Proof

Assume that \( f:\;\mathbb{P}\;\rightarrow\;\mathbb{P}\) and \( g:\;\mathbb{P}\;\rightarrow\;\mathbb{P}\) are linear mappings in the euclidean plane.

Consider the sum \( s=f+g\) of \( f\) and \( g\) .

Assume that \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^{2}\) are vectors and that \( \lambda\in\mathbb{R}\) is a scalar.

Then, because \( f\) and \( g\) are linear mappings, the following calculations may be performed.

• As the addition of vectors is commutative and associative, we have:

  \( s(\overrightarrow{u}+\overrightarrow{v}) =f(\overrightarrow{u}+\overrightarrow{v})+g(\overrightarrow{u}+\overrightarrow{v}) =(f(\overrightarrow{u})+f(\overrightarrow{v}))+(g(\overrightarrow{u})+g(\overrightarrow{v})) =(f(\overrightarrow{u})+g(\overrightarrow{u}))+(f(\overrightarrow{v})+g(\overrightarrow{v})) =s(\overrightarrow{u})+s(\overrightarrow{v})\)

• As the multiplication by a scalar is distributive on the addition of vectors, we have:

  \( s(\lambda\overrightarrow{u}) =f(\lambda\overrightarrow{u})+g(\lambda\overrightarrow{u}) =\lambda f(\overrightarrow{u})+\lambda g(\overrightarrow{u}) =\lambda(f(\overrightarrow{u})+g(\overrightarrow{u})) =\lambda s(\overrightarrow{u})\)

As this is true for any vectors \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^{2}\) and any scalar \( \lambda\in\mathbb{R}\) , \( s\) is a linear mapping.

Proof

Assume that the linear mapping \( f:\;\mathbb{P}\;\rightarrow\;\mathbb{P}\) has the matrix \( A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\) in the canonical basis.

Assume that the linear mapping \( g:\;\mathbb{P}\;\rightarrow\;\mathbb{P}\) has the matrix \( B=\begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{bmatrix}\) in the canonical basis.

Then the first column of the matrix \( S=A+B\) is equal to:

\( C_1=\begin{bmatrix} a_{11}+b_{11}\\a_{21}+b_{21} \end{bmatrix} =\begin{bmatrix} a_{11}\\a_{21} \end{bmatrix}+\begin{bmatrix} b_{11}\\b_{21} \end{bmatrix}\) .

It is the column vector of the coordinates of \( f(\overrightarrow{i})+g(\overrightarrow{i})=s(\overrightarrow{i})\) .

And the second column of the matrix \( S=A+B\) is equal to:

\( C_2=\begin{bmatrix} a_{12}+b_{12}\\a_{22}+b_{22} \end{bmatrix} =\begin{bmatrix} a_{12}\\a_{22} \end{bmatrix}+\begin{bmatrix} b_{12}\\b_{22} \end{bmatrix}\) .

It is the column vector of the coordinates of \( f(\overrightarrow{j})+g(\overrightarrow{j})=s(\overrightarrow{j})\) .

Consequently, \( S\) is the matrix of \( s\) in the canonical basis.

Proof (of the lemma 1)

Assume that \( A=\begin{bmatrix} a_{11}&a_{12}\\a_{21}&a_{22} \end{bmatrix}\) is a matrix with real elements.

1. If there exists a linear mapping \( f\) in the euclidean plane having \( A\) as its matrix in the canonical basis, then it is unique.

   Indeed, the first column \( C_1=\begin{bmatrix} a_{11}\\a_{21} \end{bmatrix}\) of \( A\) is the column vector of the cartesian coordinates of the image \( f(\overrightarrow{i})\) of the first vector \( \overrightarrow{i}\) of the canonical basis, and the second column \( C_2=\begin{bmatrix} a_{12}\\a_{22} \end{bmatrix}\) of \( A\) is the column vector of the cartesian coordinates of the image \( f(\overrightarrow{j})\) of the second vector \( \overrightarrow{j}\) of the canonical basis.

   So \( f(\overrightarrow{i})\) and \( f(\overrightarrow{j})\) are known.

   Moreover, for any vector \( \overrightarrow{u}\in\mathbb{P}\) with cartesian coordinates \( (x_{1},x_{2})\) , \( f(\overrightarrow{u})=x_{1}f(\overrightarrow{i})+x_{2}f(\overrightarrow{j})\) is also known.

   So that the knowledge of \( A\) gives the knowledge of \( f\) , that is thus unique.

2. Let’s now build \( f\) .

   Consider the vector \( \overrightarrow{I}\) with column vector of coordinates the first column \( C_1=\begin{bmatrix} a_{11}\\a_{21} \end{bmatrix}\) of \( A\) , and the vector \( \overrightarrow{J}\) with column vector of coordinates the second column \( C_2=\begin{bmatrix} a_{12}\\a_{22} \end{bmatrix}\) of \( A\) .

