If you can multiple linearly independent eigenvectors for the same eigenvalue, why do we solve for just one?
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The way I have been taught to solve for eigenvalues and eigenvectors in linear algebra is thus: compute the eigenvalues using the determinant of A-I. Then use that to compute the eigenvectors for each eigenvalue using (A-I)x = 0. The way I understood it, each eigenvector served as a basis for the space that contained all eigenvectors of that eigenvalue. However, I just learned from the interwebs that an eigenvalue can have more than one linearly independent eigenvector. So, what's the point of calculating just one of these for each eigenvalue, if there are theoretically infinitely many linearly independent eigenvectors for each eigenvalue? Why do we choose just one?
linear-algebra eigenvalues-eigenvectors
asked Dec 19 '18 at 18:39
Will BurghardWill Burghard
42
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The way I have been taught to solve for eigenvalues and eigenvectors in linear algebra is thus: compute the eigenvalues using the determinant of A-I. Then use that to compute the eigenvectors for each eigenvalue using (A-I)x = 0. The way I understood it, each eigenvector served as a basis for the space that contained all eigenvectors of that eigenvalue. However, I just learned from the interwebs that an eigenvalue can have more than one linearly independent eigenvector. So, what's the point of calculating just one of these for each eigenvalue, if there are theoretically infinitely many linearly independent eigenvectors for each eigenvalue? Why do we choose just one?
linear-algebra eigenvalues-eigenvectors
asked Dec 19 '18 at 18:39
Will BurghardWill Burghard
42
$endgroup$
add a comment |
$begingroup$
The way I have been taught to solve for eigenvalues and eigenvectors in linear algebra is thus: compute the eigenvalues using the determinant of A-I. Then use that to compute the eigenvectors for each eigenvalue using (A-I)x = 0. The way I understood it, each eigenvector served as a basis for the space that contained all eigenvectors of that eigenvalue. However, I just learned from the interwebs that an eigenvalue can have more than one linearly independent eigenvector. So, what's the point of calculating just one of these for each eigenvalue, if there are theoretically infinitely many linearly independent eigenvectors for each eigenvalue? Why do we choose just one?
linear-algebra eigenvalues-eigenvectors
asked Dec 19 '18 at 18:39
Will BurghardWill Burghard
42
$endgroup$
The way I have been taught to solve for eigenvalues and eigenvectors in linear algebra is thus: compute the eigenvalues using the determinant of A-I. Then use that to compute the eigenvectors for each eigenvalue using (A-I)x = 0. The way I understood it, each eigenvector served as a basis for the space that contained all eigenvectors of that eigenvalue. However, I just learned from the interwebs that an eigenvalue can have more than one linearly independent eigenvector. So, what's the point of calculating just one of these for each eigenvalue, if there are theoretically infinitely many linearly independent eigenvectors for each eigenvalue? Why do we choose just one?
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked Dec 19 '18 at 18:39
Will BurghardWill Burghard
42
asked Dec 19 '18 at 18:39
Will BurghardWill Burghard
42
asked Dec 19 '18 at 18:39
Will BurghardWill Burghard
42
asked Dec 19 '18 at 18:39
Will BurghardWill Burghard
42
asked Dec 19 '18 at 18:39
Will BurghardWill Burghard
42
42
add a comment |
add a comment |
2 Answers
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If you were taught that, then you were taught wrong. If $A$ is a $ntimes n$ matrix then, for each eigenvalue $lambda$ of $A$, the dimension of the space$$E_lambda={vinmathbb{R}^n,|,A.v=lambda v}$$can go from $1$ to $n$. And, of course, if it's greater than $1$, a single vector $vin E_lambda$ will not be a basis of it. If the dimension is $k$, every basis will have $k$ linearly independent eigenvectors (but, unlike what you wrote, never infinitely many).
answered Dec 19 '18 at 18:50
José Carlos SantosJosé Carlos Santos
172k22132239
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In the low-dimensional case, you start by finding the roots of the characteristic polynomial. For each root $lambda$, writing $A-lambda I=0$ gives you a matrix form of a linear system, and the solutions to this linear system represent the eigenspace for $lambda$. The dimension of the eigenspace is at most the multiplicity of the root $lambda$, which means in particular it will be finite-dimensional. And in fact there are only finitely many eigenvalues, which means there are only finitely many linearly independent eigenvectors.
