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Chapter 5: Q5.4-23E (page 267)

Verify the statements in Exercises 19–24. The matrices are square.

23. If \(B = {P^{ - 1}}AP\) and x is an eigenvector of A corresponding to an eigenvalue \(\lambda \), then \({P^{ - 1}}{\mathop{\rm x}\nolimits} \) is an eigenvector of \(B\) corresponding also to

\(\lambda \).

Short Answer

Expert verified

It is proved that \({P^{ - 1}}{\mathop{\rm x}\nolimits} \) is an eigenvector of \(B\) corresponding also to \(\lambda \).

Step by step solution

01

Show that \({P^{ - 1}}{\mathop{\rm x}\nolimits} \) is an eigenvector of \(B\) corresponding also to \(\lambda \)

When \(Ax = \lambda x\) then \({P^{ - 1}}Ax = \lambda {P^{ - 1}}{\mathop{\rm x}\nolimits} \). When \(B = {P^{ - 1}}AP\) then according to the first computations,

\(\begin{aligned}{}B\left( {{P^{ - 1}}x} \right) &= {P^{ - 1}}AP\left( {{P^{ - 1}}{\mathop{\rm x}\nolimits} } \right)\\ &= {P^{ - 1}}A{\mathop{\rm x}\nolimits} \\ &= \lambda {P^{ - 1}}{\mathop{\rm x}\nolimits} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 1 \right)\end{aligned}\)

02

Show that \({P^{ - 1}}{\mathop{\rm x}\nolimits} \) is an eigenvector of \(B\) corresponding also to \(\lambda \)

It is observed that \({P^{ - 1}}{\mathop{\rm x}\nolimits} \ne 0,\) since \({\mathop{\rm x}\nolimits} \ne 0\) and \({P^{ - 1}}\) is invertible. Therefore, the equation \(\left( 1 \right)\) demonstrates that \({P^{ - 1}}{\mathop{\rm x}\nolimits} \) is an eigenvector of \(B\) that corresponds to \(\lambda \).

Thus, it is proved that \({P^{ - 1}}{\mathop{\rm x}\nolimits} \) is an eigenvector of \(B\) corresponding also to \(\lambda \).

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Most popular questions from this chapter

Let\(D = \left\{ {{{\bf{d}}_1},{{\bf{d}}_2}} \right\}\) and \(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2}} \right\}\) be bases for vector space \(V\) and \(W\), respectively. Let \(T:V \to W\) be a linear transformation with the property that

\(T\left( {{{\bf{d}}_1}} \right) = 2{{\bf{b}}_1} - 3{{\bf{b}}_2}\), \(T\left( {{{\bf{d}}_2}} \right) = - 4{{\bf{b}}_1} + 5{{\bf{b}}_2}\)

Find the matrix for \(T\) relative to \(D\), and\(B\).

Let \(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\)be a basis for a vector space \(V\) and\(T:V \to {\mathbb{R}^2}\) be a linear transformation with the property that

\(T\left( {{x_1}{{\bf{b}}_1} + {x_2}{{\bf{b}}_2} + {x_3}{{\bf{b}}_3}} \right) = \left( {\begin{aligned}{2{x_1} - 4{x_2} + 5{x_3}}\\{ - {x_2} + 3{x_3}}\end{aligned}} \right)\)

Find the matrix for \(T\) relative to \(B\) and the standard basis for \({\mathbb{R}^2}\).

Question: Diagonalize the matrices in Exercises \({\bf{7--20}}\), if possible. The eigenvalues for Exercises \({\bf{11--16}}\) are as follows:\(\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}\); \(\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}\); \(\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}\); \(\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}\); \(\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}\); \(\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}\). For exercise \({\bf{18}}\), one eigenvalue is \(\lambda {\bf{ = 5}}\) and one eigenvector is \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

8. \(\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{1}}\\{\bf{0}}&{\bf{5}}\end{array}} \right)\)

For the matrices A in Exercises 1 through 12, find closed formulas for , where t is an arbitrary positive integer. Follow the strategy outlined in Theorem 7.4.2 and illustrated in Example 2. In Exercises 9 though 12, feel free to use technology.

1.A=1203

Suppose \({\bf{x}}\) is an eigenvector of \(A\) corresponding to an eigenvalue \(\lambda \).

a. Show that \(x\) is an eigenvector of \(5I - A\). What is the corresponding eigenvalue?

b. Show that \(x\) is an eigenvector of \(5I - 3A + {A^2}\). What is the corresponding eigenvalue?
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