- #1
JustYouAsk
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According to the spectral theorem for self-adjoint operators you can find a matrix P such that P[tex]^{-1}[/tex]AP is diagonal, i.e. P[tex]^{T}[/tex]AP (P can be shown to be orthogonal). I'm not sure what the result is if the same can be done for the following square (size n X n) and symmetric matrix of the form:
A=
[ U 0 U ]
[ 0 0 0 ]
[ U 0 U ]
where U is square matrix and 0 is a matrix of zeros.
If I am not mistaken the solution is that the columns of P are simply the eigenvectors of A? can anyone confirm this?
A=
[ U 0 U ]
[ 0 0 0 ]
[ U 0 U ]
where U is square matrix and 0 is a matrix of zeros.
If I am not mistaken the solution is that the columns of P are simply the eigenvectors of A? can anyone confirm this?