Question: Justify each answer. (T/F) If there is a nonzero vector in the kernel of a linear transformation T, then 0 is an eigenvalue of T.
Justify each answer.
(T/F) If there is a nonzero vector in the kernel of a linear transformation T, then 0 is an eigenvalue of T.
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True See the definition prior to Example 1 Let V be a vector s... View full answer
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