Formula for eigenvector corresponding to a simple eigenvalue
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In a book on numerical linear algebra (Deuflhard numerische Mathematik Band 1) there is the following exercise (translated from German):
Give a Formula (using Determinants) for an eigenvector $x in mathbb{C}^n$ corresponding to a given simple eigenvalue $lambda in mathbb{C}$ of a Matrix $A in mathbb{C}^{n times n}$.
It really should be a formula and not an algorithm, since the formula is to be used to show that the eigenvector depends on the matrix in a continuous differentiable fashion.
My problem is that any formula I think of yields $0$ instead of a proper eigenvector
Any ideas on how to get such a formula?
linear-algebra eigenvalues-eigenvectors determinant numerical-linear-algebra
asked Jan 27 '18 at 21:39
0x5390x539
1,371518
$endgroup$
add a comment |
$begingroup$
In a book on numerical linear algebra (Deuflhard numerische Mathematik Band 1) there is the following exercise (translated from German):
Give a Formula (using Determinants) for an eigenvector $x in mathbb{C}^n$ corresponding to a given simple eigenvalue $lambda in mathbb{C}$ of a Matrix $A in mathbb{C}^{n times n}$.
It really should be a formula and not an algorithm, since the formula is to be used to show that the eigenvector depends on the matrix in a continuous differentiable fashion.
My problem is that any formula I think of yields $0$ instead of a proper eigenvector
Any ideas on how to get such a formula?
linear-algebra eigenvalues-eigenvectors determinant numerical-linear-algebra
asked Jan 27 '18 at 21:39
0x5390x539
1,371518
$endgroup$
$begingroup$
When you say that "an eigenvector depends on the Matrix in a continuous differentiable fashion", it is meaningless as such because an eigenvector is defined up to a factor. The less you can do is to normalize your eigenvectors (normed to unity).
$endgroup$
– Jean Marie
Jan 27 '18 at 22:27
$begingroup$
@JeanMarie what I mean is that there exists a continuously differentiable function which maps every $A in U subset mathbb{C}^{n times n}$ to one of its eigenvectors (the fact that the eigenvalues "are continuously differentiable" was established before in the book). Obviously this function won't be unique but I don't see any other problem.
$endgroup$
– 0x539
Jan 27 '18 at 22:41
$begingroup$
I understand a little more. See similar question with answer (math.stackexchange.com/q/807144)
$endgroup$
– Jean Marie
Jan 27 '18 at 22:55
$begingroup$
see as well the second answer in (math.stackexchange.com/q/581057)
$endgroup$
– Jean Marie
Jan 28 '18 at 7:24
$begingroup$
... and (mathoverflow.net/questions/253584/…)
$endgroup$
– Jean Marie
Jan 28 '18 at 7:27
add a comment |
$begingroup$
In a book on numerical linear algebra (Deuflhard numerische Mathematik Band 1) there is the following exercise (translated from German):
Give a Formula (using Determinants) for an eigenvector $x in mathbb{C}^n$ corresponding to a given simple eigenvalue $lambda in mathbb{C}$ of a Matrix $A in mathbb{C}^{n times n}$.
It really should be a formula and not an algorithm, since the formula is to be used to show that the eigenvector depends on the matrix in a continuous differentiable fashion.
My problem is that any formula I think of yields $0$ instead of a proper eigenvector
Any ideas on how to get such a formula?
linear-algebra eigenvalues-eigenvectors determinant numerical-linear-algebra
asked Jan 27 '18 at 21:39
0x5390x539
1,371518
$endgroup$
In a book on numerical linear algebra (Deuflhard numerische Mathematik Band 1) there is the following exercise (translated from German):
Give a Formula (using Determinants) for an eigenvector $x in mathbb{C}^n$ corresponding to a given simple eigenvalue $lambda in mathbb{C}$ of a Matrix $A in mathbb{C}^{n times n}$.
It really should be a formula and not an algorithm, since the formula is to be used to show that the eigenvector depends on the matrix in a continuous differentiable fashion.
My problem is that any formula I think of yields $0$ instead of a proper eigenvector
Any ideas on how to get such a formula?
linear-algebra eigenvalues-eigenvectors determinant numerical-linear-algebra
linear-algebra eigenvalues-eigenvectors determinant numerical-linear-algebra
asked Jan 27 '18 at 21:39
0x5390x539
1,371518
asked Jan 27 '18 at 21:39
0x5390x539
1,371518
edited Dec 8 '18 at 21:22
0x539
asked Jan 27 '18 at 21:39
0x5390x539
1,371518
asked Jan 27 '18 at 21:39
0x5390x539
1,371518
asked Jan 27 '18 at 21:39
0x5390x539
1,371518
1,371518
$begingroup$
When you say that "an eigenvector depends on the Matrix in a continuous differentiable fashion", it is meaningless as such because an eigenvector is defined up to a factor. The less you can do is to normalize your eigenvectors (normed to unity).
$endgroup$
– Jean Marie
Jan 27 '18 at 22:27
$begingroup$
@JeanMarie what I mean is that there exists a continuously differentiable function which maps every $A in U subset mathbb{C}^{n times n}$ to one of its eigenvectors (the fact that the eigenvalues "are continuously differentiable" was established before in the book). Obviously this function won't be unique but I don't see any other problem.
