Eigenvalues of $A+v_1d^T$ where $Av_1 = lambda_1 v_1$ (shift of first eigenvalue)











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I have trouble to solve the following problem:




Let $Ain mathbf{R}^{ntimes n}$, $lambda_1,ldots,lambda_n$ are eigenvalues of $A$, and $A v_1=lambda_1v_1$. Let $din mathbf{R}^n$, then the eigenvalues of $A+v_1d^T$ are $lambda_1+d^Tv_1, lambda_2,ldots,lambda_n$. Note: $A$ may not be diagonalizable.




I know the following facts:





  1. $v_1d^T$ is of rank $1$.

  2. eigenvalues of $v_1d^T$ are $v_1^Td, 0,ldots, 0$.


But I still have no idea to prove this theorem. Please help me, thanks!










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    Maybe math.stackexchange.com/q/1374836/499341 will be helpful?
    – helper
    Nov 18 at 0:46

















up vote
4
down vote

favorite
1












I have trouble to solve the following problem:




Let $Ain mathbf{R}^{ntimes n}$, $lambda_1,ldots,lambda_n$ are eigenvalues of $A$, and $A v_1=lambda_1v_1$. Let $din mathbf{R}^n$, then the eigenvalues of $A+v_1d^T$ are $lambda_1+d^Tv_1, lambda_2,ldots,lambda_n$. Note: $A$ may not be diagonalizable.




I know the following facts:





  1. $v_1d^T$ is of rank $1$.

  2. eigenvalues of $v_1d^T$ are $v_1^Td, 0,ldots, 0$.


But I still have no idea to prove this theorem. Please help me, thanks!










share|cite|improve this question


















  • 1




    Maybe math.stackexchange.com/q/1374836/499341 will be helpful?
    – helper
    Nov 18 at 0:46















up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





I have trouble to solve the following problem:




Let $Ain mathbf{R}^{ntimes n}$, $lambda_1,ldots,lambda_n$ are eigenvalues of $A$, and $A v_1=lambda_1v_1$. Let $din mathbf{R}^n$, then the eigenvalues of $A+v_1d^T$ are $lambda_1+d^Tv_1, lambda_2,ldots,lambda_n$. Note: $A$ may not be diagonalizable.




I know the following facts:





  1. $v_1d^T$ is of rank $1$.

  2. eigenvalues of $v_1d^T$ are $v_1^Td, 0,ldots, 0$.


But I still have no idea to prove this theorem. Please help me, thanks!










share|cite|improve this question













I have trouble to solve the following problem:




Let $Ain mathbf{R}^{ntimes n}$, $lambda_1,ldots,lambda_n$ are eigenvalues of $A$, and $A v_1=lambda_1v_1$. Let $din mathbf{R}^n$, then the eigenvalues of $A+v_1d^T$ are $lambda_1+d^Tv_1, lambda_2,ldots,lambda_n$. Note: $A$ may not be diagonalizable.




I know the following facts:





  1. $v_1d^T$ is of rank $1$.

  2. eigenvalues of $v_1d^T$ are $v_1^Td, 0,ldots, 0$.


But I still have no idea to prove this theorem. Please help me, thanks!







matrices eigenvalues-eigenvectors






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asked Nov 18 at 0:34









sleeve chen

2,96041851




2,96041851








  • 1




    Maybe math.stackexchange.com/q/1374836/499341 will be helpful?
    – helper
    Nov 18 at 0:46
















  • 1




    Maybe math.stackexchange.com/q/1374836/499341 will be helpful?
    – helper
    Nov 18 at 0:46










1




1




Maybe math.stackexchange.com/q/1374836/499341 will be helpful?
– helper
Nov 18 at 0:46






Maybe math.stackexchange.com/q/1374836/499341 will be helpful?
– helper
Nov 18 at 0:46












