Mixing
Analytically finding minimum eigenvalue of a matrix
$begingroup$
Is there a way to find the minimum eigenvalue of a matrix analytically? (I know that the usual way of finding the eigenvalues is the characteristic polynomial).
One definition of the minimum eigenvalue is
$$
min_{,,,x\xneq 0} frac{|A x|^2}{|x|^2}
$$
For the simple matrix [ 0 1 ; -2 -3 ] I tried to find the smallest eigenvalue analytically by naming the x-vector components $x_1$ and $x_2$ and having $D = ((-2x_1 -3 x_2)^2 + x_2^2)/(x_1^2+x_2^2)$. From this I obtain the gradient $frac{partial D}{partial x_1}$, $frac{partial D}{partial x_2}$ and set it to zero solve for $x_1$ and $x_2$.
Surely enough, this system of equations does not have a solution.
linear-algebra eigenvalues-eigenvectors convex-optimization
asked Jan 9 at 0:49
divBdivB
228111
$endgroup$
add a comment |
$begingroup$
Is there a way to find the minimum eigenvalue of a matrix analytically? (I know that the usual way of finding the eigenvalues is the characteristic polynomial).
One definition of the minimum eigenvalue is
$$
min_{,,,x\xneq 0} frac{|A x|^2}{|x|^2}
$$
For the simple matrix [ 0 1 ; -2 -3 ] I tried to find the smallest eigenvalue analytically by naming the x-vector components $x_1$ and $x_2$ and having $D = ((-2x_1 -3 x_2)^2 + x_2^2)/(x_1^2+x_2^2)$. From this I obtain the gradient $frac{partial D}{partial x_1}$, $frac{partial D}{partial x_2}$ and set it to zero solve for $x_1$ and $x_2$.
Surely enough, this system of equations does not have a solution.
linear-algebra eigenvalues-eigenvectors convex-optimization
asked Jan 9 at 0:49
divBdivB
228111
$endgroup$
$begingroup$
For an eigenvector $x$ with eigenvalue $lambda$, $x^TAx=lambdalVert xrVert^2$. So it would seem that you need to divide that expression by $lVert xrVert^2$ before taking the min in order to find $lambda$.
$endgroup$
– alex.jordan
Jan 9 at 0:54
$begingroup$
You are right. I added the normalization but the system does not have a solution either. I will update my question.
$endgroup$
– divB
Jan 9 at 1:03
$begingroup$
In the most recent edit you replaced $x^TAx$ with $lVert AxrVert^2$. These are not the same. $lVert AxrVert^2$ is the same as $x^TA^TAx$.
$endgroup$
– alex.jordan
Jan 9 at 1:47
add a comment |
$begingroup$
Is there a way to find the minimum eigenvalue of a matrix analytically? (I know that the usual way of finding the eigenvalues is the characteristic polynomial).
One definition of the minimum eigenvalue is
$$
min_{,,,x\xneq 0} frac{|A x|^2}{|x|^2}
$$
For the simple matrix [ 0 1 ; -2 -3 ] I tried to find the smallest eigenvalue analytically by naming the x-vector components $x_1$ and $x_2$ and having $D = ((-2x_1 -3 x_2)^2 + x_2^2)/(x_1^2+x_2^2)$. From this I obtain the gradient $frac{partial D}{partial x_1}$, $frac{partial D}{partial x_2}$ and set it to zero solve for $x_1$ and $x_2$.
Surely enough, this system of equations does not have a solution.
linear-algebra eigenvalues-eigenvectors convex-optimization
asked Jan 9 at 0:49
divBdivB
228111
$endgroup$
Is there a way to find the minimum eigenvalue of a matrix analytically? (I know that the usual way of finding the eigenvalues is the characteristic polynomial).
One definition of the minimum eigenvalue is
$$
min_{,,,x\xneq 0} frac{|A x|^2}{|x|^2}
$$
For the simple matrix [ 0 1 ; -2 -3 ] I tried to find the smallest eigenvalue analytically by naming the x-vector components $x_1$ and $x_2$ and having $D = ((-2x_1 -3 x_2)^2 + x_2^2)/(x_1^2+x_2^2)$. From this I obtain the gradient $frac{partial D}{partial x_1}$, $frac{partial D}{partial x_2}$ and set it to zero solve for $x_1$ and $x_2$.
Surely enough, this system of equations does not have a solution.
linear-algebra eigenvalues-eigenvectors convex-optimization
linear-algebra eigenvalues-eigenvectors convex-optimization
asked Jan 9 at 0:49
divBdivB
228111
asked Jan 9 at 0:49
divBdivB
228111
edited Jan 9 at 1:32
divB
asked Jan 9 at 0:49
divBdivB
228111
asked Jan 9 at 0:49
divBdivB
228111
asked Jan 9 at 0:49
divBdivB
228111
228111
$begingroup$
For an eigenvector $x$ with eigenvalue $lambda$, $x^TAx=lambdalVert xrVert^2$. So it would seem that you need to divide that expression by $lVert xrVert^2$ before taking the min in order to find $lambda$.
$endgroup$
– alex.jordan
Jan 9 at 0:54
$begingroup$
You are right. I added the normalization but the system does not have a solution either. I will update my question.
