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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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.

booyo · Accepted Answer

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

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.

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