightarrow{I},\overrightarrow{J})\) is a basis in the plane.

Assume that \( (x,y)\in\mathbb{R_-^*}\times\mathbb{R_+^*}\) are real numbers such that \( x<0\) and \( y>0\) , and consider the vector \( \overrightarrow{v}\in\mathbb{P}\) such that \( [\overrightarrow{v}]_B=\begin{bmatrix}x\\y\end{bmatrix}\) .

Then the following geometrical facts occur.

• The projection of the end of \( \overrightarrow{v}\) on the \( X\) axis along the \( Y\) axis is at a distance \( |x|\left\|\overrightarrow{I}\right\|\) of the origin negatively along the \( X\) axis.
• The projection of the end of \( \overrightarrow{v}\) on the \( Y\) axis along the \( X\) axis is at a distance \( y\left\|\overrightarrow{J}\right\|\) of the origin positively along the \( Y\) axis.

Assume that \( B=(\overrightarrow{I},\overrightarrow{J})\) is a basis in the plane.

Assume that \( x\in\mathbb{R_-^*}\) is a real number such that \( x<0\) , and consider the vector \( \overrightarrow{v}\in\mathbb{P}\) such that \( [\overrightarrow{v}]_B=\begin{bmatrix}x\\0\end{bmatrix}\) .

Then the following geometrical facts occur.

• \( \overrightarrow{v}\) is negatively along the \( X\) axis.
• Its length is \( |x|\) times the length of \( \overrightarrow{I}\) .

Assume that \( B=(\overrightarrow{I},\overrightarrow{J})\) is a basis in the plane.

Assume that \( (x,y)\in(\mathbb{R_-^*)^{2}}\) are real numbers such that \( x<0\) and \( y<0\) , and consider the vector \( \overrightarrow{v}\in\mathbb{P}\) such that \( [\overrightarrow{v}]_B=\begin{bmatrix}x\\y\end{bmatrix}\) .

Then the following geometrical facts occur.

• The projection of the end of \( \overrightarrow{v}\) on the \( X\) axis along the \( Y\) axis is at a distance \( |x|\left\|\overrightarrow{I}\right\|\) of the origin negatively along the \( X\) axis.
• The projection of the end of \( \overrightarrow{v}\) on the \( Y\) axis along the \( X\) axis is at a distance \( |y|\left\|\overrightarrow{J}\right\|\) of the origin negatively along the \( Y\) axis.

Assume that \( B=(\overrightarrow{I},\overrightarrow{J})\) is a basis in the plane.

Assume that \( y\in\mathbb{R_-^*}\) is a real number such that \( y<0\) , and consider the vector \( \overrightarrow{v}\in\mathbb{P}\) such that \( [\overrightarrow{v}]_B=\begin{bmatrix}0\\y\end{bmatrix}\) .

Then the following geometrical facts occur.

• \( \overrightarrow{v}\) is negatively along the \( Y\) axis.
• Its length is \( |y|\) times the length of \( \overrightarrow{J}\) .

Assume that \( B=(\overrightarrow{I},\overrightarrow{J})\) is a basis in the plane.

Assume that \( (x,y)\in\mathbb{R_+^*}\times\mathbb{R_-^*}\) are real numbers such that \( x>0\) and \( y<0\) , and consider the vector \( \overrightarrow{v}\in\mathbb{P}\) such that \( [\overrightarrow{v}]_B=\begin{bmatrix}x\\y\end{bmatrix}\) .

Then the following geometrical facts occur.

• The projection of the end of \( \overrightarrow{v}\) on the \( X\) axis along the \( Y\) axis is at a distance \( x\left\|\overrightarrow{I}\right\|\) of the origin positively along the \( X\) axis.
• The projection of the end of \( \overrightarrow{v}\) on the \( Y\) axis along the \( X\) axis is at a distance \( |y|\left\|\overrightarrow{J}\right\|\) of the origin negatively along the \( Y\) axis.

Consider the aligned non zero vectors in the plane:

• \( \overrightarrow{u}_1\) with column vector of coordinates in the canonical basis \( X_1=\begin{bmatrix}2\\1\end{bmatrix}\) ,

• and \( \overrightarrow{u}_2=-\frac{1}{2}\overrightarrow{u}_1\) with column vector of coordinates in the canonical basis \( X_2=\begin{bmatrix}-1\\{-\frac{1}{2}}\end{bmatrix}\) .

