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Chapter 5: Q5.3-26E (page 267)

Question: A is a \({\bf{7}} \times {\bf{7}}\) matrix with three eigenvalues. One eigenspace is two-dimensional and one of the other eigenspaces is three-dimensional. Is it possible that A is not diagonalizable? Justify your answer.

Short Answer

Expert verified

The sum of the dimensions of the eigenspaces is 6. But the order of the matrix is \(7 \times 7\). So, matrix A is not diagonalizable.

Step by step solution

01

Write the given information

Matrix A is of an order \(7 \times 7\) that has 3 eigenvalues and one eigenspace is two-dimensional and one of the other eigenspaces is three-dimensional.

02

Find that A is diagonalizable

If the third eigenspace is only one-dimensional. The sum of the dimensions of the eigenspaces is 6. But the order of the matrix is \(7 \times 7\).

Therefore, according to theorem 7, matrix A is not diagonalizable.

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Most popular questions from this chapter

Question: Let \(A = \left( {\begin{array}{*{20}{c}}{ - 6}&{28}&{21}\\4&{ - 15}&{ - 12}\\{ - 8}&a&{25}\end{array}} \right)\). For each value of \(a\) in the set \(\left\{ {32,31.9,31.8,32.1,32.2} \right\}\), compute the characteristic polynomial of \(A\) and the eigenvalues. In each case, create a graph of the characteristic polynomial \(p\left( t \right) = \det \left( {A - tI} \right)\) for \(0 \le t \le 3\). If possible, construct all graphs on one coordinate system. Describe how the graphs reveal the changes in the eigenvalues of \(a\) changes.

Let\(\varepsilon = \left\{ {{{\bf{e}}_1},{{\bf{e}}_2},{{\bf{e}}_3}} \right\}\) be the standard basis for \({\mathbb{R}^3}\),\(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\) be a basis for a vector space \(V\) and\(T:{\mathbb{R}^3} \to V\) be a linear transformation with the property that

\(T\left( {{x_1},{x_2},{x_3}} \right) = \left( {{x_3} - {x_2}} \right){{\bf{b}}_1} - \left( {{x_1} - {x_3}} \right){{\bf{b}}_2} + \left( {{x_1} - {x_2}} \right){{\bf{b}}_3}\)

  1. Compute\(T\left( {{{\bf{e}}_1}} \right)\), \(T\left( {{{\bf{e}}_2}} \right)\) and \(T\left( {{{\bf{e}}_3}} \right)\).
  2. Compute \({\left( {T\left( {{{\bf{e}}_1}} \right)} \right)_B}\), \({\left( {T\left( {{{\bf{e}}_2}} \right)} \right)_B}\) and \({\left( {T\left( {{{\bf{e}}_3}} \right)} \right)_B}\).
  3. Find the matrix for \(T\) relative to \(\varepsilon \), and\(B\).

Question: Let \(A = \left( {\begin{array}{*{20}{c}}{.6}&{.3}\\{.4}&{.7}\end{array}} \right)\), \({v_1} = \left( {\begin{array}{*{20}{c}}{3/7}\\{4/7}\end{array}} \right)\), \({x_0} = \left( {\begin{array}{*{20}{c}}{.5}\\{.5}\end{array}} \right)\). (Note: \(A\) is the stochastic matrix studied in Example 5 of Section 4.9.)

  1. Find a basic for \({\mathbb{R}^2}\) consisting of \({{\rm{v}}_1}\) and anther eigenvector \({{\rm{v}}_2}\) of \(A\).
  2. Verify that \({{\rm{x}}_0}\) may be written in the form \({{\rm{x}}_0} = {{\rm{v}}_1} + c{{\rm{v}}_2}\).
  3. For \(k = 1,2, \ldots \), define \({x_k} = {A^k}{x_0}\). Compute \({x_1}\) and \({x_2}\), and write a formula for \({x_k}\). Then show that \({{\bf{x}}_k} \to {{\bf{v}}_1}\) as \(k\) increases.

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

15. \(\left[ {\begin{array}{*{20}{c}}4&- 7&0&2\\0&3&- 4&6\\0&0&3&{ - 8}\\0&0&0&1\end{array}} \right]\)

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

12. \(\left( {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{2}}&{\bf{2}}\\{\bf{2}}&{\bf{4}}&{\bf{2}}\\{\bf{2}}&{\bf{2}}&{\bf{4}}\end{array}} \right)\)
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