Let lambda be an eigenvalue of A, such that no eigenvector of A associated with lambda has a zero entry. Then prove that every list of n−1 columns of A−lambda I is linearly independent.

Ivan Buckley 2022-09-25 Answered
Let λ be an eigenvalue of A, such that no eigenvector of A associated with λ has a zero entry. Then prove that every list of n−1 columns of A λ I is linearly independent.
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Answers (1)

booyo
Answered 2022-09-26 Author has 6 answers
The null space of A n × n λ I is the eigenspace of λ. Since no eigenvector corresponding to λ contains a 0 entry, the basis of the eigenspace must contain a single eigenvector of A. This means the nullity of A λ I is 1, and by the rank-nullity theorem, its rank n−1.
Now recall that ( A λ I ) v is a linear combination of the columns of A λ I. This means their exists a tuple ( v 1 , v 2 , . . . , v n ) with v i 0 that satisfies v 1 C 1 + v 2 C 2 + + v n C n = 0. Since the rank of these vectors is n−1, there is one linearly dependent column C j , and since v j 0 for all j, that C j could be any column
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