Chapter 6.1, Problem 1PS

To determine

(a)

The eigenvalues of given matrices and the answer of given question.

Answer to Problem 1PS

${\lambda }_1=1and{\lambda }_2=.5$ are the required eigenvalues of matrix A.

${\lambda }_1=1and{\lambda }_2=.25$ are the required eigenvalues of matrix $A^2$.

${\lambda }_1=1and{\lambda }_2=0$ are the required eigenvalues of matrix $A^{\infty }$

Explanation of Solution

Given:

Matrices are $A=\left[\begin{array}{cc}.8& .3\\ .2& .7\end{array}\right]$, $A^2=\left[\begin{array}{cc}.70& .45\\ .30& .55\end{array}\right]$ and $A^{\infty }=\left[\begin{array}{cc}.6& .6\\ .4& .4\end{array}\right]$

Concept Used:

If T is a linear transformation such that

$T:V\left(F\right)\to V$

i.e., mapping from a vector space V over a field F into itself and $0\ne v\in V$, then v is an eigenvector of T. If T(v) is a scalar multiple of v. This condition can be written as the equation

$T\left(v\right)=\lambda v$

Where λis a scalar in the field F, known as the eigenvalue or characteristic value, or characteristic root or proper values, or latent roots associated with the eigenvector v.

If $\dim \left(V\right)<\infty$, then the linear transformation T can be represented as a square matrix A, and the vector v by a column vector which can be written as

$Av=\lambda v$

We can also say that any number such that a given matrix minus that number times the identity matrix has zero determinant, the equation formed from this operation termed as characteristic equation which have a roots thereby termed as characteristic root.

The given matrix

$A={\left[\begin{array}{ccccccc}{a}_{11}& a_{12}& a_{13}& \cdots & a_{1j}& \cdots & a_{1n}\\ a_{21}& a_{22}& a_{23}& \cdots & a_{2j}& \cdots & a_{2n}\\ a_{31}& a_{32}& a_{33}& \cdots & a_{3j}& \cdots & a_{3n}\\ ⋮& ⋮& ⋮& \cdots & ⋮& \cdots & ⋮\\ a_{i1}& a_{i2}& a_{i3}& \cdots & a_{ij}& \cdots & a_{in}\\ ⋮& ⋮& ⋮& \cdots & ⋮& \cdots & ⋮\\ a_{n1}& a_{n2}& a_{n3}& \cdots & a_{nj}& \cdots & a_{nn}\end{array}\right]}_{n\times n}={\left[a_{ij}\right]}_{n\times n}$

Has atmost n eigenvalues and thereby n eigenvector.

• Two matrices can be added or subtracted if they are of the same order.
• Two matrices can be multiplied if the number of columns of first matrix is equal to the number of rows of second matrix.

Calculation:

Given matrices are $A=\left[\begin{array}{cc}.8& .3\\ .2& .7\end{array}\right]$, $A^2=\left[\begin{array}{cc}.70& .45\\ .30& .55\end{array}\right]$ and $A^{\infty }=\left[\begin{array}{cc}.6& .6\\ .4& .4\end{array}\right]$

Since, above matrices are of order two hence it have at most two eigenvalues say ${\lambda }_1$ and ${\lambda }_2$.

Since, we know that for any given matrix minus that number times the identity matrix has zero determinant. Thus, using the given matrix A we get

$\begin{array}{l}\Rightarrow A-\lambda I=\left[\begin{array}{cc}.8& .3\\ .2& .7\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ \Rightarrow A-\lambda I=\left[\begin{array}{cc}.8-\lambda & .3\\ .2& .7-\lambda \end{array}\right]\end{array}$

and the associated characteristic equation and thereby the characteristic root of the given matrix will be obtained as

