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Chapter 5: Q5E (page 267)

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.

Short Answer

Expert verified

Yes, \(\left( {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\) is the eigenvector of the given matrix \(\left( {\begin{array}{*{20}{c}}3&7&9\\{ - 4}&{ - 5}&1\\2&4&4\end{array}} \right)\) and the eigenvalue is 0.

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 \(\left( {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\) is the eigenvector of the given matrix

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

\(A = \left( {\begin{array}{*{20}{c}}3&7&9\\{ - 4}&{ - 5}&1\\2&4&4\end{array}} \right)\), \({\bf{x}} = \left( {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\)

According to the definition of eigenvalue, \({\bf{x}} = \left( {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\) is the eigenvector of the matrix \(A\), if \(A{\bf{x}} = \lambda {\bf{x}}\).

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

\(\begin{array}{c}A{\bf{x}} = \left( {\begin{array}{*{20}{c}}3&7&9\\{ - 4}&{ - 5}&1\\2&4&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{3\left( 4 \right) + 7\left( { - 3} \right) + 9\left( 1 \right)}\\{ - 4\left( 4 \right) + \left( { - 5} \right)\left( { - 3} \right) + 1\left( 1 \right)}\\{2\left( 4 \right) + 4\left( { - 3} \right) + 4\left( 1 \right)}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{12 - 21 + 9}\\{ - 16 + 15 + 1}\\{8 - 12 + 4}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\end{array}\)

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

\(A{\bf{x}} = 0\left( {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\)

Because 0 is a multiple of every non-zero number. Hence, the given vector satisfies the condition \(A{\bf{x}} = \lambda {\bf{x}}\).

So, the vector \(\left({\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\) is the eigenvector of the given matrix \(\left( {\begin{array}{*{20}{c}}3&7&9\\{ - 4}&{ - 5}&1\\2&4&4\end{array}} \right)\).

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, \(\lambda = 0\).

So, 0 is the eigenvalue of the given matrix \(\left( {\begin{array}{*{20}{c}}3&7&9\\{ - 4}&{ - 5}&1\\2&4&4\end{array}} \right)\).

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Most popular questions from this chapter

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

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

Question: Exercises 9-14 require techniques section 3.1. Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\) determinants described prior to Exercise 15-18 in Section 3.1. [Note: Finding the characteristic polynomial of a \(3 \times 3\) matrix is not easy to do with just row operations, because the variable \(\lambda \) is involved.

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

A common misconception is that if \(A\) has a strictly dominant eigenvalue, then, for any sufficiently large value of \(k\), the vector \({A^k}{\bf{x}}\) is approximately equal to an eigenvector of \(A\). For the three matrices below, study what happens to \({A^k}{\bf{x}}\) when \({\bf{x = }}\left( {{\bf{.5,}}{\bf{.5}}} \right)\), and try to draw general conclusions (for a \({\bf{2 \times 2}}\) matrix).

a. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{{\bf{.8}}}&{\bf{0}}\\{\bf{0}}&{{\bf{.2}}}\end{aligned}} \right)\) b. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{0}}&{{\bf{.8}}}\end{aligned}} \right)\) c. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{8}}&{\bf{0}}\\{\bf{0}}&{\bf{2}}\end{aligned}} \right)\)

Question: Show that if \(A\) and \(B\) are similar, then \(\det A = \det B\).

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

5. \(\left[ {\begin{array}{*{20}{c}}2&1\\-1&4\end{array}} \right]\)
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