True or False: If 'L' is an eigenvalue for some square matrix, A, then 'L' must also be an eigenvalue for the square matrix A-1.

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BlackTom AI · Accepted Answer

First, note the setup: L is an eigenvalue of A, and we are considering the matrix A^{-1} (the inverse). The key fact is that if A is invertible and λ is an eigenvalue of A, then the eigenvalues of A^{-1} are the reciprocals 1/λ, not λ itself.

Option 1: True. Why this is incorrect: saying that L must also be an eigenvalue of A^{-1} would require L to equal 1/μ for some eigenvalue μ of A. In general, there's no reason to expect L = 1/μ for any μ; the spectrum of A and A^{-1} is related by taking reciprocals, not by preserving the same values. This choice ignores the reciprocal relationship between eigenvalues of A and those of A^{-1} and thus misstates the standard result.

Option 2: False. Why this is correct: since L is an eigenvalue of A, and A is invertible (to even talk about A^{-1}), the eigenvalues of A^{-1} are exactly the reciprocals 1/λ for each eigenvalue λ of A. Therefore, L itself would appear as an eigenvalue of A^{-1} only if L = 1/μ for some μ in the spectrum of A. In general, L is not guaranteed to equal such a reciprocal; the statement is not true in general. A concrete illustration: if A has eigenvalue L = 3, then A^{-1} has eigenvalue 1/3, not 3, so L is not an eigenvalue of A^{-1} in this case.

Additional nuance: the reasoning assumes A is invertible, which is necessary to define A^{-1}. If A were singular, A^{-1} would not exist and the statement would be moot. Even under the usual invertible condition, the direct equality of eigenvalues is not preserved; reciprocals govern the spectrum of the inverse, not the same values.

True or False: If 'L' is an eigenvalue for some square matrix, A, then 'L' must also be an eigenvalue for the square matrix A-1.

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