   Now, define the mapping \( f\) in the euclidean plane such as, for any vector \( \overrightarrow{u}\in\mathbb{P}\) with cartesian coordinates \( (x_{1},x_{2})\) , \( f(\overrightarrow{u})=x_{1}\overrightarrow{I}+x_{2}\overrightarrow{J}\) .

   Then \( f\) is a linear mapping, because of the following statements.

• For any vectors \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^{2}\) , with cartesian coordinates \( (x_{1},x_{2})\) and \( (y_{1},y_{2})\) we have:

  \( f(\overrightarrow{u}+\overrightarrow{v})=(x_{1}+y_{1})\overrightarrow{I}+(x_{2}+y_{2})\overrightarrow{J} =(x_{1}\overrightarrow{I}+y_{1}\overrightarrow{I})+(x_{2}\overrightarrow{J}+y_{2}\overrightarrow{J})\)

  \( =(x_{1}\overrightarrow{I}+x_{2}\overrightarrow{J})+(y_{1}\overrightarrow{I}+y_{2}\overrightarrow{J}) =f(\overrightarrow{u})+f(\overrightarrow{v})\)

• And for any vector \( \overrightarrow{u}\in\mathbb{P}\) with cartesian coordinates \( (x_{1},x_{2})\) , and for any scalar \( \lambda\in\mathbb{R}\) , we have:

  \( f(\lambda\overrightarrow{u})=(\lambda x_{1})\overrightarrow{I}+(\lambda x_{2})\overrightarrow{J} =\lambda (x_{1}\overrightarrow{I})+\lambda (x_{2}\overrightarrow{J})\)

  \( =\lambda (x_{1}\overrightarrow{I}+x_{2}\overrightarrow{J}) =\lambda f(\overrightarrow{u})\)

Proof (of the lemma 2)

Assume that the linear mapping \( f:\;\mathbb{P}\;\rightarrow\;\mathbb{P}\) has the matrix \( A=\begin{bmatrix} a_{11}&a_{12}\\a_{21}&a_{22} \end{bmatrix}\) in the canonical basis.

Then \( -f\) is a linear mapping, because of the following statements.

• For any vectors \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^{2}\) , with cartesian coordinates \( (x_{1},x_{2})\) and \( (y_{1},y_{2})\) we have:

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

• And for any vector \( \overrightarrow{u}\in\mathbb{P}\) with cartesian coordinates \( (x_{1},x_{2})\) , and for any scalar \( \lambda\in\mathbb{R}\) , we have:

  \( (-f)(\lambda\overrightarrow{u})=-f(\lambda\overrightarrow{u}) =-(\lambda f(\overrightarrow{u}))=\lambda (-f(\overrightarrow{u}))=\lambda (-f)(\overrightarrow{u})\)

Moreover we have:

• the first column \( C_1=\begin{bmatrix} a_{11}\\a_{21} \end{bmatrix}\) of \( A\) is the column vector of the cartesian coordinates of \( f(\overrightarrow{i})\) , so that the first column \( -C_1=\begin{bmatrix} -a_{11}\\{-a_{21}} \end{bmatrix}\) of \( -A\) is the column vector of the cartesian coordinates of \( -f(\overrightarrow{i})=(-f)(\overrightarrow{i})\) ,

• and the second column \( C_2=\begin{bmatrix} a_{12}\\a_{22} \end{bmatrix}\) of \( A\) is the column vector of the cartesian coordinates of \( f(\overrightarrow{j})\) , so that the second column \( -C_2=\begin{bmatrix} -a_{12}\\{-a_{22}} \end{bmatrix}\) of \( -A\) is the column vector of the cartesian coordinates of \( -f(\overrightarrow{j})=(-f)(\overrightarrow{j})\) .

Consequently, \( -A\) is the matrix of \( -f\) in the canonical basis.

Proof (of the corollary 2)

Assume that \( A\) , \( B\) and \( C\) are \( 2\times 2\) matrices with real elements.

Consider the linear mappings \( f\) , \( g\) and \( h\) having \( A\) , \( B\) and \( C\) as matrices in the canonical basis (cf. the lemma 1).

Then, because of the theorems 4 and 5, we have:

1. \( A+B\) is the matrix of \( f+g\) in the canonical basis, and \( B+A\) is the matrix of \( g+f=f+g\) in the canonical basis. Consequently, \( A+B=B+A\) .

2. \( (A+B)+C\) is the matrix of \( (f+g)+h\) in the canonical basis, and \( A+(B+C)\) is the matrix of \( f+(g+h)=(f+g)+h\) in the canonical basis. Consequently, \( (A+B)+C=A+(B+C)\) .

3. \( A+O_{22}\) is the matrix of \( f+o=f\) in the canonical basis, and \( O_{22}+A\) is the matrix of \( o+f=f\) in the canonical basis.

   Consequently, \( A+O_{22}=O_{22}+A=A\) .

