Question

Exercise 33 illustrates how you can use the powers of a matrix to find its dominant eigenvalue (i.e., the eigenvalue with maximal modulus), at least when this eigenvalue is real. But what about the other eigenvalues?

a. Consider a $\mathit{n}\mathit{x}\mathit{n}$matrix A with n distinct complex eigenvalues${\mathit{\lambda }}_{\mathbf{1}}\mathbf{,}{\mathit{\lambda }}_{\mathbf{2}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathit{\lambda }}_{\mathbf{n}}\mathbf{,}\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{}{\mathit{\lambda }}_{\mathbf{1}}$1 is real. Suppose you have a good (real) approximation $\mathit{\lambda }$ of ${\mathit{\lambda }}_{\mathbf{1}}$ (good in that $\mathbf{|}\mathbf{\lambda }\mathbf{-}{\mathbf{\lambda }}_{\mathbf{1}}\mathbf{|}\mathbf{<}\mathbf{|}\mathbf{\lambda }\mathbf{-}{\mathbf{\lambda }}_{\mathbf{i}}\mathbf{|}$for $\mathit{i}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{\dots }\mathbf{,}\mathit{n}\mathbf{)}$Consider the matrix $\mathit{A}\mathbf{-}\mathit{\lambda }{\mathit{I}}_{\mathbf{n}}$. What are its eigenvalues? Which has the smallest modulus? Now consider the matrix ${\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{\lambda }{\mathbf{I}}_{\mathbf{n}}\mathbf{)}}^{\mathbf{-}\mathbf{1}}$What are its eigenvalues? Which has the largest modulus? What is the relationship between the eigenvectors of A and those of ${\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{\lambda }{\mathbf{I}}_{\mathbf{n}}\mathbf{)}}^{\mathbf{-}\mathbf{1}}$? Consider higher and higher powers of .${\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{\lambda }{\mathbf{I}}_{\mathbf{n}}\mathbf{)}}^{\mathbf{-}\mathbf{1}}$How does this help you to find an eigenvector of A with eigenvalue${\mathit{\lambda }}_{\mathbf{1}}$, and${\mathit{\lambda }}_{\mathbf{1}}$itself? Use the results of Exercise 33.

b. As an example of part a, consider the matrix $\mathit{A}\mathbf{=}\mathbf{[}\begin{array}{l}\mathbf{1}\\ \mathbf{4}\\ \mathbf{7}\end{array}$$\begin{array}{c}\mathbf{2}\\ \mathbf{5}\\ \mathbf{8}\end{array}$$\begin{array}{r}\mathbf{3}\\ \mathbf{6}\\ \mathbf{10}\end{array}\mathbf{]}$$\begin{array}{c}\mathbf{2}\\ \mathbf{5}\\ \mathbf{8}\end{array}$

We wish to find the eigenvectors and eigenvalues of without using the corresponding commands on the computer (which is, after all, a “black box”). First, we find approximations for the eigenvalues by graphing the characteristic polynomial (use technology). Approximate the three real eigenvalues of to the nearest integer. One of the three eigenvalues of is negative. Find a good approximation for this eigenvalue and a corresponding eigenvector by using the procedure outlined in part a. Youare not asked to do the same for the two other eigenvalues.

Short answer

(a)If${\lambda }_i$is an eigenvalue of,$A$then${\lambda }_i-\lambda$is an eigenvalue of,$A-\lambda I$and the lowest modulo is obtained in ${\lambda }_1-\lambda$. Also, an eigenvalue of${(A-\lambda I)}^{-1}$is,$\frac{1}{{\lambda }_i-\lambda }$and the highest modulo is obtained in.$\frac{1}{{\lambda }_1-\lambda }$

(b)The approximation eigenvalues of matrix$A$is.${\lambda }_1\approx 17,{\lambda }_2\approx -1,{\lambda }_3\approx 0$

Step-by-step solution

Define matrix:

A table of numbers and symbols arranged in the form of rows and columns is known as matrix.

Find the smallest module and highest module by obtaining the eigenvalues: 

Hence, if${\lambda }_i$is an eigenvalue of$A$, then

$\det (A-{\lambda }_i{I}_n)=0\iff \det (A-\lambda I_n+\lambda I_n-{\lambda }_i{I}_n)$$=0\iff \det (A-\lambda I-({\lambda }_i-\lambda )I_n)=0$

Thus${\lambda }_i-\lambda$ is an eigenvalue of.$A-\lambda I$

Consider the given conditions, $|\lambda -{\lambda }_1|<|\lambda -{\lambda }_i|,i=2,\dots ,n$and we conclude that the eigenvalue of$A-\lambda I$with the lowest absolute value is.${\lambda }_1-\lambda$

Further, we can also conclude that the eigenvalues of ${(A-\lambda I)}^{-1}$ are$\frac{1}{{\lambda }_i-\lambda },i=1,\dots ,n$.

Hence, from the same condition, the conclusion can be made that$\frac{1}{{\lambda }_1-\lambda }$has the highest absolute value of all eigenvalues of${(A-\lambda I_n)}^{-1}$

Find the approximate eigenvalues of the given matrix: 

Consider the given matrix,

$A=[\begin{array}{l}1\\ 4\\ 7\end{array}$$\begin{array}{c}2\\ 5\\ 8\end{array}$$\begin{array}{r}3\\ 6\\ 10\end{array}]$

Solve by using the determinant,

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

$\left|\begin{array}{ccc}1-\lambda & 2& 3\\ 4& 5-\lambda & 6\\ 7& 8& 10-\lambda \end{array}\right|=0$

$(1-\lambda )(5-\lambda )(10-\lambda )+84+96-48(1-\lambda )-21(5-\lambda )-8(10.-\lambda )=0$

$-{\lambda }^3+16{\lambda }^2+12\lambda -3=0$

${\lambda }_1=16.7075,{\lambda }_2=-0.90574,{\lambda }_3=0.198247$

Thus, the approximate eigenvalues can be written as,.${\lambda }_1\approx 17,{\lambda }_2\approx -1,{\lambda }_3\approx 0$