In the high-dimensional (but still finite-dimensional) case, the principles are the same but finding the roots and solving the linear system can be impractical.
In the infinite-dimensional case, the above steps will fail (there is no characteristic polynomial for an infinite-rank operator). In fact there is no general method for finding eigenvalues or eigenvectors of an infinite-rank operator. Sometimes there are infinitely many linearly independent eigenvectors, and sometimes there are no eigenvectors at all.
answered Dec 19 '18 at 19:02
Ben WBen W
2,722818
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2 Answers
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2 Answers
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$begingroup$
If you were taught that, then you were taught wrong. If $A$ is a $ntimes n$ matrix then, for each eigenvalue $lambda$ of $A$, the dimension of the space$$E_lambda={vinmathbb{R}^n,|,A.v=lambda v}$$can go from $1$ to $n$. And, of course, if it's greater than $1$, a single vector $vin E_lambda$ will not be a basis of it. If the dimension is $k$, every basis will have $k$ linearly independent eigenvectors (but, unlike what you wrote, never infinitely many).
answered Dec 19 '18 at 18:50
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
add a comment |
$begingroup$
If you were taught that, then you were taught wrong. If $A$ is a $ntimes n$ matrix then, for each eigenvalue $lambda$ of $A$, the dimension of the space$$E_lambda={vinmathbb{R}^n,|,A.v=lambda v}$$can go from $1$ to $n$. And, of course, if it's greater than $1$, a single vector $vin E_lambda$ will not be a basis of it. If the dimension is $k$, every basis will have $k$ linearly independent eigenvectors (but, unlike what you wrote, never infinitely many).
answered Dec 19 '18 at 18:50
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
add a comment |
$begingroup$
If you were taught that, then you were taught wrong. If $A$ is a $ntimes n$ matrix then, for each eigenvalue $lambda$ of $A$, the dimension of the space$$E_lambda={vinmathbb{R}^n,|,A.v=lambda v}$$can go from $1$ to $n$. And, of course, if it's greater than $1$, a single vector $vin E_lambda$ will not be a basis of it. If the dimension is $k$, every basis will have $k$ linearly independent eigenvectors (but, unlike what you wrote, never infinitely many).
answered Dec 19 '18 at 18:50
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
If you were taught that, then you were taught wrong. If $A$ is a $ntimes n$ matrix then, for each eigenvalue $lambda$ of $A$, the dimension of the space$$E_lambda={vinmathbb{R}^n,|,A.v=lambda v}$$can go from $1$ to $n$. And, of course, if it's greater than $1$, a single vector $vin E_lambda$ will not be a basis of it. If the dimension is $k$, every basis will have $k$ linearly independent eigenvectors (but, unlike what you wrote, never infinitely many).
answered Dec 19 '18 at 18:50
José Carlos SantosJosé Carlos Santos
172k22132239
answered Dec 19 '18 at 18:50
José Carlos SantosJosé Carlos Santos
172k22132239
answered Dec 19 '18 at 18:50
José Carlos SantosJosé Carlos Santos
172k22132239
answered Dec 19 '18 at 18:50
José Carlos SantosJosé Carlos Santos
172k22132239
172k22132239
add a comment |
add a comment |
$begingroup$
In the low-dimensional case, you start by finding the roots of the characteristic polynomial. For each root $lambda$, writing $A-lambda I=0$ gives you a matrix form of a linear system, and the solutions to this linear system represent the eigenspace for $lambda$. The dimension of the eigenspace is at most the multiplicity of the root $lambda$, which means in particular it will be finite-dimensional. And in fact there are only finitely many eigenvalues, which means there are only finitely many linearly independent eigenvectors.
