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Relationship between eigenspace and column space

  1. Apr 14, 2012 #1
    Is it true that if an n by n matrix A has n-linearly independent eigenvectors, then it must also be invertible because these n-eigenvectors span n-space. But does this reasoning work the other way around: that is if A is invertible, does that imply n-linearly independent eigenvectors can be found? More generally, is there some connection between the column space and eigenspace?

    On a side note, here is an unrelated but interesting question: why is it that if the square of a matrix is the negative identity matrix (-I), then it implies that the matrix has an even number of columns and rows?

    Thanks in advance
  2. jcsd
  3. Apr 14, 2012 #2


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    What if 0 is an eigenvalue?

    No. Consider ##\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)##.

    If A^2=-I then det(A^2)=det(-I). Try to see if you can use this to answer your question.
  4. Apr 14, 2012 #3
    Thanks for your answer. The third question is completely clear to me now. But for the first and second questions, are there no connections between eigenspace and column space?
  5. Apr 16, 2012 #4
    If A =P'EP where P'P = I and E is a diagonal matrix of eigenvalues, then:

    A^-1 = P^-1* E^-1 * (P')^-1

    But P^-1 = P' and (P')^-1 = P

    So, A^-1 = P' * E^-1 * P

    So as long as the eigenvalues are non-zero, P is the eigenvector for both A and A^-1.
    Now, the column space is the row space of the transpose. Assuming again that A=P'EP then:

    A' = (P'EP)' = P'E'P = P'EP

    so the dimension of eigenspace of the column space and row space are equal. In the event that E is upper diagonal form (not a diagonal matrix), I believe a similar statement can be made.
    Last edited: Apr 16, 2012
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