Eigenvalues and Eigenvectors of a Matrix

For a square matrix A,

$\displaystyle A\vec{u}=\lambda\vec{u}$
$\vec{u}$ vectors satisfying this condition are called the eigenvectors of the matrix A, and the lambda scalar coefficients are called the eigenvalues.
An $N \times N$ matrix has different eigenvectors. However, the corresponding lambda eigenvalues for these eigenvectors may not be different.
State of a system can be expressed in terms of the eigenvectors of the system of linear equations and in terms of their eigenvalues for the measured quantities.
Create an matrix by arranging eigenvectors side by side:

$\begin{indisplay} U=\overbrace{\left[ \begin{array}{rrrr} u_{11} & u_{12} & \ld... ...ay} \right]}^{\vec{u_1}~~~~~~\vec{u_2}~~~~~\ldots~~~~~\vec{u_n}} \end{indisplay}$

When we multiply the matrix with the matrix and its inverse matrix $U^{-1}$ from both sides, we get

$\begin{indisplay} A'=U^{-1}AU=\left[ \begin{array}{rrrr} \lambda_{1} & 0 & \ldo... ...\vdots \\ 0 & 0 & \ldots & \lambda_{n} \\ \end{array} \right] \end{indisplay}$
That is, the similarity transformation with the eigenvectors matrix makes as being diagonalized and the elements on the diagonal become the eigenvalues of .
In principle, the eigenvalue problem is easy to solve. So-called characteristic equation is to be solved:

$\displaystyle det\vert A -\lambda I\vert=0$
After finding the roots of this n-degree polynomial equation, the corresponding eigenvectors can be obtained by solving the following system of equations:

$\displaystyle (A-\lambda I)\vec{v}=0$
Since this method requires determinant calculation, it is not useful for large dimensional matrices.