Consider the matrix \( P\) with columns the column vectors \( X_{1}\) and \( X_{2}\) :

• \( P=\begin{bmatrix}2&-1\\1&-\frac{1}{2}\end{bmatrix}\)

Then the following assertions hold:

• The determinant of \( P\) is \( \det(P)=2\times\left(-\frac{1}{2}\right)-1\times(-1)=-1+1=0\) .

• Consequently, \( P\) is not invertible.

• And, as \( \overrightarrow{u}_1\) and \( \overrightarrow{u}_2\) are aligned, the ordered set \( S=(\overrightarrow{u}_1,\overrightarrow{u}_2)\) is not a basis of the plane.

Consider the non aligned non zero vectors in the plane:

• \( \overrightarrow{u}_1\) with column vector of coordinates in the canonical basis \( X_1=\begin{bmatrix}2\\1\end{bmatrix}\) ,

• and \( \overrightarrow{u}_2\) with column vector of coordinates in the canonical basis \( X_2=\begin{bmatrix}1\\{-1}\end{bmatrix}\) .

Consider the matrix \( P\) with columns the column vectors \( X_{1}\) and \( X_{2}\) :

• \( P=\begin{bmatrix}2&1\\1&-1\end{bmatrix}\) .

Then the following assertions hold:

• The determinant of \( P\) is \( \det(P)=2\times(-1)-1\times 1=-2-1=-3\ne 0\) .

• Consequently, \( P\) is invertible.

• And, as \( \overrightarrow{u}_1\) and \( \overrightarrow{u}_2\) are non aligned non zero vectors, the ordered set \( S=(\overrightarrow{u}_1,\overrightarrow{u}_2)\) is a basis of the plane.

The canonical basis \( B_0=(\overrightarrow{i},\overrightarrow{j})\) is a direct orthonormal basis because of the following facts.

• The two vectors are orthogonal to each other (\( \overrightarrow{i}\bot\overrightarrow{j}\) ), that is \( \overrightarrow{i}\cdot\overrightarrow{j}=0\) .

• The two vectors are unitary: \( \left\| \overrightarrow{i} \right\|=\left\| \overrightarrow{j} \right\|=1\) .

• We go from \( \overrightarrow{i}\) to \( \overrightarrow{j}\) in the direct direction, that is the counter clockwise direction: \( \widehat{(\overrightarrow{i},\overrightarrow{j})}=+\frac{\pi}{2}\) .

The basis \( B_1=(\overrightarrow{j},\overrightarrow{i})\) is a reverse orthonormal basis because of the following facts.

• The two vectors are orthogonal to each other (\( \overrightarrow{j}\bot\overrightarrow{i}\) ), that is \( \overrightarrow{j}\cdot\overrightarrow{i}=0\) .

• The two vectors are unitary: \( \left\| \overrightarrow{j} \right\|=\left\| \overrightarrow{i} \right\|=1\) .

• We go from \( \overrightarrow{j}\) to \( \overrightarrow{i}\) in the reverse direction, that is the clockwise direction: \( \widehat{(\overrightarrow{j},\overrightarrow{i})}=-\frac{\pi}{2}\) .

Consider the vectors \( \overrightarrow{I}=\begin{bmatrix}\frac{\sqrt{3}}{2}\\\frac{1}{2}\end{bmatrix} =\begin{bmatrix}\cos\left(\frac{\pi}{6}\right)\\\sin\left(\frac{\pi}{6}\right)\end{bmatrix}\) and \( \overrightarrow{J}=\begin{bmatrix}-\frac{1}{2}\\\frac{\sqrt{3}}{2}\end{bmatrix} =\begin{bmatrix}\cos\left(\frac{2\pi}{3}\right)\\\sin\left(\frac{2\pi}{3}\right)\end{bmatrix}\) .

Then the basis \( B=(\overrightarrow{I},\overrightarrow{J})\) is a direct orthonormal basis because of the following facts.

• The two vectors are orthogonal to each other (\( \overrightarrow{I}\bot\overrightarrow{J}\) ), because \( \overrightarrow{I}\cdot\overrightarrow{J}=0\) .