In the high-dimensional (but still finite-dimensional) case, the principles are the same but finding the roots and solving the linear system can be impractical.
In the infinite-dimensional case, the above steps will fail (there is no characteristic polynomial for an infinite-rank operator). In fact there is no general method for finding eigenvalues or eigenvectors of an infinite-rank operator. Sometimes there are infinitely many linearly independent eigenvectors, and sometimes there are no eigenvectors at all.
answered Dec 19 '18 at 19:02
Ben WBen W
2,722818
$endgroup$
add a comment |
$begingroup$
In the low-dimensional case, you start by finding the roots of the characteristic polynomial. For each root $lambda$, writing $A-lambda I=0$ gives you a matrix form of a linear system, and the solutions to this linear system represent the eigenspace for $lambda$. The dimension of the eigenspace is at most the multiplicity of the root $lambda$, which means in particular it will be finite-dimensional. And in fact there are only finitely many eigenvalues, which means there are only finitely many linearly independent eigenvectors.
In the high-dimensional (but still finite-dimensional) case, the principles are the same but finding the roots and solving the linear system can be impractical.
In the infinite-dimensional case, the above steps will fail (there is no characteristic polynomial for an infinite-rank operator). In fact there is no general method for finding eigenvalues or eigenvectors of an infinite-rank operator. Sometimes there are infinitely many linearly independent eigenvectors, and sometimes there are no eigenvectors at all.
answered Dec 19 '18 at 19:02
Ben WBen W
2,722818
$endgroup$
add a comment |
$begingroup$
In the low-dimensional case, you start by finding the roots of the characteristic polynomial. For each root $lambda$, writing $A-lambda I=0$ gives you a matrix form of a linear system, and the solutions to this linear system represent the eigenspace for $lambda$. The dimension of the eigenspace is at most the multiplicity of the root $lambda$, which means in particular it will be finite-dimensional. And in fact there are only finitely many eigenvalues, which means there are only finitely many linearly independent eigenvectors.
In the high-dimensional (but still finite-dimensional) case, the principles are the same but finding the roots and solving the linear system can be impractical.
In the infinite-dimensional case, the above steps will fail (there is no characteristic polynomial for an infinite-rank operator). In fact there is no general method for finding eigenvalues or eigenvectors of an infinite-rank operator. Sometimes there are infinitely many linearly independent eigenvectors, and sometimes there are no eigenvectors at all.
answered Dec 19 '18 at 19:02
Ben WBen W
2,722818
$endgroup$
In the low-dimensional case, you start by finding the roots of the characteristic polynomial. For each root $lambda$, writing $A-lambda I=0$ gives you a matrix form of a linear system, and the solutions to this linear system represent the eigenspace for $lambda$. The dimension of the eigenspace is at most the multiplicity of the root $lambda$, which means in particular it will be finite-dimensional. And in fact there are only finitely many eigenvalues, which means there are only finitely many linearly independent eigenvectors.
In the high-dimensional (but still finite-dimensional) case, the principles are the same but finding the roots and solving the linear system can be impractical.
In the infinite-dimensional case, the above steps will fail (there is no characteristic polynomial for an infinite-rank operator). In fact there is no general method for finding eigenvalues or eigenvectors of an infinite-rank operator. Sometimes there are infinitely many linearly independent eigenvectors, and sometimes there are no eigenvectors at all.
answered Dec 19 '18 at 19:02
Ben WBen W
2,722818
answered Dec 19 '18 at 19:02
Ben WBen W
2,722818
answered Dec 19 '18 at 19:02
Ben WBen W
2,722818
answered Dec 19 '18 at 19:02
Ben WBen W
2,722818
2,722818
add a comment |
add a comment |
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