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Chapter 5: Q5.3-28E (page 267)

Question: Show that if A has n linearly independent eigenvectors, then so does \({A^T}\). [Hint: Use the diagonalization theorem.]

Short Answer

Expert verified

It is proved that the matrix \({A^T}\) is diagonalizable and the columns of matrix Q are n linearly independent eigenvectors of \({A^T}\).

Step by step solution

01

Write the diagonalization theorem

If the matrix A has n linearly independent vectors, then by the diagonalization theorem, \(A = PD{P^{ - 1}}\).

The matrix P is invertible and matrix D is diagonalizable.

02

Check if transpose of A has n linearly independent vector

Using the properties of transpose,

\[\begin{array}{c}{A^T} = {\left( {PD{P^{ - 1}}} \right)^T}\\ = {\left( {{P^{ - 1}}} \right)^T}{D^T}{P^T}\\ = {\left( {{P^T}} \right)^{ - 1}}D{P^T}\\ = QD{Q^{ - 1}}\end{array}\]

Here, \(Q = {\left( {{P^T}} \right)^{ - 1}}\). Thus, the matrix \({A^T}\) is diagonalizable and the columns of matrix Q are n linearly independent eigenvectors of \({A^T}\).

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Most popular questions from this chapter

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: 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.

Let\(G = \left( {\begin{aligned}{*{20}{c}}A&X\\{\bf{0}}&B\end{aligned}} \right)\). Use formula\(\left( {\bf{1}} \right)\)for the determinant in section\({\bf{5}}{\bf{.2}}\)to explain why\(\det G = \left( {\det A} \right)\left( {\det B} \right)\). From this, deduce that the characteristic polynomial of\(G\)is the product of the characteristic polynomials of\(A\)and\(B\).

Question: In Exercises \({\bf{1}}\) and \({\bf{2}}\), let \(A = PD{P^{ - {\bf{1}}}}\) and compute \({A^{\bf{4}}}\).

1. \(P{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{7}}\\{\bf{2}}&{\bf{3}}\end{array}} \right)\), \(D{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{2}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}\end{array}} \right)\)

Question: In Exercises \({\bf{5}}\) and \({\bf{6}}\), the matrix \(A\) is factored in the form \(PD{P^{ - {\bf{1}}}}\). Use the Diagonalization Theorem to find the eigenvalues of \(A\) and a basis for each eigenspace.

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