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Chapter 5: Q6E (page 267)

Question: Is \(\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\) an eigenvector of\(\left){\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)\)? If so, find the eigenvalue.

Short Answer

Expert verified

Yes, \(\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\) is the eigenvector of the given matrix \(\left( {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)\) , and the eigenvalue is \( - 2\).

Step by step solution

01

Definition of eigenvector

If there exists a non-zero vector \({\bf{x}}\) that satisfies \(A{\bf{x}} = \lambda {\bf{x}}\) for some scaler \(\lambda \), then \({\bf{x}}\) be the eigenvector of an \(n \times n\) matrix \(A\), and if \(A{\bf{x}} = \lambda {\bf{x}}\) exists, then scaler \(\lambda \) is the eigenvalue of the matrix.

02

Determine whether \(\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\) is the eigenvector of the given matrix

Denote the given matrix by \(A\) and the given vector by \({\bf{x}}\).

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

According to the definition of eigenvalue, \({\bf{x}} = \left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\) is the eigenvector of the matrix \(A\), if \(A{\bf{x}} = \lambda {\bf{x}}\).

Find the product of \(A\), and \({\bf{x}}\).

\(\begin{array}{c}A{\bf{x}} = \left( {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{3\left( 1 \right) + 6\left( { - 2} \right) + 7\left( 1 \right)}\\{3\left( 1 \right) + 3\left( { - 2} \right) + 7\left( 1 \right)}\\{5\left( 1 \right) + 6\left( { - 2} \right) + 5\left( 1 \right)}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{3 - 12 + 7}\\{3 - 6 + 7}\\{5 - 12 + 5}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 2}\\4\\{ - 2}\end{array}} \right)\end{array}\)

The obtained matrix in the form of the given vector can be written as:

\(A{\bf{x}} = \left( { - 2} \right)\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\)

So, the vector \(\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\) is the eigenvector of the given matrix \(\left( {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)\).

03

Determine the eigenvalue 

As the given vector satisfies the condition \(A{\bf{x}} = \lambda {\bf{x}}\). Which implies that \(\lambda \) is the eigenvalue of the given matrix. So, \(\lambda = - 2\).

So, \( - 2\) is the eigenvalue of the given matrix \(\left( {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)\).

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Most popular questions from this chapter

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

13. \(A = \left( {\begin{array}{*{20}{c}}3&{ - 2}&8\\0&5&{ - 2}\\0&{ - 4}&3\end{array}} \right)\)

Question: Construct a random integer-valued \(4 \times 4\) matrix \(A\).

  1. Reduce \(A\) to echelon form \(U\) with no row scaling, and use \(U\) in formula (1) (before Example 2) to compute \(\det A\). (If \(A\) happens to be singular, start over with a new random matrix.)
  2. Compute the eigenvalues of \(A\) and the product of these eigenvalues (as accurately as possible).
  3. List the matrix \(A\), and, to four decimal places, list the pivots in \(U\) and the eigenvalues of \(A\). Compute \(\det A\) with your matrix program, and compare it with the products you found in (a) and (b).

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 \(\det A \le {\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2}\).

Question: For the matrices in Exercises 15-17, list the eigenvalues, repeated according to their multiplicities.

17. \(\left[ {\begin{array}{*{20}{c}}3&0&0&0&0\\- 5&1&0&0&0\\3&8&0&0&0\\0&- 7&2&1&0\\- 4&1&9&- 2&3\end{array}} \right]\)

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

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