4. Because of the lemma 2, \( -A\) is the matrix of \( -f\) in the canonical basis.

   Consequently, \( A+(-A)\) is the matrix of \( f+(-f)=o\) in the canonical basis, and \( (-A)+A\) is the matrix of \( (-f)+f=o\) in the canonical basis.

   Consequently, \( A+(-A)=(-A)+A=O_{22}\) .

Proof (of the theorem 5)

Assume that \( f\) , \( g\) and \( h\) are linear mappings in the euclidean plane.

Assume that \( \overrightarrow{u}\in\mathbb{P}\) is a vector in the euclidean plane.

Then the following calculations may be performed.

1. Because of the commutativity of the addition of vectors, we have:

   \( (f+g)(\overrightarrow{u})=f(\overrightarrow{u})+g(\overrightarrow{u}) =g(\overrightarrow{u})+f(\overrightarrow{u}) =(g+f)(\overrightarrow{u})\) .

   As this is true for any vector \( \overrightarrow{u}\) , \( f+g=g+f\) .

2. Because of the associativity of the addition of vectors, we have:

   \( ((f+g)+h)(\overrightarrow{u})=(f+g)(\overrightarrow{u})+h(\overrightarrow{u})\)

   \( =(f(\overrightarrow{u})+g(\overrightarrow{u}))+h(\overrightarrow{u}) =f(\overrightarrow{u})+(g(\overrightarrow{u})+h(\overrightarrow{u}))\)

   \( =f(\overrightarrow{u})+(g+h)(\overrightarrow{u}) =(f+(g+h))(\overrightarrow{u})\) .

   As this is true for any vector \( \overrightarrow{u}\) , \( (f+g)+h=f+(g+h)\) .

3. As the null vextor \( \overrightarrow{0}\) is neutral for the addition of vectors, we have:

   \( (f+o)(\overrightarrow{u})=f(\overrightarrow{u})+\overrightarrow{0}=f(\overrightarrow{u})\) .

   And also:

   \( (o+f)(\overrightarrow{u})=\overrightarrow{0}+f(\overrightarrow{u})=f(\overrightarrow{u})\) .

   As this is true for any vector \( \overrightarrow{u}\) , \( f+o=o+f=f\) .

4. \( -f\) is a linear mapping because of the lemma 2.

   Moreover, because the opposite \( -\overrightarrow{v}\) of a vector \( \overrightarrow{v}\) is such that

   \( \overrightarrow{v}+(-\overrightarrow{v})=(-\overrightarrow{v})+\overrightarrow{v}=\overrightarrow{0}\) , we have:

   \( (f+(-f))(\overrightarrow{u})=f(\overrightarrow{u})+(-f)(\overrightarrow{u}) =f(\overrightarrow{u})+(-f(\overrightarrow{u}))=\overrightarrow{0}\) ,

   and:

   \( ((-f)+f)(\overrightarrow{u})=(-f)(\overrightarrow{u})+f(\overrightarrow{u}) =(-f(\overrightarrow{u}))+f(\overrightarrow{u})=\overrightarrow{0}\)

Proof

Assume that \( f:\;\mathbb{P}\;\rightarrow\;\mathbb{P}\) and \( g:\;\mathbb{P}\;\rightarrow\;\mathbb{P}\) are linear mappings in the euclidean plane.

Consider the difference \( d=f-g\) of \( f\) and \( g\) .

Assume that \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^{2}\) are vectors and that \( \lambda\in\mathbb{R}\) is a scalar.

Then, because \( f\) and \( -g\) are linear mappings, the following calculations may be performed.

• As the addition of vectors is commutative and associative and as \( \overrightarrow{u}_{1}-\overrightarrow{u}_{2}=\overrightarrow{u}_{1}+(-\overrightarrow{u}_{2})\) , we have:

  \( d(\overrightarrow{u}+\overrightarrow{v}) =f(\overrightarrow{u}+\overrightarrow{v})-g(\overrightarrow{u}+\overrightarrow{v})\)

  \( =f(\overrightarrow{u}+\overrightarrow{v})+(-g(\overrightarrow{u}+\overrightarrow{v}))\)

  \( =(f(\overrightarrow{u})+f(\overrightarrow{v}))+((-g)(\overrightarrow{u})+g(\overrightarrow{v}))\)

  \( =(f(\overrightarrow{u})+f(\overrightarrow{v}))+((-g)(\overrightarrow{u})+(-g)(\overrightarrow{v}))\)

  \( =(f(\overrightarrow{u})+f(\overrightarrow{v}))+((-g(\overrightarrow{u}))+(-g(\overrightarrow{v})))\)

  \( =(f(\overrightarrow{u})+(-g(\overrightarrow{u})))+(f(\overrightarrow{v})+(-g(\overrightarrow{v})))\)

  \( =(f(\overrightarrow{u})-g(\overrightarrow{u}))+(f(\overrightarrow{v})-g(\overrightarrow{v})) =d(\overrightarrow{u})+d(\overrightarrow{v})\)

