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No. Say your matrix is called A, then a number e is an eigenvalue of A exactly when A-eI is singular, where I is the identity matrix of the same dimensions as A. A-eI is singular exactly when (A-eI)T is singular, but (A-eI)T=AT-(eI)T =AT-eI. Therefore we can conclude that e is an eigenvalue of A exactly when it is an eigenvalue of AT.
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