• The two vectors are unitary, because \( \left\| \overrightarrow{I} \right\|=\left\| \overrightarrow{J} \right\|=1\) .

• We go from \( \overrightarrow{i}\) to \( \overrightarrow{j}\) in the direct direction, because

  \( \widehat{(\overrightarrow{I},\overrightarrow{J})}= \widehat{(\overrightarrow{i},\overrightarrow{J})}-\widehat{(\overrightarrow{i},\overrightarrow{I})} =\frac{2\pi}{3}-\frac{\pi}{6}=+\frac{\pi}{2}\) .

Consider the vectors \( \overrightarrow{I}=\begin{bmatrix}\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\end{bmatrix} =\begin{bmatrix}\cos\left(\frac{\pi}{4}\right)\\\sin\left(\frac{\pi}{4}\right)\end{bmatrix}\) and \( \overrightarrow{J}=\begin{bmatrix}\frac{\sqrt{2}}{2}\\{-\frac{\sqrt{2}}{2}}\end{bmatrix} =\begin{bmatrix}\cos\left(-\frac{\pi}{4}\right)\\\sin\left(-\frac{\pi}{4}\right)\end{bmatrix}\) .

Then the basis \( B=(\overrightarrow{I},\overrightarrow{J})\) is a reverse orthonormal basis because of the following facts.

• The two vectors are orthogonal to each other (\( \overrightarrow{I}\bot\overrightarrow{J}\) ), because \( \overrightarrow{I}\cdot\overrightarrow{J}=0\) .

• The two vectors are unitary, because \( \left\| \overrightarrow{I} \right\|=\left\| \overrightarrow{J} \right\|=1\) .

• We go from \( \overrightarrow{i}\) to \( \overrightarrow{j}\) in the reverse direction, because

  \( \widehat{(\overrightarrow{I},\overrightarrow{J})}= \widehat{(\overrightarrow{i},\overrightarrow{J})}-\widehat{(\overrightarrow{i},\overrightarrow{I})} =-\frac{\pi}{4}-\frac{\pi}{4}=-\frac{\pi}{2}\) .

Assume that \( B=(\overrightarrow{I},\overrightarrow{J})\) is a direct orthonormal basis and consider the angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{I})}\) .

Then the coordinates of \( \overrightarrow{I}\) and \( \overrightarrow{J}\) in the canonical basis are the following:

• \( [\overrightarrow{I}]_{B_0}=\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}\) ,

• and \( [\overrightarrow{J}]_{B_0}=\begin{bmatrix} \cos\left(\theta+\frac{\pi}{2}\right)\\ \sin\left(\theta+\frac{\pi}{2}\right)\end{bmatrix}= \begin{bmatrix}-\sin(\theta)\\\cos(\theta)\end{bmatrix}\) .

Consequently, the following assertions are fulfilled.

• The transition matrix from the basis \( B\) to the canonical basis is equal to \( P=\begin{bmatrix} \cos(\theta)&-\sin(\theta)\\ \sin(\theta)&\cos(\theta) \end{bmatrix}\) , the matrix of the rotation of angle \( \theta\) .

• And the transition matrix from the canonical basis to the basis \( B\) is equal to:

  \( P^{-1} =\frac{1}{\cos^{2}(\theta)+\sin^{2}(\theta)} \begin{bmatrix} \cos(\theta)&\sin(\theta)\\{-\sin(\theta)}&\cos(\theta) \end{bmatrix} =\begin{bmatrix} \cos(\theta)&\sin(\theta)\\{-\sin(\theta)}&\cos(\theta) \end{bmatrix}\)

  It is the matrix of the rotation of angle \( -\theta\) .

Assume that \( B=(\overrightarrow{I},\overrightarrow{J})\) is a reverse orthonormal basis and consider the angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{I})}\) .

Then the coordinates of \( \overrightarrow{I}\) and \( \overrightarrow{J}\) in the canonical basis are the following:

• \( [\overrightarrow{I}]_{B_0}=\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}\) ,

• and \( [\overrightarrow{J}]_{B_0}=\begin{bmatrix} \cos\left(\theta-\frac{\pi}{2}\right)\\ \sin\left(\theta-\frac{\pi}{2}\right)\end{bmatrix}= \begin{bmatrix}\sin(\theta)\\{-\cos(\theta)}\end{bmatrix}\) .