• As the multiplication by a scalar is distributive on the addition of vectors, we have:

  \( d(\lambda\overrightarrow{u}) =f(\lambda\overrightarrow{u})-g(\lambda\overrightarrow{u}) =f(\lambda\overrightarrow{u})+(-g)(\lambda\overrightarrow{u}) =\lambda f(\overrightarrow{u})+\lambda (-g)(\overrightarrow{u}))\)

  \( =\lambda(f(\overrightarrow{u})+(-g)(\overrightarrow{u})) =\lambda(f(\overrightarrow{u})+(-g(\overrightarrow{u}))) =\lambda(f(\overrightarrow{u})-g(\overrightarrow{u})) =\lambda d(\overrightarrow{u})\)

As this is true for any vectors \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^{2}\) and any scalar \( \lambda\in\mathbb{R}\) , \( d\) is a linear mapping.

Proof

Assume that the linear mapping \( f:\;\mathbb{P}\;\rightarrow\;\mathbb{P}\) has the matrix \( A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\) in the canonical basis.

Assume that the linear mapping \( g:\;\mathbb{P}\;\rightarrow\;\mathbb{P}\) has the matrix \( B=\begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{bmatrix}\) in the canonical basis.

Then the first column of the matrix \( D=A-B\) is equal to:

\( C_1=\begin{bmatrix} a_{11}-b_{11}\\a_{21}-b_{21} \end{bmatrix} =\begin{bmatrix} a_{11}\\a_{21} \end{bmatrix}-\begin{bmatrix} b_{11}\\b_{21} \end{bmatrix}\) .

It is the column vector of the coordinates of \( f(\overrightarrow{i})-g(\overrightarrow{i})=d(\overrightarrow{i})\) .

And the second column of the matrix \( D=A-B\) is equal to:

\( C_2=\begin{bmatrix} a_{12}-b_{12}\\a_{22}-b_{22} \end{bmatrix} =\begin{bmatrix} a_{12}\\a_{22} \end{bmatrix}-\begin{bmatrix} b_{12}\\b_{22} \end{bmatrix}\) .

It is the column vector of the coordinates of \( f(\overrightarrow{j})-g(\overrightarrow{j})=d(\overrightarrow{j})\) .

Consequently, \( D\) is the matrix of \( d\) in the canonical basis.

Proof (of the corollary 3)

Assume that \( A\) is a \( 2\times 2\) matrix with real elements.

Consider the linear mapping \( f\) having \( A\) as matrix in the canonical basis (cf. the lemma 1).

Note that the linear mapping having \( O_{22}\) as matrix in the canonical basis is the null mapping \( o\)

Then, because of the theorems 7 and 8, and the lemma 2, we have:

1. \( A-O_{22}\) is the matrix of \( f-o=f\) in the canonical basis, and \( f\) has the matrix \( A\) in the canonical basis. Consequently, \( A-O_{22}=A\) .

2. \( O_{22}-A\) is the matrix of \( o-f=-f\) in the canonical basis, and \( -f\) has the matrix \( -A\) in the canonical basis. Consequently, \( O_{22}-A=-A\) .

3. \( A-A\) is the matrix of \( f-f=o\) in the canonical basis, and \( o\) has the matrix \( O_{22}\) in the canonical basis. Consequently, \( A-A=O_{22}\) .

Proof (of the theorem 8)

Assume that \( f\) is a linear mapping in the euclidean plane.

Assume that \( \overrightarrow{u}\in\mathbb{P}\) is a vector in the euclidean plane.

Then the following calculations may be performed.

1. Because the null vector is neutral for the addition of vectors and is equal to its opposite, we have:

   \( (f-o)(\overrightarrow{u})=f(\overrightarrow{u})-\overrightarrow{0} =f(\overrightarrow{u})+(-\overrightarrow{0}) =f(\overrightarrow{u})-\overrightarrow{0}=f(\overrightarrow{u})\) .

   As this is true for any vector \( \overrightarrow{u}\) , \( f-o=f\) .

2. Because the null vector is neutral for the addition of vectors, we have:

   \( (o-f)(\overrightarrow{u})=\overrightarrow{0}-f(\overrightarrow{u}) =\overrightarrow{0}+(-f(\overrightarrow{u})) =\overrightarrow{0}+(-f)(\overrightarrow{u})=(-f)(\overrightarrow{u})\) .

   As this is true for any vector \( \overrightarrow{u}\) , \( o-f=-f\) .

3. Because a vector minus itsels is equal to the null vector, we have:

   \( (f-f)(\overrightarrow{u})=f(\overrightarrow{u})-f(\overrightarrow{u}) =\overrightarrow{0}=o(\overrightarrow{u})\) .

   As this is true for any vector \( \overrightarrow{u}\) , \( f-f=o\) .

Proof (of the corollary 4)

Assume that \( A\) and \( B\) are \( 2\times 2\) matrices with real elements.

