Mixing
Spectral decomposition of an operator
$begingroup$
Given the operator:
$$begin{pmatrix}
i & 0 & - 4\
0 & - 3i & 0\
2 & 0 & - i
end{pmatrix}$$
Now $det(lambda1-A)=(z+3i)(z^2+9)=(z+3i)^2(z-3i)$,
I know a theorem that says that this three prepositions are equivalent:
Eigenvectors of A form a basis,
There are no spectral operators in spectral decomposition of A, $mathbb{E} $,
R(z) has only poles of order one.
A satisfies an algebric equation with all zeroes of order one ("simple roots").
Now to find the spectral representation I calculate the operator $R(z)=frac{1}{z1-A}$:
$$R(z)=begin{pmatrix}
frac{z+i} {(z+3i)(z-3i)} & 0 & frac{-4}{(z+3i)(z-3i)}\
0 & frac{1}{z+3i} & 0\
frac{2}{(z+3i)(z-3i)} & 0 & frac{z-i} {(z+3i)(z-3i)}
end{pmatrix} $$
And find that there are no spectral operators $mathbb{E} $ respecting 1 and 3 of the theorem, but the $det$ has not only zeroes of order one. How is this possible? How should I interpret the point 4 of the theorem?
Thanks you all
operator-theory determinant diagonalization spectra
asked Jan 3 at 10:09
pter26pter26
317111
$endgroup$
add a comment |
$begingroup$
Given the operator:
$$begin{pmatrix}
i & 0 & - 4\
0 & - 3i & 0\
2 & 0 & - i
end{pmatrix}$$
Now $det(lambda1-A)=(z+3i)(z^2+9)=(z+3i)^2(z-3i)$,
I know a theorem that says that this three prepositions are equivalent:
Eigenvectors of A form a basis,
There are no spectral operators in spectral decomposition of A, $mathbb{E} $,
R(z) has only poles of order one.
A satisfies an algebric equation with all zeroes of order one ("simple roots").
Now to find the spectral representation I calculate the operator $R(z)=frac{1}{z1-A}$:
$$R(z)=begin{pmatrix}
frac{z+i} {(z+3i)(z-3i)} & 0 & frac{-4}{(z+3i)(z-3i)}\
0 & frac{1}{z+3i} & 0\
frac{2}{(z+3i)(z-3i)} & 0 & frac{z-i} {(z+3i)(z-3i)}
end{pmatrix} $$
And find that there are no spectral operators $mathbb{E} $ respecting 1 and 3 of the theorem, but the $det$ has not only zeroes of order one. How is this possible? How should I interpret the point 4 of the theorem?
Thanks you all
operator-theory determinant diagonalization spectra
asked Jan 3 at 10:09
pter26pter26
317111
$endgroup$
add a comment |
$begingroup$
Given the operator:
$$begin{pmatrix}
i & 0 & - 4\
0 & - 3i & 0\
2 & 0 & - i
end{pmatrix}$$
Now $det(lambda1-A)=(z+3i)(z^2+9)=(z+3i)^2(z-3i)$,
I know a theorem that says that this three prepositions are equivalent:
Eigenvectors of A form a basis,
There are no spectral operators in spectral decomposition of A, $mathbb{E} $,
R(z) has only poles of order one.
A satisfies an algebric equation with all zeroes of order one ("simple roots").
Now to find the spectral representation I calculate the operator $R(z)=frac{1}{z1-A}$:
$$R(z)=begin{pmatrix}
frac{z+i} {(z+3i)(z-3i)} & 0 & frac{-4}{(z+3i)(z-3i)}\
0 & frac{1}{z+3i} & 0\
frac{2}{(z+3i)(z-3i)} & 0 & frac{z-i} {(z+3i)(z-3i)}
end{pmatrix} $$
And find that there are no spectral operators $mathbb{E} $ respecting 1 and 3 of the theorem, but the $det$ has not only zeroes of order one. How is this possible? How should I interpret the point 4 of the theorem?
Thanks you all
operator-theory determinant diagonalization spectra
asked Jan 3 at 10:09
pter26pter26
317111
$endgroup$
Given the operator:
$$begin{pmatrix}
i & 0 & - 4\
0 & - 3i & 0\
2 & 0 & - i
end{pmatrix}$$
Now $det(lambda1-A)=(z+3i)(z^2+9)=(z+3i)^2(z-3i)$,
I know a theorem that says that this three prepositions are equivalent:
Eigenvectors of A form a basis,
There are no spectral operators in spectral decomposition of A, $mathbb{E} $,
R(z) has only poles of order one.
A satisfies an algebric equation with all zeroes of order one ("simple roots").
Now to find the spectral representation I calculate the operator $R(z)=frac{1}{z1-A}$:
$$R(z)=begin{pmatrix}
frac{z+i} {(z+3i)(z-3i)} & 0 & frac{-4}{(z+3i)(z-3i)}\
0 & frac{1}{z+3i} & 0\
frac{2}{(z+3i)(z-3i)} & 0 & frac{z-i} {(z+3i)(z-3i)}
end{pmatrix} $$
And find that there are no spectral operators $mathbb{E} $ respecting 1 and 3 of the theorem, but the $det$ has not only zeroes of order one. How is this possible? How should I interpret the point 4 of the theorem?
Thanks you all
operator-theory determinant diagonalization spectra
operator-theory determinant diagonalization spectra
asked Jan 3 at 10:09
pter26pter26
317111
asked Jan 3 at 10:09
pter26pter26
317111
edited Jan 3 at 10:46
pter26
asked Jan 3 at 10:09
pter26pter26
317111
asked Jan 3 at 10:09
pter26pter26
317111
asked Jan 3 at 10:09
pter26pter26
317111
317111
add a comment |
add a comment |
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