/*! 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} Q5.4-24E Verify the statements in Exercis... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

24. If \(A\) and \(B\) are similar, then they have the same rank. (Hint: Refer to supplementary Exercises 13 and 14 for Chapter 4.)

Short Answer

Expert verified

It is proved that \(A\) and \(B\) have the same rank.

Step by step solution

01

Rank of the matrix

It is known that when \(P\) is aninvertible \(m \times m\) matrix, then \({\mathop{\rm rank}\nolimits} A = {\mathop{\rm rank}\nolimits} B\) .

Also, when \(Q\) is invertible, then \({\mathop{\rm rank}\nolimits} AQ = {\mathop{\rm rank}\nolimits} A\).

02

Show that \(A\) and \(B\) have the same rank 

When \(A = PB{P^{ - 1}}\) then \({\mathop{\rm rank}\nolimits} A = {\mathop{\rm rank}\nolimits} P\left( {B{P^{ - 1}}} \right) = {\mathop{\rm rank}\nolimits} B{P^{ - 1}}\) , according to the above statement. Moreover, \({\mathop{\rm rank}\nolimits} B{P^{ - 1}} = {\mathop{\rm rank}\nolimits} B\), according to the above statement, because \({P^{ - 1}}\) is invertible.

Therefore, \({\mathop{\rm rank}\nolimits} A = {\mathop{\rm rank}\nolimits} B\) .

Thus, it is proved that \(A\) and \(B\) have the same rank.

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

Question: Is \(\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)\)? If so, find the eigenvalue.

(M)The MATLAB command roots\(\left( p \right)\) computes the roots of the polynomial equation \(p\left( t \right) = {\bf{0}}\). Read a MATLAB manual, and then describe the basic idea behind the algorithm for the roots command.

Question 19: Let \(A\) be an \(n \times n\) matrix, and suppose A has \(n\) real eigenvalues, \({\lambda _1},...,{\lambda _n}\), repeated according to multiplicities, so that \(\det \left( {A - \lambda I} \right) = \left( {{\lambda _1} - \lambda } \right)\left( {{\lambda _2} - \lambda } \right) \ldots \left( {{\lambda _n} - \lambda } \right)\) . Explain why \(\det A\) is the product of the n eigenvalues of A. (This result is true for any square matrix when complex eigenvalues are considered.)

Let\(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\) be a basis for a vector space\(V\). Find \(T\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right)\) when \(T\) isa linear transformation from \(V\) to \(V\) whose matrix relative to \(B\) is

\({\left( T \right)_B} = \left( {\begin{aligned}0&{}&{ - 6}&{}&1\\0&{}&5&{}&{ - 1}\\1&{}&{ - 2}&{}&7\end{aligned}} \right)\)

Let \(A{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{{a_{{\bf{11}}}}}&{{a_{{\bf{12}}}}}\\{{a_{{\bf{21}}}}}&{{a_{{\bf{22}}}}}\end{aligned}} \right)\). Recall from Exercise \({\bf{25}}\) in Section \({\bf{5}}{\bf{.4}}\) that \({\rm{tr}}\;A\) (the trace of \(A\)) is the sum of the diagonal entries in \(A\). Show that the characteristic polynomial of \(A\) is \({\lambda ^2} - \left( {{\rm{tr}}A} \right)\lambda + \det A\). Then show that the eigenvalues of a \({\bf{2 \times 2}}\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2}\).

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.