Consider the linear mappings \( f\) and \( g\) having respectively \( A\) and \( B\) as matrices in the canonical basis (cf. the lemma 1).

Then, because of the theorems 4, 7 and 9, we have:

1. \( A-B\) is the matrix of \( f-g\) in the canonical basis, and \( A+(-B)\) is the matrix of \( f+(-g)=f-g\) in the canonical basis. Consequently, \( A-B=A+(-B)\) .

2. Because of the previous item, the corollary 3 and the associativity of the addition of the \( 2\times 2\) matrices with real elements, we have:

1. \( (A+B)-B=(A+B)+(-B)=A+(B+(-B))=A+O_{22}=A\) ,

2. and \( (A-B)+B=(A+(-B))+B=A+((-B)+B)=A+O_{22}=A\)

Proof (of the theorem 9)

Assume that \( f\) and \( g\) are linear mappings in the euclidean plane.

Assume that \( \overrightarrow{u}\in\mathbb{P}\) is a vector in the euclidean plane.

Then the following calculations may be performed.

1. Because subtracting a vector is adding its opposite,

   \( (f-g)(\overrightarrow{u})=f(\overrightarrow{u})-g(\overrightarrow{u}) =f(\overrightarrow{u})+(-g(\overrightarrow{u})) =f(\overrightarrow{u})+(-g)(\overrightarrow{u}) =(f+(-g))(\overrightarrow{u})\) .

   As this is true for any vector \( \overrightarrow{u}\) , \( f-g=f+(-g)\) .

2. Because of the previous item, the theorem 8 and the associativity of the addition of the linear mappings, we have:

1. \( (f+g)-g=(f+g)+(-g)=f+(g+(-g))=f+o=f\)

2. \( (f-g)+g=(f+(-g))+g)=f+((-g)+g)=f+o=f\)

Proof

Assume that \( f:\;\mathbb{P}\;\rightarrow\;\mathbb{P}\) is a linear mapping in the euclidean plane and that \( \lambda\in\mathbb{R}\) is a real number.

Consider the product \( p=\lambda f\) of \( f\) by \( \lambda\) .

Assume that \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^{2}\) are vectors and that \( \mu\in\mathbb{R}\) is a scalar.

Then, because \( f\) is a linear mappings, the following calculations may be performed.

• As the multiplication of a vector by a scalar is distributive on the addition of vectors, we have:

  \( p(\overrightarrow{u}+\overrightarrow{v})=\lambda f(\overrightarrow{u}+\overrightarrow{v}) =\lambda(f(\overrightarrow{u})+f(\overrightarrow{v}))\)

  \( =\lambda f(\overrightarrow{u})+\lambda f(\overrightarrow{v}) =p(\overrightarrow{u})+p(\overrightarrow{v})\)

• Because of the associativity law of the multiplication of a vector by a scalar, we have:

  \( p(\mu\overrightarrow{u})=\lambda f(\mu\overrightarrow{u}) =\lambda (\mu f(\overrightarrow{u})) =(\lambda\mu) f(\overrightarrow{u})=(\mu\lambda) f(\overrightarrow{u})\)

  \( =\mu(\lambda f(\overrightarrow{u}))=\mu p(\overrightarrow{u}))\)

As this is true for any vectors \( (\overrightarrow{u},\overrightarrow{v})\in\mathbb{P}^{2}\) and any scalar \( \mu\in\mathbb{R}\) , \( p\) is a linear mapping.

Proof

Assume that the linear mapping \( f:\;\mathbb{P}\;\rightarrow\;\mathbb{P}\) has the matrix \( A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\) in the canonical basis.

Assume that \( \lambda\in\mathbb{R}\) is a real number.

Then the first column of the matrix \( P=\lambda A\) is equal to:

\( C_1=\begin{bmatrix} \lambda a_{11}\\\lambda a_{21} \end{bmatrix} =\lambda \begin{bmatrix} a_{11}\\a_{21} \end{bmatrix}\) .

It is the column vector of the coordinates of \( \lambda f(\overrightarrow{i})=p(\overrightarrow{i})\) .

And the second column of the matrix \( P=\lambda A\) is equal to:

\( C_2=\begin{bmatrix} \lambda a_{12}\\\lambda a_{22} \end{bmatrix} =\lambda \begin{bmatrix} a_{12}\\a_{22} \end{bmatrix}\) .

It is the column vector of the coordinates of \( \lambda f(\overrightarrow{j})=p(\overrightarrow{j})\) .

Consequently, \( P\) is the matrix of \( p\) in the canonical basis.

Proof (of the corollary 5)

Assume that \( A\) is a \( 2\times 2\) matrix with real elements.

Consider the linear mapping \( f\) having \( A\) as matrix in the canonical basis (cf. the lemma 1).

Note that the linear mapping having \( O_{22}\) as matrix in the canonical basis is the null mapping \( o\)

Then, because of the theorems 11 and 12, we have:

1. \( 0\times A\) is the matrix of \( 0\times f=o\) in the canonical basis, and \( o\) has the matrix \( O_{22}\) in the canonical basis. Consequently, \( 0\times A=O_{22}\) .