$\begin{array}{l}\Rightarrow \left|A-\lambda I\right|=0\\ \Rightarrow \left|\begin{array}{cc}.8-\lambda & .3\\ .2& .7-\lambda \end{array}\right|=0\\ \Rightarrow \left(.8-\lambda \right)\left(.7-\lambda \right)-.06=0\\ \Rightarrow .56-.8\lambda -.7\lambda +{\lambda }^2-.06=0\\ \Rightarrow {\lambda }^2-1.5\lambda +.50=0\\ \Rightarrow {\lambda }^2-.5\lambda -\lambda +.5=0\\ \Rightarrow \left(\lambda -1\right)\left(\lambda -.5\right)=0\\ \Rightarrow \lambda =1\text{ or }\lambda =.5\\ \Rightarrow \lambda =1,.5\text{ are the required eigenvalues of matrix }A.\end{array}$

Now, using the given matrix $B=A^2=\left[\begin{array}{cc}.70& .45\\ .30& .55\end{array}\right]$ we get

$\begin{array}{l}\Rightarrow B-\lambda I=\left[\begin{array}{cc}.70& .45\\ .30& .55\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ \Rightarrow B-\lambda I=\left[\begin{array}{cc}.70-\lambda & .45\\ .30& .55-\lambda \end{array}\right]\end{array}$

and the associated characteristic equation and thereby the characteristic root of the given matrix will be obtained as

$\begin{array}{l}\Rightarrow \left|B-\lambda I\right|=0\\ \Rightarrow \left|\begin{array}{cc}.70-\lambda & .45\\ .30& .55-\lambda \end{array}\right|=0\\ \Rightarrow \left(.70-\lambda \right)\left(.55-\lambda \right)-\text{.135}=0\\ \Rightarrow \text{.385}-.55\lambda -.70\lambda +{\lambda }^2-.135=0\\ \Rightarrow {\lambda }^2-1.25\lambda +.25=0\\ \Rightarrow {\lambda }^2-\lambda -.25\lambda +.25=0\\ \Rightarrow \lambda \left(\lambda -1\right)-.25\left(\lambda -1\right)=0\\ \Rightarrow \left(\lambda -1\right)\left(\lambda -.25\right)=0\\ \Rightarrow \left(\lambda -1\right)=0\text{or }\left(\lambda -.25\right)=0\\ \Rightarrow \lambda =1\text{ or }\lambda =.25\\ \Rightarrow \lambda =1,.25\text{ are the required eigenvalues of matrix }B=A^2\end{array}$

Let $C=A^{\infty }=\left[\begin{array}{cc}.6& .6\\ .4& .4\end{array}\right]$

Now, using the given matrix C we get

$\begin{array}{l}\Rightarrow C-\lambda I=\left[\begin{array}{cc}.6& .6\\ .4& .4\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ \Rightarrow C-\lambda I=\left[\begin{array}{cc}.6-\lambda & .6\\ .4& .4-\lambda \end{array}\right]\end{array}$

and the associated characteristic equation and thereby the characteristic root of the given matrix will be obtained as

$\begin{array}{l}\Rightarrow \left|C-\lambda I\right|=0\\ \Rightarrow \left|\begin{array}{cc}.6-\lambda & .6\\ .4& .4-\lambda \end{array}\right|=0\\ \Rightarrow \left(.6-\lambda \right)\left(.4-\lambda \right)-\text{.24}=0\\ \Rightarrow \text{.24}-.4\lambda -.6\lambda +{\lambda }^2-.24=0\\ \Rightarrow {\lambda }^2-\lambda =0\\ \Rightarrow \lambda \left(\lambda -1\right)=0\\ \Rightarrow \lambda =0\text{or }\left(\lambda -1\right)=0\\ \Rightarrow \lambda =0\text{ or }\lambda =1\\ \Rightarrow \lambda =1,0\text{ are the required eigenvalues of matrix }C=A^{\infty }\end{array}$

As we have

$A=\left[\begin{array}{cc}.8& .3\\ .2& .7\end{array}\right]$

Let us reduce the matrix A in its echelon form.

