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

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

Short Answer

Expert verified

The eigenvalues of \(A\) are \( - 1\), \( - 1\), \( - 5\), \(6\).

Step by step solution

01

Step 1: Find the characteristic polynomial of \(U\)

Assume \(G = \left( {\begin{array}{*{20}{c}}U&X\\0&V\end{array}} \right)\).

Then we have,

\(\begin{aligned}{c}A &= \left( {\begin{aligned}{*{20}{c}}1&5&{ - 6}&{ - 7}\\2&4&5&2\\0&0&{ - 7}&{ - 4}\\0&0&3&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}U&X\\0&V\end{aligned}} \right)\end{aligned}\)

On comparison we get,

\(U = \left( {\begin{aligned}{*{20}{c}}1&5\\2&4\end{aligned}} \right)\),\(V = \left( {\begin{aligned}{*{20}{c}}{ - 7}&{ - 4}\\3&1\end{aligned}} \right)\)

Now characteristic polynomial of \(U\) are as shown below:

\(\begin{aligned}{c}\det \left( {\begin{aligned}{*{20}{c}}{1 - \lambda }&5\\2&{4 - \lambda }\end{aligned}} \right) &= \left( {1 - \lambda } \right)\left( {4 - \lambda } \right) - 10\\ &= {\lambda ^2} - 5\lambda - 6\\ &= \left( {\lambda + 1} \right)\left( {\lambda - 6} \right)\end{aligned}\)

02

Step 2: Find the characteristic polynomial of \(V\)

\(\begin{aligned}{c}\det \left( {\begin{aligned}{*{20}{c}}{ - 7 - \lambda }&{ - 4}\\3&{1 - \lambda }\end{aligned}} \right) &= \left( { - 7 - \lambda } \right)\left( {1 - \lambda } \right) + 12\\ &= {\lambda ^2} + 6\lambda + 5\\ &= \left( {\lambda + 1} \right)\left( {\lambda + 5} \right)\end{aligned}\)

On multiplication we get the characteristic polynomial of \(A\).

\(\begin{aligned}{c}A &= \left( {\lambda + 1} \right)\left( {\lambda - 6} \right)\left( {\lambda + 1} \right)\left( {\lambda + 5} \right)\\ &= {\left( {\lambda + 1} \right)^2}\left( {\lambda + 5} \right)\left( {\lambda - 6} \right)\end{aligned}\)

Thus, the eigenvalues of \(A\) are \( - 1\), \( - 1\), \( - 5\), \(6\).

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Most popular questions from this chapter

Question: Is \(\left( {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\) an eigenvector of \(\left( {\begin{array}{*{20}{c}}3&7&9\\{ - 4}&{ - 5}&1\\2&4&4\end{array}} \right)\)? If so, find the eigenvalue.

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?

Question: In Exercises 21 and 22, \(A\) and \(B\) are \(n \times n\) matrices. Mark each statement True or False. Justify each answer.

  1. If \(A\) is \(3 \times 3\), with columns \({{\rm{a}}_1}\), \({{\rm{a}}_2}\), and \({{\rm{a}}_3}\), then \(\det A\) equals the volume of the parallelepiped determined by \({{\rm{a}}_1}\), \({{\rm{a}}_2}\), and \({{\rm{a}}_3}\).
  2. \(\det {A^T} = \left( { - 1} \right)\det A\).
  3. The multiplicity of a root \(r\) of the characteristic equation of \(A\) is called the algebraic multiplicity of \(r\) as an eigenvalue of \(A\).
  4. A row replacement operation on \(A\) does not change the eigenvalues.

Question: Is \(\lambda = - 2\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right)\)? Why or why not?

Suppose \(A\) is diagonalizable and \(p\left( t \right)\) is the characteristic polynomial of \(A\). Define \(p\left( A \right)\) as in Exercise 5, and show that \(p\left( A \right)\) is the zero matrix. This fact, which is also true for any square matrix, is called the Cayley-Hamilton theorem.

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