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Q5.2-18E

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Linear Algebra and its Applications
Found in: Page 267
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Question 18: It can be shown that the algebraic multiplicity of an eigenvalue \(\lambda \) is always greater than or equal to the dimension of the eigenspace corresponding to \(\lambda \). Find \(h\) in the matrix \(A\) below such that the eigenspace for \(\lambda = 5\) is two-dimensional:

\[A = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&3&h&0\\0&0&5&4\\0&0&0&1\end{array}} \right]\]

It is observed that the system requires two free variables for a two-dimensional eigenspace. This occurs only when \(h = 6\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition of the characteristic polynomial

The eigenvalue of an \(n \times n\) matrix \(A\) is a scalar \(\lambda \) such that \(\lambda \) satisfies the characteristic equation \(\det \left( {A - \lambda I} \right) = 0\).

When \(A\) is a \(n \times n\) matrix, \(\det \left( {A - \lambda I} \right)\) is the characteristic polynomial of \(A\) which is the polynomial of degree \(n\).

In particular, the multiplicity of an eigenvalue \(\lambda \) represents its multiplication as a root of the characteristic equation.

Step 2: Determine h in matrix A

Write the matrix in the \[\left( {A - 5I} \right)\] form, as shown below.

\[\begin{array}{c}\left( {{\rm{A}} - 5I} \right) = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&3&{\rm{h}}&0\\0&0&5&4\\0&0&0&1\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}5&0&0&0\\0&5&0&0\\0&0&5&0\\0&0&0&5\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&{ - 2}&{\rm{h}}&0\\0&0&0&4\\0&0&0&{ - 4}\end{array}} \right]\end{array}\]

Write the matrix as an augmented matrix \(\left( {A - 5I} \right)x = 0\), as shown below.

\( \sim \left[ {\begin{array}{*{20}{c}}0&{ - 2}&6&{ - 1}&0\\0&{ - 2}&{\rm{h}}&0&0\\0&0&0&4&0\\0&0&0&{ - 4}&0\end{array}} \right]\)

Apply row operation in the augmented matrix:

At row 2, subtract row 1 from row 2. At row 4, multiply row 4 by \( - 1\), as shown below.

\( \sim \left[ {\begin{array}{*{20}{c}}0&{ - 2}&6&{ - 1}&0\\0&0&{{\rm{h}} - 6}&1&0\\0&0&0&4&0\\0&0&0&4&0\end{array}} \right]\)

At row 4, multiply row 3 by 1 and subtract it from row 4, as shown below.

\( \sim \left[ {\begin{array}{*{20}{c}}0&{ - 2}&6&{ - 1}&0\\0&0&{{\rm{h}} - 6}&1&0\\0&0&0&4&0\\0&0&0&0&0\end{array}} \right]\)

At row 1, multiply row 1 by \( - \frac{1}{2}\) and at row 3, multiply row 3 by \(\frac{1}{4}\), as shown below.

\( \sim \left[ {\begin{array}{*{20}{c}}0&1&{ - 3}&1&0\\0&0&{{\rm{h}} - 6}&1&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]\)

At row 1, subtract row 3 from row 1, and at row 2, subtract row 3 from row 2, as shown below.

\( \sim \left[ {\begin{array}{*{20}{c}}0&1&{ - 3}&0&0\\0&0&{{\rm{h}} - 6}&0&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]\)

The above system requires two free variables for a two-dimensional eigenspace. This occurs only when \(h = 6\).

Thus, the value of \(h\) in the matrix is 6.

Most popular questions for Math Textbooks

Use Exercise 12 to find the eigenvalues of the matrices in Exercises 13 and 14.

14. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{5}}&{{\bf{ - 6}}}&{{\bf{ - 7}}}\\{\bf{2}}&{\bf{4}}&{\bf{5}}&{\bf{2}}\\{\bf{0}}&{\bf{0}}&{{\bf{ - 7}}}&{{\bf{ - 4}}}\\{\bf{0}}&{\bf{0}}&{\bf{3}}&{\bf{1}}\end{array}} \right)\)

Show that \(I - A\) is invertible when all the eigenvalues of \(A\) are less than 1 in magnitude. (Hint: What would be true if \(I - A\) were not invertible?)

For the Matrices A find real closed formulas for the trajectory x→(t+1)=Ax→(t) where x (0)→=[01]

A=[43-34]

Consider an invertible n × n matrix A such that the zero state is a stable equilibrium of the dynamical system x→(t+1)=ATx→(t) What can you say about the stability of the systems.

x→(t+1)=ATx→(t)

Define \(T:{{\rm P}_2} \to {\mathbb{R}^3}\) by \(T\left( {\bf{p}} \right) = \left( {\begin{aligned}{{\bf{p}}\left( { - 1} \right)}\\{{\bf{p}}\left( 0 \right)}\\{{\bf{p}}\left( 1 \right)}\end{aligned}} \right)\).

  1. Find the image under\(T\)of\({\bf{p}}\left( t \right) = 5 + 3t\).
  2. Show that \(T\) is a linear transformation.
  3. Find the matrix for \(T\) relative to the basis \(\left\{ {1,t,{t^2}} \right\}\)for \({{\rm P}_2}\)and the standard basis for \({\mathbb{R}^3}\).
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