In that test, you will discover in advance the diagonalization of a matrix through its eigenvalues and eigenvectors.

You will do it for a symmetric matrix, and will discover that you may choose an orthonormal basis where the matrix is diagonal.

In other terms, the transition matrix in that case is a unitary matrix.

From now on, \( (a,b,d)\in\mathbb{R}^3\) are real numbers and you will consider the symmetric matrix \( A=\begin{bmatrix} a&b\\ b&d \end{bmatrix}\) .

You will assume that \( b\ne 0\) , because if \( b=0\) the matrix is already a diagonal matrix.

By definition, an eigenvalue of \( A\) is a scalar\( \lambda\) such that their exists a non zero column vector \( X\) so that \( AX=\lambda X\) .

Question 1: Write the \( 2\times 2\) system of linear equations in the coordinates \( (x,y)\) of \( X\) equivalent to \( AX=\lambda X\) .

Question 2: Transform that system in a system with 0’s on the right hands of the equations.

Question 3: Calculate the determinant \( P(\lambda)\) of the last system.

Develop it as a polynomial in \( \lambda\) .

Question 4: Because of the definition of the eigenvalues, \( \lambda\) is one of them if and only if the system obtained in the question 2 has a solution other than the trivial solution \( (0,0)\) .

For that to be obtained, its determinant \( P(\lambda)\) must be null.

Calculate the discriminant of the quadratic equation \( P(\lambda)=0\) .

Transform it as a sum of square.

Question 5: Deduce that the matrix \( A\) has two distinct eigenvalues \( \lambda_1\) and \( \lambda_2\) . Use the fact that \( b\ne 0\) .

An eigenvector of \( A\) for the eigenvalue \( \lambda_i\) is a non zero solution of the system obtained in the question 2 for \( \lambda=\lambda_i\) .

Question 6:

1. Write the systems for \( \lambda=\lambda_1\) .

2. Find a non zero column solution \( X_1=\begin{bmatrix} x_1\\ y_1 \end{bmatrix}\) of the first equation.

3. prove that it is also a solution of the second equation.

TIPS: Remember that \( b\ne 0\) and that \( P(\lambda_1)=0\) .

Question 7:

1. Write the systems for \( \lambda=\lambda_2\) .

2. Find a non zero column solution \( X_2=\begin{bmatrix} x_2\\ y_2 \end{bmatrix}\) of the second equation.

3. prove that it is also a solution of the first equation.

TIPS: Remember that \( b\ne 0\) and that \( P(\lambda_2)=0\) .

Question 9: Give the matrix \( D\) of the linear mapping with matrix \( A\) in the canonical basis in the basis \( B\) of the vectors of column vectors of coordinates \( X_1\) and \( X_2\) .

TIP: Use the facy that \( AX_i=\lambda_{i} X_i\) for \( i=1,2\) .

We have found a basis \( B\) with orthogonal vectors in which the matrix \( D\) of the linear mapping having \( A\) as matrix in the canonical basis is a diagonal matrix.

We have thus ’diagonalized’ the matrix \( A\) , and:

• the diagonal elements od \( D\) are the eigenvalues of \( A\) ,

• and the vectors of the basis \( B\) are eigenvectors of these eigenvalues in the same order.

Moreover, if we divide each vector of the basis \( B\) by its norm, we obtain an orthonormal basis \( B'\) in which the matrix of the linear mapping corresponding to \( A\) is the same diagonal matrix \( D\) .

Indeed, if we multiply an eigenvector for the eigenvalue \( \lambda_i\) by a non zero scalar, we obtain another eigenvector for \( \lambda_i\) .

We may thus diagonalize the matrix \( A\) in an orthonormal basis.

And the transition matrix \( U\) from the orthonormal basis \( B'\) to the canonical basis is a unitary matrix.

Consequently, we have \( D=U^{-1}AU=U^TAU\) , so that \( A=UDU^T\) , where:

• \( D\) is a diagonal matrix,

• and \( U\) is a unitary matrix.

It is the spectral decomposition in dimension 2.