finding eigenvalues

This example investigates eigenvalues and the similarity transformation usedto diagonalize matrices. We seek the eigenvalues of the matrix $A$ below. Afterward,we can transform this matrix into a diagonal matrix which has many useful applications.

Here, we need to solve the corresponding matrix equation;

or

$$AX=\lambda X$$

rearranging gives

$$AX-\lambda X=0$$

or

$$(A-\lambda I)X=0$$

We seek the values for $\lambda$ and $X$.First, we need to solve the characteristic equation of $A$. We do this by finding$det(A-\lambda I)$. First, calculating $A-\lambda I$ gives;

Next, calculating $det(A-\lambda I)$ yields

$$det(A-\lambda I)={(2-\lambda )}^2-1={\lambda }^2-4\lambda +3=(\lambda -1)(\lambda -3)=0$$

Substituting $\lambda =1$ into $(A-\lambda I)X$ gives…

$$\{\begin{array}{c}\hfill x_1+x_2=0\hfill \\ \hfill x_1+x_2=0\hfill \end{array}$$

so that $x_2=-x_1$ and the corresponding eigenvector is

$$\left(\begin{array}{c}\hfill t\hfill \\ \hfill -t\hfill \end{array}\right)=t\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \end{array}\right)$$

where $t\ne 0.$
Substituting $\lambda =3$ gives…

$$\{\begin{array}{c}\hfill -x_1+x_2=0\hfill \\ \hfill x_1-x_2=0\hfill \end{array}$$

so that $x_2=x_1$ and the corresponding eigenvector is

$$\left(\begin{array}{c}\hfill t\hfill \\ \hfill t\hfill \end{array}\right)=t\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right)$$

where $t\ne 0.$
Finally, to diagonalize $A$ we let the eigenvectors be the columns of a new matrix

and then since our eigenvectors are linearly independent we can also find;

then we create a diagonal matrix as follows…

Computing powers of $A$ is a very useful application of $D$. Solving for $A$ lets us compute powers of $A$

$$A=PD{P}^{-1}$$

so that

$$A^n=P{D}^n{P}^{-1}$$

or