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Math Help - Spectral Decomposition of a Matrix

  1. #1
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    Spectral Decomposition of a Matrix

    A= \begin{bmatrix} 7 & 3 & 3 &2 \\ 0 & 1 & 2 & -4 \\ -8 &<br />
   -4& -5 &0 \\ 2& 1 & 2 & 3\end{bmatrix}<br />

    Let T be the linear Operator in  R_n that has the given matrix A relative to the standard basis
     E_n.
    Find the Spectral Decomposition of T.

    I found that characteristic polynomial is (x-3)^2 (x+1)(x-1)
    So  T=P_1-P_2+3P_3 , where P are projections.

    Trying to find E's such that A=x_1E_1+x_2E_2+x_3E_3=E_1-E_2+3E_3
    E are Projections matrix relative to  E_3

    Thanks in advance


    Last edited by firebio; April 17th 2010 at 03:11 PM.
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  2. #2
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    Quote Originally Posted by firebio View Post
    A= \begin{bmatrix} 7 & 3 & 3 &2 \\ 0 & 1 & 2 & -4 \\ -8 &<br />
   -4& -5 &0 \\ -2 & 1 & 2 & 3\end{bmatrix}<br />

    Let T be the linear Operator in  R_n that has the given matrix A relative to the standard basis  E_n.
    Find the Spectral Decomposition of T.

    I found that characteristic polynomial is (x-3)^2 (x+1)(x-1)
    So  T=P_1-P_2+3P_3 , where P are projections.

    Trying to find E's such that A=x_1E_1+x_2E_2+x_3E_3=E_1-E_2+3E_3
    E are Projections matrix relative to  E_3
    This problem is made harder by the fact that (x-3)^2 (x+1)(x-1) is not the characteristic polynomial of the given matrix. I suspect that the matrix should be \begin{bmatrix} 7 & 3 & 3 &2 \\ 0 & 1 & 2 & -4 \\ -8 &<br />
   -4& -5 &0 \\ 2 & 1 & 2 & 3\end{bmatrix} (with a +2 in the bottom left corner). That matrix at least appears to have the right eigenvalues.

    The key fact that you need to solve the problem is that if \mathbf{v} = (w,x,y,z) is a unit vector (I'll write vectors as rows rather than columns, to save space), then the projection onto the one-dimensional space spanned by \mathbf{v} has matrix  \begin{bmatrix}w^2&wx&wy&wz\\ wx&x^2&xy&xz\\ wy&xy&y^2&yz\\ wz&xz&yz&z^2\end{bmatrix} (with respect to the standard basis).
    For the matrix \begin{bmatrix} 7 & 3 & 3 &2 \\ 0 & 1 & 2 & -4 \\ -8 &<br />
   -4& -5 &0 \\ 2 & 1 & 2 & 3\end{bmatrix}, a unit eigenvector corresponding to the eigenvalue 1 is \tfrac1{\sqrt5}(1,-2,0,0). The projection P_1 onto the eigenspace of the eigenvalue x=1 is therefore \frac15\!\begin{bmatrix}1&-2&0&0\\ -2&4&0&0\\ 0&0&0&0\\ 0&0&0&0\end{bmatrix}.

    The projection P_{-1} onto the one-dimensional eigenspace of the eigenvalue x=1 is calculated in a similar way.

    For the two-dimensional eigenspace of the repeated eigenvalue x=3, there are various possible methods to find the projection P_3 onto it.

    Method 1 (the slow but safe method). Find two orthonormal eigenvectors for the eigenvalue x=3, form the projections onto the one-dimensional spaces spanned by each of them, then add those two projections to get P_3.

    Method 2 (the lazy way). Use the fact that sum of P_1, P_{-1} and P_3 must be the identity matrix. Since you have already found P_1 and P_{-1}, you can just write down P_3 = I-P_1-P_{-1}.

    Method 3 (the belt and braces method). Use Method 1 to find P_3. Then use Method 2 to check that your answer is right.

    The advantage of Method 2 is that it is quick. The disadvantage is that it assumes that the eigenspace of the eigenvalue x=3 is two-dimensional. There is in principle no guarantee of this. If the eigenvalue x=3 only has one linearly independent eigenvector then the matrix A will not have a spectral decomposition at all.
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