# finding eigenvalues

This example investigates eigenvalues and the similarity transformation used to 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.

 $A=\left(\begin{array}[]{cc}2&1\\ 1&2\end{array}\right)$

Here, we need to solve the corresponding matrix equation;

 $\left(\begin{array}[]{cc}2&1\\ 1&2\end{array}\right)\left(\begin{array}[]{c}x_{1}\\ x_{2}\end{array}\right)=\lambda\left(\begin{array}[]{c}x_{1}\\ x_{2}\end{array}\right)$

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;

 $A-\lambda I=\left(\begin{array}[]{cc}2-\lambda&1\\ 1&2-\lambda\end{array}\right)$

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…

 $\left\{\begin{array}[]{c}x_{1}+x_{2}=0\\ x_{1}+x_{2}=0\end{array}\right.$

so that $x_{2}=-x_{1}$ and the corresponding eigenvector is

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

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

 $\left\{\begin{array}[]{c}-x_{1}+x_{2}=0\\ x_{1}-x_{2}=0\end{array}\right.$

so that $x_{2}=x_{1}$ and the corresponding eigenvector is

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

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

 $P=\left(\begin{array}[]{cc}1&-1\\ 1&1\end{array}\right)$

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

 $P^{-1}=\frac{1}{2}\left(\begin{array}[]{cc}1&1\\ -1&1\end{array}\right)$

then we create a diagonal matrix as follows…

 $D=P^{-1}AP=\left(\begin{array}[]{cc}1&0\\ 0&3\end{array}\right)$

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

 $A=PDP^{-1}$

so that

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

or

 $A^{n}=P\left(\begin{array}[]{cc}1^{n}&0\\ 0&3^{n}\end{array}\right)P^{-1}$

Title finding eigenvalues FindingEigenvalues 2013-03-22 15:52:35 2013-03-22 15:52:35 PrimeFan (13766) PrimeFan (13766) 6 PrimeFan (13766) Example msc 15A18