Consequently, the following assertions are fulfilled.

• The transition matrix from the basis \( B\) to the canonical basis is equal to \( P=\begin{bmatrix} \cos(\theta)&\sin(\theta)\\ \sin(\theta)&-\cos(\theta) \end{bmatrix}\) .

• And the transition matrix from the canonical basis to the basis \( B\) is equal to:

  \( P^{-1} =\frac{1}{\cos^{2}(\theta)+\sin^{2}(\theta)} \begin{bmatrix} \cos(\theta)&\sin(\theta)\\\sin(\theta)&-\cos(\theta) \end{bmatrix} =\begin{bmatrix} \cos(\theta)&\sin(\theta)\\\sin(\theta)&-\cos(\theta) \end{bmatrix} =P\)

  \( P\) is thus idempotent, and \( P^{2}=I\) the identity matrix.

This is because \( \left\| \overrightarrow{u}\right\|=\sqrt{\overrightarrow{u}\cdot\overrightarrow{u}}\) .

We shall apply the theorem 10 to the formula \( \overrightarrow{u}\cdot\overrightarrow{v}= \left\| \overrightarrow{u} \right\| \left\| \overrightarrow{v} \right\|\cos(\theta)\) .

Assume that \( (\overrightarrow{u},\overrightarrow{v})\in(\mathbb{P}^*)^2\) are non zero vectors and consider their angle \( \theta=\widehat{(\overrightarrow{u},\overrightarrow{v})}\) .

Consider the orthonormal basis \( B=(\overrightarrow{I},\overrightarrow{J})\) such that:

• \( \overrightarrow{I}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) ,

• and, if \( \overrightarrow{I}=\begin{bmatrix}I_1\\I_2\end{bmatrix}\) , then \( \overrightarrow{J}=\begin{bmatrix}-I_2\\I_1\end{bmatrix}\) .

Then, in the basis \( B\) , we have:

• \( [\overrightarrow{u}]_B=\begin{bmatrix}\left\| \overrightarrow{u} \right\|\\ \\0\end{bmatrix}\) ,

• and \( [\overrightarrow{v}]_B=\begin{bmatrix} \left\| \overrightarrow{v} \right\|\cos(\theta)\\ \\ \left\| \overrightarrow{v}\right\|\sin(\theta)\end{bmatrix}\) .

Consequently, \( \overrightarrow{u}\cdot\overrightarrow{v}= \left\| \overrightarrow{u} \right\| \left\| \overrightarrow{v} \right\|\cos(\theta)\) . QED

Assume that \( \alpha\in\mathbb{R}\) is a real number, and consider the rotation \( \rho_{\alpha}\) of angle \( \alpha\) .

Then the matrix of \( \rho_{\alpha}\) in the canonical basis \( B_{0}\) is \( A_{\alpha}= \begin{bmatrix}\cos(\alpha)&-\sin(\alpha)\\ \sin(\alpha)&\cos(\alpha) \end{bmatrix}\) .

Assume that \( B=(\overrightarrow{I},\overrightarrow{J})\) is a direct orthonormal basis, and consider the angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{I})}\) between the first vectors of the bases \( B_{0}\) and \( B\) .

Then the transition matrix from the basis \( B\) to the canonical basis \( B_{0}\) is \( P=\begin{bmatrix} \cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta) \end{bmatrix}\) , and its inverse is \( P^{-1}=\begin{bmatrix} \cos(\theta)&\sin(\theta)\\{-\sin(\theta)}&\cos(\theta) \end{bmatrix}\) .

Consequently, the matrix of \( \rho_{\alpha}\) in the basis \( B\) is equal to \( B_{\alpha}=P^{-1}A_{\alpha}P\) .

Let’s prove that \( B_{\alpha}=A_{\alpha}\) .

