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Eigenvalues of a rank 1 matrix?

  1. Mar 31, 2013 #1
    How come a square matrix has eigenvalues of 0 and the trace of the matrix?
    Is there any other proof other than just solving det(A-λI)=0?
     
  2. jcsd
  3. Apr 1, 2013 #2
    I might argue something like the following: By row operations, a rank 1 matrix may be reduced to a matrix with only the first row being nonzero. The eigenvectors of such a matrix may be chosen to be the ordinary Euclidian basis, in which the eigenvalues become zero's and the 11-component of this reduced matrix. As row operations are invertible, the trace is unchanged, and thus this nonzero eigenvalue equals the trace of the original matrix.

    Afterthought: But that is probably erroneous, because even though the row operations are indeed invertible, they do not generally preserve the trace. So the last part of my argument fails.

    A better argument seems to be the following: For a rank k matrix there exists a basis in which k of its columns are nonzero, the other ones being zero. The transformation between bases may be chosen to be orthogonal, thus preserving the trace.
     
    Last edited: Apr 1, 2013
  4. Apr 1, 2013 #3

    jbunniii

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    We assume ##A## is an ##n \times n## rank one matrix. If ##n > 1##, any rank one matrix is singular. Therefore ##\lambda = 0## is an eigenvalue: for an eigenvector, just take any nonzero ##v## such that ##Av = 0##.

    So let's see if there are any nonzero eigenvalues.

    If ##A## is a rank one matrix, then all of its columns are scalar multiples of each other. Thus we may write ##A = xy^T## where ##x## and ##y## are nonzero ##n \times 1## vectors.

    If ##\lambda## is an eigenvalue of ##A##, then there is a nonzero vector ##v## such that ##Av = \lambda v##. This means that ##(xy^T)v = \lambda v##. By associativity, we may rewrite the left hand side as ##x(y^T v) = \lambda v##.

    Note that ##y^T v## is a scalar, and of course ##\lambda## is also a scalar. If we assume ##\lambda \neq 0##, then this means that ##v## is a scalar multiple of ##x##: specifically, ##v = x(y^T v)/\lambda##.

    Therefore ##x## itself is an eigenvector associated with ##\lambda##, so we have ##x(y^T x) = \lambda x##, or equivalently, ##x(\lambda - y^T x) = 0##. As ##x## is nonzero, this forces ##\lambda = y^T x##.

    All that remains is to recognize that ##y^T x = \sum_{n = 1}^{N} x_n y_n## is the trace of ##A = xy^T##.
     
  5. Apr 1, 2013 #4

    jbunniii

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    By the way, note that this does not necessarily mean that ##A## has two distinct eigenvalues. The trace may well be zero, for example
    $$A = \begin{bmatrix}
    1 & 1 \\
    -1 & -1
    \end{bmatrix}$$
    is a rank one matrix whose only eigenvalue is 0.
     
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