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Let A be an n x n matrix and let I be the n x n identity matrix. Show that if A? = I and A + I, then A= -1 is an eigenvalue of A.

Let A be an n x n matrix and let I be the n x n identity matrix. Show that if A? = I and A + I, then A= -1 is an eigenvalue of A.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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for this problem, it cannot be assumed that A is diagonal. In particular, A2 = A does not imply A = 0 or A = I, and A2 = I does not imply A = ±I.

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Let A be an n x n matrix and let I be the n x n identity matrix. Show that if A? = I and A + I, then A= -1 is an eigenvalue of A.
Transcribed Image Text:Let A be an n x n matrix and let I be the n x n identity matrix. Show that if A? = I and A + I, then A= -1 is an eigenvalue of A.
The proofs use one or more of the following: The minimal polynomial of A is the monic polynomial of lowest degree with A as a zero. In particular, ma(A) = 0. - If f(t) is a polynomial such that f(A) = 0, then ma(t) is a factor of f(t). The eigenvalues of A are precisely the roots of ma(t). Note that the characteristic polynomial is not listed. In fact, we have much less in- formation about A4(t) than we do about ma(t), as we do not even know the size of A.
Transcribed Image Text:The proofs use one or more of the following: The minimal polynomial of A is the monic polynomial of lowest degree with A as a zero. In particular, ma(A) = 0. - If f(t) is a polynomial such that f(A) = 0, then ma(t) is a factor of f(t). The eigenvalues of A are precisely the roots of ma(t). Note that the characteristic polynomial is not listed. In fact, we have much less in- formation about A4(t) than we do about ma(t), as we do not even know the size of A.
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