\( B_{\alpha}=\begin{bmatrix} \cos(\theta)&\sin(\theta)\\{-\sin(\theta)}&\cos(\theta) \end{bmatrix} \begin{bmatrix}\cos(\alpha)&{-\sin(\alpha)} \\\sin(\alpha)&\cos(\alpha) \end{bmatrix} \begin{bmatrix} \cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta) \end{bmatrix}\)

\( =\begin{bmatrix} \cos(\theta)&\sin(\theta)\\{-\sin(\theta)}&\cos(\theta) \end{bmatrix}\) \( \begin{bmatrix} \cos(\alpha)\cos(\theta)-\sin(\alpha)\sin(\theta)& -\cos(\alpha)\sin(\theta)-\sin(\alpha)\cos(\theta) \\ \sin(\alpha)\cos(\theta)+\cos(\alpha)\sin(\theta)& -\sin(\alpha)\sin(\theta) +\cos(\alpha)\cos(\theta) \end{bmatrix}\)

\( =\begin{bmatrix} \cos(\theta)&\sin(\theta)\\{-\sin(\theta)}&\cos(\theta) \end{bmatrix} \begin{bmatrix} \cos(\alpha+\theta)&-\sin(\alpha+\theta) \\ \sin(\alpha+\theta)&\cos(\alpha+\theta) \end{bmatrix}\)

\( =\begin{bmatrix} \cos(\theta)\cos(\alpha+\theta) +\sin(\theta)\sin(\alpha+\theta)& -\cos(\theta)\sin(\alpha+\theta) +\sin(\theta)\cos(\alpha+\theta) \\ -\sin(\theta)\cos(\alpha+\theta) +\cos(\theta)\sin(\alpha+\theta)& \sin(\theta)\sin(\alpha+\theta) +\cos(\theta)\cos(\alpha+\theta) \end{bmatrix}\)

\( =\begin{bmatrix} \cos(\theta-(\alpha+\theta)) & \sin(\theta-(\alpha+\theta) \\ \sin((\alpha+\theta)-\theta) & \cos(\theta-(\alpha+\theta)) \end{bmatrix}\)

\( =\begin{bmatrix}\cos(-\alpha)&\sin(-\alpha) \\\sin(\alpha)&\cos(-\alpha) \end{bmatrix} =\begin{bmatrix}\cos(\alpha)&-\sin(\alpha) \\\sin(\alpha)&\cos(\alpha) \end{bmatrix}=A_{\alpha}\) QED

Consider the canonical basis \( B_0=(\overrightarrow{i},\overrightarrow{j})\) , and consider the bisector \( (B)\) of the axes, of equation \( y=x\) in \( B_{0}\) .

Consider the reflexion \( \sigma_0\) in the line \( (B)\) .

Then \( \sigma_0\) exchanges \( \overrightarrow{i}\) and \( \overrightarrow{j}\) .

Consequently, the matrix of \( \sigma_0\) in the canonical basis is \( S_0=\begin{bmatrix}0&1\\1&0 \end{bmatrix}\) .

Consider the straight line \( (D)\) directed by the unit vector \( \overrightarrow{u}_0\) , of angle \( \theta\) with the vector \( \overrightarrow{i}\) .

Consider the basis \( B=(\overrightarrow{I},\overrightarrow{J})\) where:

• \( \overrightarrow{I}\) is the unit vector with angle \( \theta-\frac{\pi}{4}\) with the vector \( \overrightarrow{i}\) ,

• and \( \overrightarrow{J}\) is the unit vector with angle \( \theta+\frac{\pi}{4}\) with the vector \( \overrightarrow{i}\)

Then the basis \( B=(\overrightarrow{I},\overrightarrow{J})\) is a direct orthonormal basis, and the transition matrix from the basis \( B\) to the canonical basis is \( P=\begin{bmatrix} \cos(\theta-\frac{\pi}{4})&-\sin(\theta-\frac{\pi}{4})\\ \sin(\theta-\frac{\pi}{4})&\cos(\theta-\frac{\pi}{4}) \end{bmatrix}\) , with inverse \( P^{-1}=\begin{bmatrix} \cos(\theta-\frac{\pi}{4})&\sin(\theta-\frac{\pi}{4})\\ -\sin(\theta-\frac{\pi}{4})&\cos(\theta-\frac{\pi}{4}) \end{bmatrix}\) .

Moreover, the line \( (D)\) is the bisector of the axes of the orthonormal basis \( B=(\overrightarrow{I},\overrightarrow{J})\) , and consequently the matrix, in the basis \( B\) , of the reflection \( \sigma\) in the line \( (D)\) is \( S_0=\begin{bmatrix}0&1\\1&0 \end{bmatrix}\) .