Operate $R_2\to R_2+\left(\frac{-1}{4}\right)R_1$, we get

$\sim \left[\begin{array}{cc}.8& .3\\ 0& \frac{2.5}{4}\end{array}\right]$

Operate $R_1\to R_1+\left(\frac{-1.2}{2.5}\right)R_2$, we get

$\sim \left[\begin{array}{cc}.8& 0\\ 0& \frac{2.5}{4}\end{array}\right]$

Operate $R_1\to \left(\frac{1}{.8}\right)R_1;R_2\to \left(\frac{4}{2.5}\right)R_2$, we get

$\sim \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow \lambda =1,1\text{ are the required eigenvalues of reduced form matrix }A.$

Conclusion:

As can be seen that eigenvalues of increasing power of A is equal to the power of eigenvalues of A. Different powers of A have different eigenvalues. Eigenvalues of reduced form of matrix is different from its original form.

To determine

(b)

Why the matrix having eigenvalues zero, does not changed by the steps of elimination?

Explanation of Solution

Concept Used:

If T is a linear transformation such that

$T:V\left(F\right)\to V$

i.e., mapping from a vector space V over a field F into itself and $0\ne v\in V$, then v is an eigenvector of T. If T(v) is a scalar multiple of v. This condition can be written as the equation

$T\left(v\right)=\lambda v$

Where λis a scalar in the field F, known as the eigenvalue or characteristic value, or characteristic root or proper values, or latent roots associated with the eigenvector v.

If $\dim \left(V\right)<\infty$, then the linear transformation T can be represented as a square matrix A, and the vector v by a column vector which can be written as

$Av=\lambda v$

We can also say that any number such that a given matrix minus that number times the identity matrix has zero determinant, the equation formed from this operation termed as characteristic equation which have a roots thereby termed as characteristic root.

The given matrix

$A={\left[\begin{array}{ccccccc}{a}_{11}& a_{12}& a_{13}& \cdots & a_{1j}& \cdots & a_{1n}\\ a_{21}& a_{22}& a_{23}& \cdots & a_{2j}& \cdots & a_{2n}\\ a_{31}& a_{32}& a_{33}& \cdots & a_{3j}& \cdots & a_{3n}\\ ⋮& ⋮& ⋮& \cdots & ⋮& \cdots & ⋮\\ a_{i1}& a_{i2}& a_{i3}& \cdots & a_{ij}& \cdots & a_{in}\\ ⋮& ⋮& ⋮& \cdots & ⋮& \cdots & ⋮\\ a_{n1}& a_{n2}& a_{n3}& \cdots & a_{nj}& \cdots & a_{nn}\end{array}\right]}_{n\times n}={\left[a_{ij}\right]}_{n\times n}$ has at most n eigenvalues.

Consider a matrix $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$

Since, above matrices are of order two hence it have at most two eigenvalues say ${\lambda }_1$ and ${\lambda }_2$.

Since, we know that for any given matrix minus that number times the identity matrix has zero determinant. Thus, using the given matrix A we get

$\begin{array}{l}\Rightarrow A-\lambda I=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ \Rightarrow A-\lambda I=\left[\begin{array}{cc}-\lambda & 1\\ 0& -\lambda \end{array}\right]\end{array}$

and the associated characteristic equation and thereby the characteristic root of the given matrix will be obtained as

$\begin{array}{l}\Rightarrow \left|A-\lambda I\right|=0\\ \Rightarrow \left|\begin{array}{cc}-\lambda & 1\\ 0& -\lambda \end{array}\right|=0\\ \Rightarrow {\lambda }^2-0=0\\ \Rightarrow {\lambda }^2=0\\ \Rightarrow \lambda =0\text{ are the required eigenvalues of matrix }A.\end{array}$

The given matrix $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ cannot be reduce in its echelon form.

Hence, the given matrix always have zero eigenvalues and which cannot be avoided even after elimination process on the matrix.