2. \( 1\times A\) is the matrix of \( 1\times f=f\) in the canonical basis, and \( f\) has the matrix \( A\) in the canonical basis. Consequently, \( 1\times A=A\) .

Proof (of the theorem 12)

Assume that \( f\) is a linear mapping in the euclidean plane.

Assume that \( \overrightarrow{u}\in\mathbb{P}\) is a vector in the euclidean plane.

Then the following calculations may be performed.

1. Because \( 0\) multipliplied by any vector is equal to the null vector, we have:

   \( (0\times f)(\overrightarrow{u})=0\times (f(\overrightarrow{u})) \overrightarrow{0}=o(\overrightarrow{u})\) .

   As this is true for any vector \( \overrightarrow{u}\) , \( 0\times f=o\) .

2. Because \( 1\) multipliplied by any vector is equal to that vector, we have:

   \( (1\times f)(\overrightarrow{u})=1\times (f(\overrightarrow{u})) =f(\overrightarrow{u})\) .

   As this is true for any vector \( \overrightarrow{u}\) , \( 1\times f=f\) .

Proof (of the corollary 4)

Assume that \( A\) and \( B\) are \( 2\times 2\) matrices with real elements, and that \( (\lambda,\mu)\in\mathbb{R}^2\) are real numbers.

Consider the linear mappings \( f\) and \( g\) having respectively \( A\) and \( B\) as matrices in the canonical basis (cf. the lemma 1).

Then, because of the theorems 4, 7 and 11, we have:

1. \( (\lambda+\mu) A\) is the matrix of \( (\lambda+\mu) f\) in the canonical basis, and \( \lambda A+\mu A\) is the matrix of \( \lambda f+\mu f=(\lambda+\mu) f\) in the canonical basis. Consequently, \( (\lambda+\mu) A=\lambda A+\mu A\) .

2. \( \lambda(A+B)\) is the matrix of \( \lambda(f+g)\) in the canonical basis, and \( \lambda A + \lambda B=\lambda(f+g)\) is the matrix of \( \lambda f + \lambda g\) in the canonical basis. Consequently, \( \lambda(A+B)=\lambda A + \lambda B\) .

3. \( \lambda(\mu A)\) is the matrix of \( \lambda(\mu f)\) in the canonical basis, and \( (\lambda \mu) A\) is the matrix of \( (\lambda \mu) f=\lambda(\mu f)\) in the canonical basis. Consequently, \( \lambda(\mu A)=(\lambda\mu) A\) .

Proof (of the theorem 9)

Assume that \( f\) and \( g\) are linear mappings in the euclidean plane, and that \( (\lambda,\mu)\in\mathbb{R}^2\) are real numbers.

Assume that \( \overrightarrow{u}\in\mathbb{P}\) is a vector in the euclidean plane.

Then the following calculations may be performed.

1. Because of the first distributivity law in the euclidean plane \( \mathbb{P}\) ,

   \( ((\lambda+\mu) f)(\overrightarrow{u})=(\lambda+\mu)f(\overrightarrow{u}) =\lambda f(\overrightarrow{u})+\mu f(\overrightarrow{u}) =(\lambda f)(\overrightarrow{u})+(\mu f)(\overrightarrow{u})\) .

   As this is true for any vector \( \overrightarrow{u}\) , \( (\lambda+\mu) f=\lambda f+\mu f\) .

2. Because of the second distributivity law in the euclidean plane \( \mathbb{P}\) ,

   \( (\lambda(f+ g))(\overrightarrow{u})=\lambda(f+g)(\overrightarrow{u}) =\lambda (f(\overrightarrow{u})+g(\overrightarrow{u})) =\lambda f(\overrightarrow{u})+\lambda g(\overrightarrow{u})) =(\lambda f)(\overrightarrow{u})+(\lambda g)(\overrightarrow{u})\) .

   As this is true for any vector \( \overrightarrow{u}\) , \( \lambda(f+ g)=\lambda f + \lambda g\) .

3. Because of the associativity law in the vector plane \( \mathbb{P}\) ,

   \( (\lambda(\mu f))(\overrightarrow{u})=\lambda(\mu f)(\overrightarrow{u}) =\lambda (\mu f(\overrightarrow{u}))=(\lambda \mu) f(\overrightarrow{u} =((\lambda \mu) f)(\overrightarrow{u}\) .

   As this is true for any vector \( \overrightarrow{u}\) , \( \lambda(\mu f)=(\lambda\mu) f\) .

Proof

Assume that \( (\lambda,\mu)\in\mathbb{R}^2\) are real numbers, and consider the homotheties \( h_{\lambda}\) and \( h_{\mu}\) of factors \( \lambda\) and \( \mu\) .

Assume that \( \overrightarrow{u}\in\mathbb{P}\) is a vector in the euclidean plane.