1 Answer
1






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1
down vote



accepted










Sketch: Use Jordan decomposition. WLOG, assume
begin{align}
A = UJU^{-1}
end{align}

where
begin{align}
J =
begin{pmatrix}
lambda& 1 & 0 & dots & dots & 0\
0 & lambda & 1 & 0 & dots & vdots\
vdots & 0 & ddots & ddots & dots &vdots\
vdots & vdots & ddots& ddots & 1 & vdots\
vdots & dots& dots & 0 & lambda & 1\
0 & dots & dots & dots & 0 & lambda
end{pmatrix}
end{align}

Since $Ue_1 = v_1$ then we see that
begin{align}
A+v_1d^T =& U(J+U^{-1}v_1d^TU)U^{-1}\
=& U(J+e_1(U^Td)^T)U^{-1}.
end{align}

Note that
begin{align}
e_1(U^Td)^T =
begin{pmatrix}
1 \
0 \
vdots\
0
end{pmatrix}
begin{pmatrix}
v_1^Td & * &dots & *
end{pmatrix}
=
begin{pmatrix}
v_1^Td & * &dots & *\
0 & 0 & dots & 0\
vdots & vdots & dots & vdots\
0 & 0 & dots & 0
end{pmatrix}.
end{align}






share|cite|improve this answer





















  • But your assumption implies all eigenvalues are the same? then it may not be general.
    – sleeve chen
    Nov 18 at 5:44






  • 1




    Note the general decomposition consists of many blocks of the above form.
    – Jacky Chong
    Nov 18 at 5:46













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Sketch: Use Jordan decomposition. WLOG, assume
begin{align}
A = UJU^{-1}
end{align}

where
begin{align}
J =
begin{pmatrix}
lambda& 1 & 0 & dots & dots & 0\
0 & lambda & 1 & 0 & dots & vdots\
vdots & 0 & ddots & ddots & dots &vdots\
vdots & vdots & ddots& ddots & 1 & vdots\
vdots & dots& dots & 0 & lambda & 1\
0 & dots & dots & dots & 0 & lambda
end{pmatrix}
end{align}

Since $Ue_1 = v_1$ then we see that
begin{align}
A+v_1d^T =& U(J+U^{-1}v_1d^TU)U^{-1}\
=& U(J+e_1(U^Td)^T)U^{-1}.
end{align}

Note that
begin{align}
e_1(U^Td)^T =
begin{pmatrix}
1 \
0 \
vdots\
0
end{pmatrix}
begin{pmatrix}
v_1^Td & * &dots & *
end{pmatrix}
=
begin{pmatrix}
v_1^Td & * &dots & *\
0 & 0 & dots & 0\
vdots & vdots & dots & vdots\
0 & 0 & dots & 0
end{pmatrix}.
end{align}






share|cite|improve this answer





















  • But your assumption implies all eigenvalues are the same? then it may not be general.
    – sleeve chen
    Nov 18 at 5:44






  • 1




    Note the general decomposition consists of many blocks of the above form.
    – Jacky Chong
    Nov 18 at 5:46

















up vote
1
down vote



accepted










Sketch: Use Jordan decomposition. WLOG, assume
begin{align}
A = UJU^{-1}
end{align}

where
begin{align}
J =
begin{pmatrix}
lambda& 1 & 0 & dots & dots & 0\
0 & lambda & 1 & 0 & dots & vdots\
vdots & 0 & ddots & ddots & dots &vdots\
vdots & vdots & ddots& ddots & 1 & vdots\
vdots & dots& dots & 0 & lambda & 1\
0 & dots & dots & dots & 0 & lambda
end{pmatrix}
end{align}

Since $Ue_1 = v_1$ then we see that
begin{align}
A+v_1d^T =& U(J+U^{-1}v_1d^TU)U^{-1}\
=& U(J+e_1(U^Td)^T)U^{-1}.
end{align}

Note that
begin{align}
e_1(U^Td)^T =
begin{pmatrix}
1 \
0 \
vdots\
0
end{pmatrix}
begin{pmatrix}
v_1^Td & * &dots & *
end{pmatrix}
=
begin{pmatrix}
v_1^Td & * &dots & *\
0 & 0 & dots & 0\
vdots & vdots & dots & vdots\
0 & 0 & dots & 0
end{pmatrix}.
end{align}






share|cite|improve this answer





















  • But your assumption implies all eigenvalues are the same? then it may not be general.
    – sleeve chen
    Nov 18 at 5:44






