Conference rusure::math

I'll try and answer the easy part (my Linear Algebra is pretty feeble):

If one triangular half of a matrix is zero, and the main diagonal is zero,
then either the top or bottom row is all zero; since the determinant is a
sum of products, and each product has one entry from every row of the matrix,
the determinant will be zero.

							Richie

    Continuing from .-1, if A is the matrix, I is the identity matrix,
    and you solve for the eigenvalues x by
    
                            det(A - xI) = 0
    
    you end up with polynomial x^n = 0.  So all of the eigenvalues
    are zero in this case.
    
    [it's been a long time, someone should check this]
    
    Dan

    There's nothing to prevent *one* of the 'distinct' eigenvalues from
    being zero, so a matrix with distinct eigenvalues can be singular.

    If a matrix is diagonal, then the eigenvalues are along the major
    diagonal already; so if the major diagonal is all zero then all
    the eigenvalues are zero too.  Such a matrix is called "nilpotent"
    since raising it to a power will always result in a null matrix
    after some point.

    Note that it is entirely possible for the eigenvectors of a matrix
    to not span the whole vector space the matrix acts on.  This in fact
    happens in the nilpotent case you mentioned.  Additional vectors
    must be added to 'flesh out' the eigenspace; they're called
    principal vectors if I recall correctly.

    There exist various canonical forms that exhibit the underlying
    structure of matrices, and the upper triangular form is one of them.

    - Jim

I agree with the answers above.

As long as we're on the subject, a friend of mine the other day was
looking at a paper on recursively defined matrices and noted that the
author qualified his proof with the statement that his oeprations only
worked on "matrices of full rank."  This subject used to be a hobby of
mine, but I can't for the life of me recall what "full rank" means.
Can somebody in the notesfile define "full rank" for a matrix?

Thanks,
Dave

    My recollection:
    
    Take an m by n matrix, with m >= n, and use it as a linear
    transformation of all n-vectors into m-vectors.  If the subspace
    of m-dimensional space occupied by the transformed n-vectors
    is n-dimensional, then the matrix is of full rank.  For an n by n
    matrix I think it means nonsingular (invertible).
    
    Dan

I just realized that the questions in .0 didn't get completely
answered, and I agreed a bit hastily.

No, having one of the triangles in a matrix be 0 does not imply
that the eigenvalues are 0.  The eigenvalues of any upper or
lower diagonal matrix are the main diagonal elements.  To see
this, note that

  a x x  L = a,b,c since (L1-a)(L2-b)(L3-c) = 0
  0 b x
  0 0 c

If you get n distinct eigenvalues, then the matrix is diagonalizable,
so, of course, the inverse exists. The determinant must be nonzero 
for the matrix to be invertible.

Dave

    Re .-1:
    
    The question in .0 about zero eigenvalues came after stipulating
    that the triangular matrix had all zeroes on the diagonal, too.
    
    A triangular matrix with one zero on the diagonal may still have
    distinct eigenvalues and yet is not invertible.
    
    Dan

  Yes, of course.  Reading the question helps.

  thanks,
  Dave