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Chapter 5: Q19E (page 267)

Question: For \(A = \left( {\begin{array}{*{20}{c}}1&2&3\\1&2&3\\1&2&3\end{array}} \right)\), find the eigenvalues, with no calculation. Justify your answer.

Short Answer

Expert verified

Eigenvalues of the given matrix is 0, as the determinant of the matrix is 0, so it is non-invertible, which imply \(A{\bf{x}} = 0\) has a nontrivial solution.

Step by step solution

01

Definition

Eigenvalue: Let \(\lambda \) is a scaler, \(A\) is an \(n \times n\) matrix and \({\bf{x}}\) is an eigenvector corresponding to \(\lambda \), \(\lambda \) is said to an eigenvalue of the matrix \(A\) if there exists a nontrivial solution \({\bf{x}}\) of \(A{\bf{x}} = \lambda {\bf{x}}\).

02

Find Eigenvalues 

The given matrix is \(A = \left( {\begin{array}{*{20}{c}}1&2&3\\1&2&3\\1&2&3\end{array}} \right)\).

As per the definition of eigenvalues, if \(A{\bf{x}} = \lambda {\bf{x}}\) has a nontrivial solution, then \(\lambda \) is the eigenvalue, and the matrix \(A\) is non-invertible if and only if \(A{\bf{x}} = 0\) has nontrivial solution and 0 is the eigenvalue.

Here, the matrix is invertible if its determinant is not 0. So, check, the matrix is invertible or not by checking its determinant.

The determinant of the determinant is 0, as all the rows of the matrix are identical. Which imply the matrix is non-invertible. Hence, the eigenvalue will be 0.

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Most popular questions from this chapter

Let\(B = \left\{ {{{\bf{b}}_{\bf{1}}},{{\bf{b}}_{\bf{2}}},{{\bf{b}}_{\bf{3}}}} \right\}\) and \(D = \left\{ {{{\bf{d}}_{\bf{1}}},{{\bf{d}}_{\bf{2}}}} \right\}\) be bases for vector space \(V\) and \(W\), respectively. Let \(T:V \to W\) be a linear transformation with the property that

\(T\left( {{{\bf{b}}_1}} \right) = 3{{\bf{d}}_1} - 5{{\bf{d}}_2}\), \(T\left( {{{\bf{b}}_2}} \right) = - {{\bf{d}}_1} + 6{{\bf{d}}_2}\), \(T\left( {{{\bf{b}}_3}} \right) = 4{{\bf{d}}_2}\)

Find the matrix for \(T\) relative to \(B\), and\(D\).

Apply the results of Exercise \({\bf{15}}\) to find the eigenvalues of the matrices \(\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{2}}\\{\bf{2}}&{\bf{1}}&{\bf{2}}\\{\bf{2}}&{\bf{2}}&{\bf{1}}\end{aligned}} \right)\) and \(\left( {\begin{aligned}{*{20}{c}}{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}\end{aligned}} \right)\).

Define\(T:{{\rm P}_3} \to {\mathbb{R}^4}\)by\(T\left( {\bf{p}} \right) = \left( {\begin{aligned}{{\bf{p}}\left( { - 3} \right)}\\{{\bf{p}}\left( { - 1} \right)}\\{{\bf{p}}\left( 1 \right)}\\{{\bf{p}}\left( 3 \right)}\end{aligned}} \right)\).

  1. Show that \(T\) is a linear transformation.
  2. Find the matrix for \(T\) relative to the basis \(\left\{ {1,t,{t^2},{t^3}} \right\}\)for \({{\rm P}_3}\)and the standard basis for \({\mathbb{R}^4}\).

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

7. \(\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{6}}&{{\bf{ - 1}}}\end{array}} \right)\)

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

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