  • 1




    Note the general decomposition consists of many blocks of the above form.
    – Jacky Chong
    Nov 18 at 5:46















up vote
1
down vote



accepted







up vote
1
down vote



accepted






Sketch: Use Jordan decomposition. WLOG, assume
begin{align}
A = UJU^{-1}
end{align}

where
begin{align}
J =
begin{pmatrix}
lambda& 1 & 0 & dots & dots & 0\
0 & lambda & 1 & 0 & dots & vdots\
vdots & 0 & ddots & ddots & dots &vdots\
vdots & vdots & ddots& ddots & 1 & vdots\
vdots & dots& dots & 0 & lambda & 1\
0 & dots & dots & dots & 0 & lambda
end{pmatrix}
end{align}

Since $Ue_1 = v_1$ then we see that
begin{align}
A+v_1d^T =& U(J+U^{-1}v_1d^TU)U^{-1}\
=& U(J+e_1(U^Td)^T)U^{-1}.
end{align}

Note that
begin{align}
e_1(U^Td)^T =
begin{pmatrix}
1 \
0 \
vdots\
0
end{pmatrix}
begin{pmatrix}
v_1^Td & * &dots & *
end{pmatrix}
=
begin{pmatrix}
v_1^Td & * &dots & *\
0 & 0 & dots & 0\
vdots & vdots & dots & vdots\
0 & 0 & dots & 0
end{pmatrix}.
end{align}






share|cite|improve this answer












Sketch: Use Jordan decomposition. WLOG, assume
begin{align}
A = UJU^{-1}
end{align}

where
begin{align}
J =
begin{pmatrix}
lambda& 1 & 0 & dots & dots & 0\
0 & lambda & 1 & 0 & dots & vdots\
vdots & 0 & ddots & ddots & dots &vdots\
vdots & vdots & ddots& ddots & 1 & vdots\
vdots & dots& dots & 0 & lambda & 1\
0 & dots & dots & dots & 0 & lambda
end{pmatrix}
end{align}

Since $Ue_1 = v_1$ then we see that
begin{align}
A+v_1d^T =& U(J+U^{-1}v_1d^TU)U^{-1}\
=& U(J+e_1(U^Td)^T)U^{-1}.
end{align}

Note that
begin{align}
e_1(U^Td)^T =
begin{pmatrix}
1 \
0 \
vdots\
0
end{pmatrix}
begin{pmatrix}
v_1^Td & * &dots & *
end{pmatrix}
=
begin{pmatrix}
v_1^Td & * &dots & *\
0 & 0 & dots & 0\
vdots & vdots & dots & vdots\
0 & 0 & dots & 0
end{pmatrix}.
end{align}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 18 at 2:15









Jacky Chong

17.3k21027




17.3k21027












  • But your assumption implies all eigenvalues are the same? then it may not be general.
    – sleeve chen
    Nov 18 at 5:44






  • 1




    Note the general decomposition consists of many blocks of the above form.
    – Jacky Chong
    Nov 18 at 5:46




















  • But your assumption implies all eigenvalues are the same? then it may not be general.
    – sleeve chen
    Nov 18 at 5:44






  • 1




    Note the general decomposition consists of many blocks of the above form.
    – Jacky Chong
    Nov 18 at 5:46


















But your assumption implies all eigenvalues are the same? then it may not be general.
– sleeve chen
Nov 18 at 5:44




But your assumption implies all eigenvalues are the same? then it may not be general.
– sleeve chen
Nov 18 at 5:44




1




1




Note the general decomposition consists of many blocks of the above form.
– Jacky Chong
Nov 18 at 5:46






Note the general decomposition consists of many blocks of the above form.
– Jacky Chong
Nov 18 at 5:46




















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