Basic question of linear algebra

  • Thread starter td21
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  • #1
td21
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I know the rank nullity theorem.
But can i say that the vectors in the column space of A and the vectors in the nullspace of A are linearly independent?
Thanks and this is not a hw.
 

Answers and Replies

  • #2
td21
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i think the nilpotent matrix may say its not true, any thought?
 
  • #3
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no, you don't know that vectors Tv will belong in your nullspace or not (in general , but you might have an invariant transformation, or a projection). You know that v is linearly independent from your nullspace vectors if Tv is not zero
 
  • #4
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Well, I don't know if this is what you refer to, but the relation between A

and A^T seen as maps is that the nullspace of A^T is the orthogonal complement

of Im(A) (the image set of A, or, if A represents a map L:V-->W: {Ax: x in V} )

and the image of A^T is the orthogonal complement of N(A), the nullspace of A.
 
  • #5
td21
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Well, I don't know if this is what you refer to, but the relation between A

and A^T seen as maps is that the nullspace of A^T is the orthogonal complement

of Im(A) (the image set of A, or, if A represents a map L:V-->W: {Ax: x in V} )

and the image of A^T is the orthogonal complement of N(A), the nullspace of A.

Well what i mean is that are the vectors in R(A) and N(A) linearly independent.
 
  • #6
Deveno
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for a linear transformation T:U-->V, Im(T) = T(U) and ker(T) = null(T) are subspaces of two different vector spaces. T(U) is a subspace of V, ker(T) is a subspace of U. so it doesn't even make sense to say that the vectors of T(U) and ker(T) are "independent", they aren't even in the same vector space.

Even if T:V-->V, you can have Tv be in ker(T), for example if v is in the null space of T^2 (in fact with nilpotent matrices of order 2, this is exactly what happens).
 
Last edited:
  • #7
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yea, I misread row space and column space, not nullspace. But then , like Deveno said,
your question would only make sense for a map T:V-->V.
 

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