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Chapter 5: Q26E (page 267)

Question: Show that if \({A^{\bf{2}}}\) is the zero matrix, then the only eigenvalue of A is 0.

Short Answer

Expert verified

If the matrix \({A^2}\) is zero, then each eigenvalue of A is zero.

Step by step solution

01

Write the given information

For matrix A, the matrix \({A^2}\) is 0.

02

Check for the eigenvalue of \({A^{ - {\bf{1}}}}\)

As \({A^2}\) is a zero matrix, so the equation \(A{\bf{x}} = \lambda {\bf{x}}\) and \(x \ne 0\) can be represented as;

\(\begin{array}{c}{A^2}{\bf{x}} = A\left( {A{\bf{x}}} \right)\\ = A\left( {\lambda {\bf{x}}} \right)\\ = \lambda A{\bf{x}}\\ = {\lambda ^2}{\bf{x}}\end{array}\)

As x is nonzero, so \(\lambda \) must be zero.

Therefore, the eigenvalue of A is zero.

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Most popular questions from this chapter

Question: In Exercises \({\bf{3}}\) and \({\bf{4}}\), use the factorization \(A = PD{P^{ - {\bf{1}}}}\) to compute \({A^k}\) where \(k\) represents an arbitrary positive integer.

3. \(\left( {\begin{array}{*{20}{c}}a&0\\{3\left( {a - b} \right)}&b\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}a&0\\0&b\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\{ - 3}&1\end{array}} \right)\)

Suppose \(A = PD{P^{ - 1}}\), where \(P\) is \(2 \times 2\) and \(D = \left( {\begin{array}{*{20}{l}}2&0\\0&7\end{array}} \right)\)

a. Let \(B = 5I - 3A + {A^2}\). Show that \(B\) is diagonalizable by finding a suitable factorization of \(B\).

b. Given \(p\left( t \right)\) and \(p\left( A \right)\) as in Exercise 5 , show that \(p\left( A \right)\) is diagonalizable.

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.

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

[M]Repeat Exercise 25 for \[A{\bf{ = }}\left[ {\begin{array}{*{20}{c}}{{\bf{ - 8}}}&{\bf{5}}&{{\bf{ - 2}}}&{\bf{0}}\\{{\bf{ - 5}}}&{\bf{2}}&{\bf{1}}&{{\bf{ - 2}}}\\{{\bf{10}}}&{{\bf{ - 8}}}&{\bf{6}}&{{\bf{ - 3}}}\\{\bf{3}}&{{\bf{ - 2}}}&{\bf{1}}&{\bf{0}}\end{array}} \right]\].

Let \({\bf{u}}\) be an eigenvector of \(A\) corresponding to an eigenvalue \(\lambda \), and let \(H\) be the line in \({\mathbb{R}^{\bf{n}}}\) through \({\bf{u}}\) and the origin.

  1. Explain why \(H\) is invariant under \(A\) in the sense that \(A{\bf{x}}\) is in \(H\) whenever \({\bf{x}}\) is in \(H\).
  2. Let \(K\) be a one-dimensional subspace of \({\mathbb{R}^{\bf{n}}}\) that is invariant under \(A\). Explain why \(K\) contains an eigenvector of \(A\).
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