Mixing
Explanation required for observations regarding Determinant and Condition Number
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I'm currently working on an eigenvalue problem of the form $$Azeta=lambda Bzeta$$ where all the elements of $A$ and $B$ are only functions of a variable $k$. Now, I can obtain values for $lambda$ using the eig() function in MATLAB, but I wanted to do an arc-length continuation program to plot $lambda$ as a function of $k$, so I needed a function such that $$F(k,lambda(k))=0$$ My initial thought was to use $F=det(A-lambda B)$, but I was surprised to find that this value turned out to be extremely high (I got ~$10^{171}$ for one case) regardless of whether the input value of $lambda$ was correct (i.e. the ones provided by the eig() function) or not. Later I realized that this was because $(A-lambda B)$ was ill-conditioned, but upon further investigation, I found that the correct value of $lambda$ for a given $k$ corresponded to a local maxima in the 1-norm condition number for $(A-lambda B)$. Is there an explanation for this? Is there a different choice of $F(k,lambda(k))$ that would help me out?
Some additional information:
$bullet$ Both $lambda$ and $k$ are positive real numbers.
$bullet$ Both $A$ and $B$ are sparse matrices, and their general form can be seen here (page 54).
linear-algebra eigenvalues-eigenvectors
edited Jan 31 at 7:50
YuiTo Cheng
2,2734937
asked Jan 31 at 5:46
Daniel HarperDaniel Harper
155
$endgroup$
add a comment |
$begingroup$
I'm currently working on an eigenvalue problem of the form $$Azeta=lambda Bzeta$$ where all the elements of $A$ and $B$ are only functions of a variable $k$. Now, I can obtain values for $lambda$ using the eig() function in MATLAB, but I wanted to do an arc-length continuation program to plot $lambda$ as a function of $k$, so I needed a function such that $$F(k,lambda(k))=0$$ My initial thought was to use $F=det(A-lambda B)$, but I was surprised to find that this value turned out to be extremely high (I got ~$10^{171}$ for one case) regardless of whether the input value of $lambda$ was correct (i.e. the ones provided by the eig() function) or not. Later I realized that this was because $(A-lambda B)$ was ill-conditioned, but upon further investigation, I found that the correct value of $lambda$ for a given $k$ corresponded to a local maxima in the 1-norm condition number for $(A-lambda B)$. Is there an explanation for this? Is there a different choice of $F(k,lambda(k))$ that would help me out?
Some additional information:
$bullet$ Both $lambda$ and $k$ are positive real numbers.
$bullet$ Both $A$ and $B$ are sparse matrices, and their general form can be seen here (page 54).
linear-algebra eigenvalues-eigenvectors
edited Jan 31 at 7:50
YuiTo Cheng
2,2734937
asked Jan 31 at 5:46
Daniel HarperDaniel Harper
155
$endgroup$
add a comment |
$begingroup$
I'm currently working on an eigenvalue problem of the form $$Azeta=lambda Bzeta$$ where all the elements of $A$ and $B$ are only functions of a variable $k$. Now, I can obtain values for $lambda$ using the eig() function in MATLAB, but I wanted to do an arc-length continuation program to plot $lambda$ as a function of $k$, so I needed a function such that $$F(k,lambda(k))=0$$ My initial thought was to use $F=det(A-lambda B)$, but I was surprised to find that this value turned out to be extremely high (I got ~$10^{171}$ for one case) regardless of whether the input value of $lambda$ was correct (i.e. the ones provided by the eig() function) or not. Later I realized that this was because $(A-lambda B)$ was ill-conditioned, but upon further investigation, I found that the correct value of $lambda$ for a given $k$ corresponded to a local maxima in the 1-norm condition number for $(A-lambda B)$. Is there an explanation for this? Is there a different choice of $F(k,lambda(k))$ that would help me out?
Some additional information:
$bullet$ Both $lambda$ and $k$ are positive real numbers.
$bullet$ Both $A$ and $B$ are sparse matrices, and their general form can be seen here (page 54).
linear-algebra eigenvalues-eigenvectors
edited Jan 31 at 7:50
YuiTo Cheng
2,2734937
asked Jan 31 at 5:46
Daniel HarperDaniel Harper
155
$endgroup$
I'm currently working on an eigenvalue problem of the form $$Azeta=lambda Bzeta$$ where all the elements of $A$ and $B$ are only functions of a variable $k$. Now, I can obtain values for $lambda$ using the eig() function in MATLAB, but I wanted to do an arc-length continuation program to plot $lambda$ as a function of $k$, so I needed a function such that $$F(k,lambda(k))=0$$ My initial thought was to use $F=det(A-lambda B)$, but I was surprised to find that this value turned out to be extremely high (I got ~$10^{171}$ for one case) regardless of whether the input value of $lambda$ was correct (i.e. the ones provided by the eig() function) or not. Later I realized that this was because $(A-lambda B)$ was ill-conditioned, but upon further investigation, I found that the correct value of $lambda$ for a given $k$ corresponded to a local maxima in the 1-norm condition number for $(A-lambda B)$. Is there an explanation for this? Is there a different choice of $F(k,lambda(k))$ that would help me out?
Some additional information:
$bullet$ Both $lambda$ and $k$ are positive real numbers.
$bullet$ Both $A$ and $B$ are sparse matrices, and their general form can be seen here (page 54).
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
edited Jan 31 at 7:50
YuiTo Cheng
2,2734937
asked Jan 31 at 5:46
Daniel HarperDaniel Harper
155
edited Jan 31 at 7:50
YuiTo Cheng
2,2734937
asked Jan 31 at 5:46
Daniel HarperDaniel Harper
155
edited Jan 31 at 7:50
YuiTo Cheng
2,2734937
edited Jan 31 at 7:50
YuiTo Cheng
2,2734937
edited Jan 31 at 7:50
YuiTo Cheng
2,2734937
2,2734937
asked Jan 31 at 5:46
Daniel HarperDaniel Harper
155
asked Jan 31 at 5:46
Daniel HarperDaniel Harper
155
asked Jan 31 at 5:46
Daniel HarperDaniel Harper
155
155
add a comment |
add a comment |
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