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Chapter 5: Q23E (page 267)

Question: Explain why a \({\bf{2}} \times {\bf{2}}\) matrix can have at most two distinct eigenvalues. Explain why an \(n \times n\) matrix can have at most n distinct eigenvalues.

Short Answer

Expert verified

A matrix \(2 \times 2\) matrix A were to have three distinct eigenvalues. This is impossible because the vectors all belong to a two-dimensional vector space and the set of vectors is linearly dependent.

If the vectors belong to n-dimensional vectors space, then p cannot exceed n.

Step by step solution

01

Write an explanation for the matrix to have two distinct eigenvalues

According to Theorem 2, a matrix \(2 \times 2\) matrix A was to have three distinct eigenvalues. This is impossible because the vectors all belong to a two-dimensional vector space and the set of vectors is linearly dependent.

02

Write an explanation for the matrix to have n distinct eigenvalues

If \(n \times n\) the matrix has \(p\) distinct values, there would be a linearly independent set of p eigenvectors.

Therefore, these vectors belong to n-dimensional vectors space, p cannot exceed n.

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Most popular questions from this chapter

a. Let \(A\) be a diagonalizable \(n \times n\) matrix. Show that if the multiplicity of an eigenvalue \(\lambda \) is \(n\), then \(A = \lambda I\).

b. Use part (a) to show that the matrix \(A =\left({\begin{aligned}{*{20}{l}}3&1\\0&3\end{aligned}}\right)\) is not diagonalizable.

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.

For the matrix A, find real closed formulas for the trajectoryx→(t+1)=Ax¯(t)where x→=[01]. Draw a rough sketch

A=[15-27]

Show that \(I - A\) is invertible when all the eigenvalues of \(A\) are less than 1 in magnitude. (Hint: What would be true if \(I - A\) were not invertible?)

Question 19: Let \(A\) be an \(n \times n\) matrix, and suppose A has \(n\) real eigenvalues, \({\lambda _1},...,{\lambda _n}\), repeated according to multiplicities, so that \(\det \left( {A - \lambda I} \right) = \left( {{\lambda _1} - \lambda } \right)\left( {{\lambda _2} - \lambda } \right) \ldots \left( {{\lambda _n} - \lambda } \right)\) . Explain why \(\det A\) is the product of the n eigenvalues of A. (This result is true for any square matrix when complex eigenvalues are considered.)
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