Then because of the associativity law in the vector plane \( \mathbb{P}\) , we have:

\( (h_{\lambda}\circ h_{\mu})(\overrightarrow{u})=h_{\lambda}(h_{\mu}(\overrightarrow{u})) =\lambda(\mu\overrightarrow{u})=(\lambda\mu)\overrightarrow{u} =h_{\lambda\mu}(\overrightarrow{u})\) .

As this is true for any vector \( \overrightarrow{u}\) , \( h_{\lambda}\circ h_{\mu}=h_{\lambda\mu}\) .

Proof

Assume that \( (\theta,\mu)\in\mathbb{R}^2\) are real numbers, and consider the rotations \( \rho_{\theta}\) and \( \rho_{\mu}\) of angles \( \theta\) and \( \mu\) .

Assume that \( \overrightarrow{u}\in\mathbb{P}\) is a vector in the euclidean plane, and consider the corresponding complex number \( z_{u}\) .

Then the following assertions hold:

• \( \rho_{\mu}(\overrightarrow{u})\) corresponds to the complex number \( e^{i\mu}z_{u}\) .

• Because of the associativity of the multiplication in \( \mathbb{C}\) (that will be proved in the last section of the course), and because \( e^{i\theta}e^{i\mu}=e^{i(\theta+\mu)}\) , \( (\rho_{\theta}\circ\rho_{\mu})(\overrightarrow{u})=\rho_{\theta}(\rho_{\mu}(\overrightarrow{u}))\) corresponds to the complex number: \( e^{i\theta}(e^{i\mu}z_{u})=(e^{i\theta}e^{i\mu})z_{u}=e^{i(\theta+\mu)}z_{u}\) , that corresponds to the vector \( \rho_{(\theta+\mu)}(\overrightarrow{u})\) .

As this is true for any vector \( \overrightarrow{u}\) , \( \rho_{\theta}\circ\rho_{\mu}=\rho_{(\theta+\mu)}\) .

Proof

Assume that \( f\) is a linear mapping in the euclidean plane, and consider the null mapping \( o\) in the euclidean plane.

Assume that \( \overrightarrow{u}\in\mathbb{P}\) is a vector in the euclidean plane.

Then the following assertions hold:

• By definition of the null mapping \( o\) , \( (o\circ f)(\overrightarrow{u})=o(f(\overrightarrow{u}))=\overrightarrow{0}=o(\overrightarrow{u})\) .

  As this is true for any vector \( \overrightarrow{u}\) , \( o\circ f=o\) .

• And by definition of the null mapping \( o\) , together with the fact that \( f(\overrightarrow{0})=\overrightarrow{0}\) , \( (f\circ o)(\overrightarrow{u})=f(o(\overrightarrow{u}))=f(\overrightarrow{0})=\overrightarrow{0}=o(\overrightarrow{u})\) .

  As this is true for any vector \( \overrightarrow{u}\) , \( f\circ o=o\) .

Proof

Assume that \( f\) is a linear mapping in the euclidean plane, and consider the identity mapping \( \text{Id}_{\mathbb{P}}\) in the euclidean plane.

Assume that \( \overrightarrow{u}\in\mathbb{P}\) is a vector in the euclidean plane.

Then the following assertions hold:

• By definition of the identity mapping \( \text{Id}_{\mathbb{P}}\) , \( (\text{Id}_{\mathbb{P}}\circ f)(\overrightarrow{u})=\text{Id}_{\mathbb{P}}(f(\overrightarrow{u})) =f(\overrightarrow{u})\) .

  As this is true for any vector \( \overrightarrow{u}\) , \( \text{Id}_{\mathbb{P}}\circ f=f\) .

• And by definition of the identity mapping \( \text{Id}_{\mathbb{P}}\) again,

  \( (f\circ \text{Id}_{\mathbb{P}})(\overrightarrow{u})=f(\text{Id}_{\mathbb{P}}(\overrightarrow{u})) =f(\overrightarrow{u})\) .

  As this is true for any vector \( \overrightarrow{u}\) , \( f\circ \text{Id}_{\mathbb{P}}=f\) .

Proof (of the theorem 24)

Assume that \( f\;:\;\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping with matrix \( A=\begin{bmatrix} a&b\\c&d \end{bmatrix}\) in the canonical basis, and consider the determinant \( \det(A)=ad-bc\) of the matrix \( A\) .

Then the kernel of \( f\) is the set of vectors with cartesian coordinates solutions of the \( 2\times 2\) linear system:

1. Assume that \( A=O_{22}\) .

   Then \( f=o\) the null mapping, so that \( \ker(f)=\mathbb{P}\) .

   Moreover, \( f\) is not invertible as it is equal to \( o\) .

2. Assume that \( A\ne O_{22}\) and \( \det{A}=0\) .

   Then \( a\) , \( b\) , \( c\) and \( d\) are not zero together, and the solution of the sytem 1 is the following.

