/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q2SE Show that if \({\bf{x}}\) is an ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that if \({\bf{x}}\) is an eigenvector of the matrix product \(AB\) and \(B{\rm{x}} \ne 0\), then \(B{\rm{x}}\) is an eigenvector of\(BA\).

Short Answer

Expert verified

It is proved that if \({\rm{x}}\) is an eigenvector of \(AB\) then \(B{\rm{x}}\) is an eigenvector of \(BA\).

Step by step solution

01

Definition of matrix and Eigen Vectors

A matrix(plural matrices) is a array or table of,, or, arranged in rows and columns, which is used to represent a or a property of such an object.

A vector \(v\)is called an eigenvector of corresponding to an eigenvalue of a matrix \(A\) if it satisfies \(A{\rm{v}} = \lambda {\rm{v}}\).

02

Find the eigenvector of matrix product

Since \({\rm{x}}\) is an eigenvector of \(AB\), hence there exist an eigenvalue \(\lambda \), such that

\(\left( {AB} \right){\rm{x}} = \lambda {\rm{x}}\).

We can rewrite this equality as \(A(B{\rm{x}}) = \lambda {\rm{x}}\).

Multiplying both sides by \(B\) from the left we will have,

\(\begin{array}{c}BA\left( {B{\rm{x}}} \right) &= B\left( {\lambda {\rm{x}}} \right)\\\left( {BA} \right)\left( {B{\rm{x}}} \right) &= \lambda \left( {B{\rm{x}}} \right)\end{array}\)

It is given that\(B{\rm{x}} \ne 0\), so by the definition \(B{\rm{x}}\) is an eigenvector of \(BA\) (\(\lambda \) is an eigenvalue).

Therefore,

\[\begin{aligned}{c}AB{\rm{x}} &= \lambda {\rm{x}}\\BA\left( {B{\rm{x}}} \right) &= B\left( {\lambda {\rm{x}}} \right)\\BA\left( {B{\rm{x}}} \right){\rm{ }} &= \lambda \left( {B{\rm{x}}} \right)\end{aligned}\]

It is proved that if \({\rm{x}}\) is an eigenvector of \(AB\) then \(B{\rm{x}}\) is an eigenvector of \(BA\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

[M]Repeat Exercise 25 for \[A{\bf{ = }}\left[ {\begin{array}{*{20}{c}}{{\bf{ - 8}}}&{\bf{5}}&{{\bf{ - 2}}}&{\bf{0}}\\{{\bf{ - 5}}}&{\bf{2}}&{\bf{1}}&{{\bf{ - 2}}}\\{{\bf{10}}}&{{\bf{ - 8}}}&{\bf{6}}&{{\bf{ - 3}}}\\{\bf{3}}&{{\bf{ - 2}}}&{\bf{1}}&{\bf{0}}\end{array}} \right]\].

Let\(T:{{\rm P}_2} \to {{\rm P}_3}\) be a linear transformation that maps a polynomial \({\bf{p}}\left( t \right)\) into the polynomial \(\left( {t + 5} \right){\bf{p}}\left( t \right)\).

  1. Find the image of\({\bf{p}}\left( t \right) = 2 - t + {t^2}\).
  2. Show that \(T\) is a linear transformation.
  3. Find the matrix for \(T\) relative to the bases \(\left\{ {1,t,{t^2}} \right\}\) and \(\left\{ {1,t,{t^2},{t^3}} \right\}\).

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?

In Exercises \({\bf{3}}\) and \({\bf{4}}\), use the factorization \(A = PD{P^{ - {\bf{1}}}}\) to compute \({A^k}\) where \(k\) represents an arbitrary positive integer.

4. \(\left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\)

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 \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

14. \(\left( {\begin{array}{*{20}{c}}4&0&{ - 2}\\2&5&4\\0&0&5\end{array}} \right)\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.