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Q5.2-6E

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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: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

6. \(\left[ {\begin{array}{*{20}{c}}3&- 4\\4&8\end{array}} \right]\)

Characteristic polynomial: \({\lambda ^2} - 11\lambda + 40\).

\(A\) has no real Eigen values.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Find the characteristic polynomial

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

It is given that \(A = \left[ {\begin{array}{*{20}{c}}3&- 4\\4&8\end{array}} \right]\) and \(I = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\) is identity matrix. Find the matrix\(\left( {A - \lambda I} \right)\) as shown below:

\[\begin{array}A - \lambda I = \left[ {\begin{array}{*{20}{c}}3&- 4\\4&8\end{array}} \right] - \lambda \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 4}\\4&{8 - \lambda }\end{array}} \right]\end{array}\]

Now, calculate the determinant of the matrix \(\left( {A - \lambda I} \right)\) as shown below:

\[\begin{array}det\left( {A - \lambda I} \right) = det\left[ {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 4}\\4&{8 - \lambda }\end{array}} \right]\\ = \left( {8 - \lambda } \right)\left( {3 - {\rm{\lambda }}} \right) + 16\\ = {\lambda ^2} - 11\lambda + 24 + 16\\ = {\lambda ^2} - 11\lambda + 40\end{array}\]

So, the characteristic polynomial of is \({\lambda ^2} - 11\lambda + 40\).

Step 2: Describe the characteristic equation

To find the eigenvalues of the matrix, we must calculate all the scalars such that \(\left( {A - \lambda I} \right)x = 0\) has a non-trivial solution which is equivalent to finding all such that the matrix \(\left( {A - \lambda I} \right)\)is not invertible, that is, when determinant of \(\left( {A - \lambda I} \right)\)is zero.

Thus, the eigenvalues of \(A\) are the solutions of the characteristic equation\(\det \left( {A - \lambda I} \right) = 0\). So, find the characteristic equation \(\det \left( {A - \lambda I} \right) = 0\).

\[\begin{array}det\left[ {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 4}\\4&{8 - \lambda }\end{array}} \right] = 0\\\left( {8 - \lambda } \right)\left( {3 - {\rm{\lambda }}} \right) + 16 = 0\\{\lambda ^2} - 11\lambda + 24 + 16 = 0\\{\lambda ^2} - 11\lambda + 40 = 0\end{array}\]

Step 3: Find roots of characteristic equation

For the quadratic equation, \(a{x^2} + bx + c = 0\) , the general solution is given as\(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} \;\;}}{{2a}}\).

Thus, the solution of the characteristic equation \[{\lambda ^2} - 11\lambda + 40 = 0\] is obtained as follows:

\[\begin{array}{\lambda ^2} - 11\lambda + 40 = 0\\\lambda = \frac{{ - \left( { - 11} \right) \pm \sqrt {{{\left( { - 11} \right)}^2} - 4\left( {40} \right)} }}{2}\\ = \frac{{ - 11 \pm \sqrt { - 39} }}{2}\end{array}\]

The eigenvalues of \(A\) are complex. So, \(A\) has no real Eigen values. This implies that no real vector \(x\)in \({\mathbb{R}^2}\)satisfies the characteristic equation \(\left( {A - \lambda I} \right)x = 0\).

Most popular questions for Math Textbooks

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 \(\left\{ {1,t,{t^2},{t^3}} \right\}\).

Exercises 19–23 concern the polynomial \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + ... + {a_{n - {\bf{1}}}}{t^{n - {\bf{1}}}} + {t^n}\) and \(n \times n\) matrix \({C_p}\) called the companion matrix of \(p\): \({C_p} = \left[ {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right]\).

22. Let \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + {a_{\bf{2}}}{t^{\bf{2}}} + {t^{\bf{3}}}\), and let \(\lambda \) be a zero of \(p\).

  1. Write the companion matrix for \(p\).
  2. Explain why \({\lambda ^{\bf{3}}} = - {a_{\bf{0}}} - {a_{\bf{1}}}\lambda - {a_{\bf{2}}}{\lambda ^{\bf{2}}}\), and show that \(\left( {{\bf{1}},\lambda ,{\lambda ^2}} \right)\) is an eigenvector of the companion matrix for \(p\).

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

A=[43-34]

Consider the growth of a lilac bush. The state of this lilac bush for several years (at year’s end) is shown in the accompanying sketch. Let n(t) be the number of new branches (grown in the year t) and a(t) the number of old branches. In the sketch, the new branches are represented by shorter lines. Each old branch will grow two new branches in the following year. We assume that no branches ever die.

(a) Find the matrix A such that [nt+1at+1]=A[ntat]

(b) Verify that [11] and [2-1] are eigenvectors of A. Find the associated eigenvalues.

(c) Find closed formulas for n(t) and a(t).

Question: Construct a random integer-valued \(4 \times 4\) matrix \(A\), and verify that \(A\) and \({A^T}\) have the same characteristic polynomial (the same eigenvalues with the same multiplicities). Do \(A\) and \({A^T}\) have the same eigenvectors? Make the same analysis of a \(5 \times 5\) matrix. Report the matrices and your conclusions.

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