Let’s denote \( S\) the matrix in the canonical basis of the reflection \( \sigma\) in the line \( (D)\) .

Then \( S_0=P^{-1}SP\) , so that \( S=PS_0P^{-1}\) .

Let’s prove that \( S=\begin{bmatrix} -\cos(2\theta)&\sin(2\theta)\\ \sin(2\theta)&\cos(2\theta) \end{bmatrix}\) .

\( S=PS_0P^{-1}\)

\( =\begin{bmatrix} \cos(\theta-\frac{\pi}{4})&\sin(\theta-\frac{\pi}{4})\\ -\sin(\theta-\frac{\pi}{4})&\cos(\theta-\frac{\pi}{4}) \end{bmatrix} \begin{bmatrix}0&1\\1&0 \end{bmatrix} \begin{bmatrix} \cos(\theta-\frac{\pi}{4})&-\sin(\theta-\frac{\pi}{4})\\ \sin(\theta-\frac{\pi}{4})&\cos(\theta-\frac{\pi}{4}) \end{bmatrix}\)

\( =\begin{bmatrix} \cos(\theta-\frac{\pi}{4})&\sin(\theta-\frac{\pi}{4})\\ -\sin(\theta-\frac{\pi}{4})&\cos(\theta-\frac{\pi}{4}) \end{bmatrix} \begin{bmatrix} \sin(\theta-\frac{\pi}{4})&\cos(\theta-\frac{\pi}{4})\\ \cos(\theta-\frac{\pi}{4})&-\sin(\theta-\frac{\pi}{4}) \end{bmatrix}\)

Let’s denote \( \alpha=\theta-\frac{\pi}{4}\) .

Then we have:

\( S=\begin{bmatrix} \cos(\alpha)&\sin(\alpha)\\ -\sin(\alpha)&\cos(\alpha)) \end{bmatrix} \begin{bmatrix} \sin(\alpha)&\cos(\alpha)\\ \cos(\alpha)&-\sin(\alpha) \end{bmatrix}\)

\( =\begin{bmatrix} 2\cos(\alpha)\sin(\alpha)&\cos^2(\alpha)-\sin^2(\alpha)\\ \cos^2(\alpha)-\sin^2(\alpha)&-2\cos(\alpha)\sin(\alpha) \end{bmatrix}\)

\( =\begin{bmatrix} \sin(2\alpha)&\cos(2\alpha)\\ \cos(2\alpha)&-\sin(2\alpha) \end{bmatrix}\)

But \( 2\alpha=2\theta-\frac{\pi}{2}\) , so that:

\( S=\begin{bmatrix} \sin(2\theta-\frac{\pi}{2})&\cos(2\theta-\frac{\pi}{2})\\ \cos(2\theta-\frac{\pi}{2})&-\sin(2\theta-\frac{\pi}{2}) \end{bmatrix} =\begin{bmatrix} -\cos(2\theta)&\sin(2\theta)\\ \sin(2\theta)&\cos(2\theta) \end{bmatrix}\) QED

Assume now that the line \( (D)\) has equation \( ax+by=0\) .

Then it is directed by the unit vector \( \overrightarrow{u_0} =\begin{bmatrix} \frac{b}{\sqrt{a^2+b^2}}\\ -\frac{a}{\sqrt{a^2+b^2}}\end{bmatrix}\) , and the angle \( \theta\) is such that \( \overrightarrow{u_0}=\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}\) .

Consequently, \( \cos(\theta)=\frac{b}{\sqrt{a^2+b^2}}\) and \( \sin(\theta)=-\frac{a}{\sqrt{a^2+b^2}}\) .

But with the trigonometric formulae:

\( S=\begin{bmatrix} -\cos^2(\theta)+\sin^2(\theta)&2\sin(\theta)\cos(\theta)\\ 2\sin(\theta)\cos(\theta)&\cos^2(\theta)-\sin^2(\theta) \end{bmatrix}\) .

So that \( S=\begin{bmatrix} \frac{-{a^2+b^2}}{a^2+b^2}&\frac{2ab}{a^2+b^2}\\ \\ \frac{2ab}{a^2+b^2}&\frac{{a^2-b^2}}{a^2+b^2} \end{bmatrix}\) .

QED