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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: 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 18: It can be shown that the algebraic multiplicity of an eigenvalue \(\lambda \) is always greater than or equal to the dimension of the eigenspace corresponding to \(\lambda \). Find \(h\) in the matrix \(A\) below such that the eigenspace for \(\lambda = 5\) is two-dimensional:

\[A = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&3&h&0\\0&0&5&4\\0&0&0&1\end{array}} \right]\]

Question: Is \(\lambda = 2\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right)\)? Why or why not?

The trace of a square matrix \(A\) is the sum of the diagonal entries in A and is denoted by \({\mathop{\rm tr}\nolimits} A\). It can be verified that \({\mathop{\rm tr}\nolimits} \left( {FG} \right) = {\mathop{\rm tr}\nolimits} \left( {GF} \right)\) for any \(n \times n\) matrices Fand G. Show that if A and B are similar, then \({\mathop{\rm tr}\nolimits} A = {\mathop{\rm tr}\nolimits} B\).

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

15. \(\left( {\begin{array}{*{20}{c}}{\bf{7}}&{\bf{4}}&{{\bf{16}}}\\{\bf{2}}&{\bf{5}}&{\bf{8}}\\{{\bf{ - 2}}}&{{\bf{ - 2}}}&{{\bf{ - 5}}}\end{array}} \right)\)
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