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

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Consider the statement and the two possible responses to evaluate.
Option 1: True. When L is an eigenvalue of A, there exists a nonzero vector v such that Av = Lv. Taking transposes, we have v^T A^T = L v^T, which shows that L is also an eigenvalue of A^T with corresponding eigenvector v (or, more precisely, the transpose relationship confirms that the eigenvalues of A and A^T coincide). A more formal justification uses the fact that det(A − λI) = det(A^T − λI), so the characteristic polynomials are identical and their roots (the eigenvalues) match. Therefore, L being an eigenvalue of A implies L is an eigenvalue of A^T.
Option 2: False. This would claim that the statement sometimes fails, but the underlying spectral property actually guarantees identical eigenvalues for A and its transpose, due to the equality of characteristic polynomials. Misconceptions here might involve thinking about eigenvectors changing under transpose, but eigenvalues themselves remain the same since det(A − λI) = det(A^T − λI) holds for all λ. Thus this option contradicts the established result and is incorrect.
Option 3: If presented, would typically explore edge cases or special fields, but in standard linear algebra over the real or complex numbers, the eigenvalue correspondence between A and A^T is robust, so no exception undermines the general statement. Any claim suggesting a general failure would be unfounded.
Option 4: Any reasoning that relies on a mismatch between eigenvalues of A and A^T due to, say, non-square contexts or defective matrices would be misguided here since we are dealing with square matrices and the transpose preserves the eigenvalue set. Such arguments misinterpret the invariant nature of eigenvalues under transposition.

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

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