$endgroup$
– 0x539
Jan 27 '18 at 22:41
$begingroup$
I understand a little more. See similar question with answer (math.stackexchange.com/q/807144)
$endgroup$
– Jean Marie
Jan 27 '18 at 22:55
$begingroup$
see as well the second answer in (math.stackexchange.com/q/581057)
$endgroup$
– Jean Marie
Jan 28 '18 at 7:24
$begingroup$
... and (mathoverflow.net/questions/253584/…)
$endgroup$
– Jean Marie
Jan 28 '18 at 7:27
add a comment |
$begingroup$
When you say that "an eigenvector depends on the Matrix in a continuous differentiable fashion", it is meaningless as such because an eigenvector is defined up to a factor. The less you can do is to normalize your eigenvectors (normed to unity).
$endgroup$
– Jean Marie
Jan 27 '18 at 22:27
$begingroup$
@JeanMarie what I mean is that there exists a continuously differentiable function which maps every $A in U subset mathbb{C}^{n times n}$ to one of its eigenvectors (the fact that the eigenvalues "are continuously differentiable" was established before in the book). Obviously this function won't be unique but I don't see any other problem.
$endgroup$
– 0x539
Jan 27 '18 at 22:41
$begingroup$
I understand a little more. See similar question with answer (math.stackexchange.com/q/807144)
$endgroup$
– Jean Marie
Jan 27 '18 at 22:55
$begingroup$
see as well the second answer in (math.stackexchange.com/q/581057)
$endgroup$
– Jean Marie
Jan 28 '18 at 7:24
$begingroup$
... and (mathoverflow.net/questions/253584/…)
$endgroup$
– Jean Marie
Jan 28 '18 at 7:27
$begingroup$
When you say that "an eigenvector depends on the Matrix in a continuous differentiable fashion", it is meaningless as such because an eigenvector is defined up to a factor. The less you can do is to normalize your eigenvectors (normed to unity).
$endgroup$
– Jean Marie
Jan 27 '18 at 22:27
$begingroup$
When you say that "an eigenvector depends on the Matrix in a continuous differentiable fashion", it is meaningless as such because an eigenvector is defined up to a factor. The less you can do is to normalize your eigenvectors (normed to unity).
$endgroup$
– Jean Marie
Jan 27 '18 at 22:27
$begingroup$
@JeanMarie what I mean is that there exists a continuously differentiable function which maps every $A in U subset mathbb{C}^{n times n}$ to one of its eigenvectors (the fact that the eigenvalues "are continuously differentiable" was established before in the book). Obviously this function won't be unique but I don't see any other problem.
$endgroup$
– 0x539
Jan 27 '18 at 22:41
$begingroup$
@JeanMarie what I mean is that there exists a continuously differentiable function which maps every $A in U subset mathbb{C}^{n times n}$ to one of its eigenvectors (the fact that the eigenvalues "are continuously differentiable" was established before in the book). Obviously this function won't be unique but I don't see any other problem.
$endgroup$
– 0x539
Jan 27 '18 at 22:41
$begingroup$
I understand a little more. See similar question with answer (math.stackexchange.com/q/807144)
$endgroup$
– Jean Marie
Jan 27 '18 at 22:55
$begingroup$
I understand a little more. See similar question with answer (math.stackexchange.com/q/807144)
$endgroup$
– Jean Marie
Jan 27 '18 at 22:55
$begingroup$
see as well the second answer in (math.stackexchange.com/q/581057)
$endgroup$
– Jean Marie
Jan 28 '18 at 7:24
$begingroup$
see as well the second answer in (math.stackexchange.com/q/581057)
$endgroup$
– Jean Marie
Jan 28 '18 at 7:24
$begingroup$
... and (mathoverflow.net/questions/253584/…)
$endgroup$
– Jean Marie
Jan 28 '18 at 7:27
$begingroup$
... and (mathoverflow.net/questions/253584/…)
$endgroup$
– Jean Marie
Jan 28 '18 at 7:27
add a comment |
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$begingroup$
When you say that "an eigenvector depends on the Matrix in a continuous differentiable fashion", it is meaningless as such because an eigenvector is defined up to a factor. The less you can do is to normalize your eigenvectors (normed to unity).
$endgroup$
– Jean Marie
Jan 27 '18 at 22:27
$begingroup$
@JeanMarie what I mean is that there exists a continuously differentiable function which maps every $A in U subset mathbb{C}^{n times n}$ to one of its eigenvectors (the fact that the eigenvalues "are continuously differentiable" was established before in the book). Obviously this function won't be unique but I don't see any other problem.
$endgroup$
– 0x539
Jan 27 '18 at 22:41
$begingroup$
I understand a little more. See similar question with answer (math.stackexchange.com/q/807144)
$endgroup$
– Jean Marie
Jan 27 '18 at 22:55
$begingroup$
see as well the second answer in (math.stackexchange.com/q/581057)
$endgroup$
– Jean Marie
Jan 28 '18 at 7:24
$begingroup$
... and (mathoverflow.net/questions/253584/…)
$endgroup$
– Jean Marie
Jan 28 '18 at 7:27