• If \( a\) and \( b\) are not zero together, it is the straight line of equation \( ax+by=0\) .

• And if \( a=b=0\) , it is the straight line of equation \( cx+dy=0\) .

3. Assume that \( \det{A}\ne 0\) .

   Then the system 1 has a unique solution and, as \( (0,0)\) is a solution, it is the solution and \( \ker(f)=\{\overrightarrow{0}\}\) .

   Moreover, for any vector \( \overrightarrow{v}\) with cartesian coordinates \( (X,Y)\) , the vectors \( \overrightarrow{u})\) such that \( f(\overrightarrow{u}=\overrightarrow{v}\) are the vectors with cartesian coordinates solutions of the \( 2\times 2\) linear system:

   \[ \begin{equation}\left\{\begin{matrix} ax&+&by&=&X\\ cx&+&dy&=&Y \end{matrix}\right.\end{equation} \]

   (2)

As \( \det{A}\ne 0\) , that system has a unique solution, so that \( f\) is one-to-one from \( \mathbb{P}\) onto \( \mathbb{P}\) .

Consequently, \( f\) is invertible.

Proof (of the theorem 25)

Assume that \( f\;:\;\mathbb{P}\rightarrow\mathbb{P}\) is a linear mapping with matrix \( A=\begin{bmatrix} a&b\\c&d \end{bmatrix}\) in the canonical basis, and consider the determinant \( \det(A)=ad-bc\) of the matrix \( A\) .

Then the range of \( f\) is the set of vectors with cartesian coordinates \( (X,Y)\) such that the following \( 2\times 2\) linear system has at least one solution.

1. Assume that \( A=O_{22}\) .

   Then \( f=o\) the null mapping, so that \( \text{range}(f)=\{\overrightarrow{0}\}\) .

   Moreover, \( f\) is not invertible as it is equal to \( o\) .

2. Assume that \( A\ne O_{22}\) and \( \det{A}=0\) .

   Then \( a\) , \( b\) , \( c\) and \( d\) are not zero together, and that \( ad=bc\) , the system 3 has an infinity of solutions under the following conditions.

• If \( a\ne 0\) and \( c\ne 0\) , then \( \frac{d}{c}=\frac{b}{a}\) , that we denote \( k\) .

  The system 3 becomes then:

  \[ \begin{equation} \left\{\begin{matrix} &a(x&+&ky)&=&X\\ &c(x&+&ky)&=&Y \end{matrix}\right. \end{equation} \]

  (4)

  Consequently, \( cX-aY=0\) .

  Reversely, if \( cX-aY=0\) , then \( cX=aY\) the first equation of 5 is equivalent to \( ca(x+ky)=cX\) and the second equation of 5 is equivalent to \( ac(x+ky)=aY\) , that is the same equation.

  Consequently, the range of \( f\) is the straight line of equation

  \( cx-ay=0\) .

• If \( a\ne 0\) and \( c= 0\) , then \( d=0\) because \( ad=bc=0\) .

  The system 3 becomes then:

  \[ \begin{equation} \left\{\begin{matrix} &ax&=&X\\ &0&=&Y \end{matrix}\right. \end{equation} \]

  (5)

  Consequently, \( Y=0\) .

  Reversely, if \( Y=0\) , then the first equation of 5 is equivalent to \( ax=X\) , that has a solution in \( x\) because \( a\ne 0\) , and the second equation of 5 is equivalent to \( 0=0\) , that is always true.

  Consequently, the range of \( f\) is the \( x\) axis.

• If \( a=c=0\) , \( b\ne 0\) and \( d\ne 0\) and the system 3 becomes:

  \[ \left\{\begin{matrix} &by&=&X\\&dy&=&Y \end{matrix}\right. \]

  That system has at least one solution if and only if \( \frac{X}{b}=\frac{Y}{d}\) , that is equivalent to \( dX-bY=0\) .

  Consequently, the range of \( f\) is the straight line of equation

  \( dx-by=0\) .

• If \( a=c=b=0\) , then \( d\ne 0\) and the system 3 becomes:

  \[ \left\{\begin{matrix} &0&=&X\\&dy&=&Y \end{matrix}\right. \]

  Consequently, \( X=0\) .

  Reversely, if \( X=0\) , then the first equation of 5 is equivalent to \( 0=0\) , that is always true, and the second equation of 5 is equivalent to \( dy=Y\) , that has a solution in \( y\) because \( d\ne 0\) .

  Consequently, the range of \( f\) is the \( y\) axis.

• Moreover, under the conditions \( A\ne O_{22}\) and \( \det{A}=0\) , \( f\) is not invertible, as we saw in the proof of the theorem 24.

3. Assume that \( \det{A}\ne 0\) .

   Then the system 3 has a unique solution for any \( (X,Y)\) , and \( \text{range}(f)=\mathbb{P}\) .

   Moreover, under the condition \( \det{A}\ne 0\) , \( f\) is invertible, as we saw in the proof of the theorem 24.