$endgroup$
– divB
Jan 9 at 1:03
$begingroup$
In the most recent edit you replaced $x^TAx$ with $lVert AxrVert^2$. These are not the same. $lVert AxrVert^2$ is the same as $x^TA^TAx$.
$endgroup$
– alex.jordan
Jan 9 at 1:47
add a comment |
$begingroup$
For an eigenvector $x$ with eigenvalue $lambda$, $x^TAx=lambdalVert xrVert^2$. So it would seem that you need to divide that expression by $lVert xrVert^2$ before taking the min in order to find $lambda$.
$endgroup$
– alex.jordan
Jan 9 at 0:54
$begingroup$
You are right. I added the normalization but the system does not have a solution either. I will update my question.
$endgroup$
– divB
Jan 9 at 1:03
$begingroup$
In the most recent edit you replaced $x^TAx$ with $lVert AxrVert^2$. These are not the same. $lVert AxrVert^2$ is the same as $x^TA^TAx$.
$endgroup$
– alex.jordan
Jan 9 at 1:47
$begingroup$
For an eigenvector $x$ with eigenvalue $lambda$, $x^TAx=lambdalVert xrVert^2$. So it would seem that you need to divide that expression by $lVert xrVert^2$ before taking the min in order to find $lambda$.
$endgroup$
– alex.jordan
Jan 9 at 0:54
$begingroup$
For an eigenvector $x$ with eigenvalue $lambda$, $x^TAx=lambdalVert xrVert^2$. So it would seem that you need to divide that expression by $lVert xrVert^2$ before taking the min in order to find $lambda$.
$endgroup$
– alex.jordan
Jan 9 at 0:54
$begingroup$
You are right. I added the normalization but the system does not have a solution either. I will update my question.
$endgroup$
– divB
Jan 9 at 1:03
$begingroup$
You are right. I added the normalization but the system does not have a solution either. I will update my question.
$endgroup$
– divB
Jan 9 at 1:03
$begingroup$
In the most recent edit you replaced $x^TAx$ with $lVert AxrVert^2$. These are not the same. $lVert AxrVert^2$ is the same as $x^TA^TAx$.
$endgroup$
– alex.jordan
Jan 9 at 1:47
$begingroup$
In the most recent edit you replaced $x^TAx$ with $lVert AxrVert^2$. These are not the same. $lVert AxrVert^2$ is the same as $x^TA^TAx$.
$endgroup$
– alex.jordan
Jan 9 at 1:47
add a comment |
1 Answer
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$begingroup$
Your "definition" of minimum eigenvalue has several problems. For one thing, it would imply that every matrix has a real and positive eigenvalue, which is not true in general.
Also, dont' forget a simple sanity check: your matrix property should apply for a $1times 1$ matrix... and it clearly doesn't.
What you actually mean, I guess, is : if $M$ is hermitian and positive definite (which implies that its eigenvalues are real and positive), then letting $lambda_1$ be its smallest eigenvalue, we have
$$lambda_1 = min_{x ne 0} frac{x^t M x }{x^t x} $$
Further, if we decompose $M=A^t A$ the above gives
$$lambda_1 = min_{x ne 0} frac{x^t A^t A x }{x^t x} =min_{,,,x\xneq 0} frac{|A x|^2}{|x|^2}$$
But, bear in mind, here $lambda_1$ is an eigenvalue of $M$, not of $A$.
To answer the question in the title: there cannot be a general analytical procedure for computing the smallest eingenvalue of a matrix, because the problem maps to finding the smallest root of a polinomial, and for degrees five and above there's no analytical procedure (in the usual sense of the expression) for doing that (Abel-Ruffini theorem)
On the other side, if the matrix is $2times 2$, then it's trivial. But you would't use the Rayleigh quotient for that. This is only useful for computing numerically the smallest (or biggest) eigenvalue, by an iterative method.
answered Jan 9 at 2:09
leonbloyleonbloy
40.7k645107
$endgroup$
add a comment |
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$begingroup$
Your "definition" of minimum eigenvalue has several problems. For one thing, it would imply that every matrix has a real and positive eigenvalue, which is not true in general.
Also, dont' forget a simple sanity check: your matrix property should apply for a $1times 1$ matrix... and it clearly doesn't.
What you actually mean, I guess, is : if $M$ is hermitian and positive definite (which implies that its eigenvalues are real and positive), then letting $lambda_1$ be its smallest eigenvalue, we have
$$lambda_1 = min_{x ne 0} frac{x^t M x }{x^t x} $$
Further, if we decompose $M=A^t A$ the above gives
$$lambda_1 = min_{x ne 0} frac{x^t A^t A x }{x^t x} =min_{,,,x\xneq 0} frac{|A x|^2}{|x|^2}$$
But, bear in mind, here $lambda_1$ is an eigenvalue of $M$, not of $A$.
To answer the question in the title: there cannot be a general analytical procedure for computing the smallest eingenvalue of a matrix, because the problem maps to finding the smallest root of a polinomial, and for degrees five and above there's no analytical procedure (in the usual sense of the expression) for doing that (Abel-Ruffini theorem)
On the other side, if the matrix is $2times 2$, then it's trivial. But you would't use the Rayleigh quotient for that. This is only useful for computing numerically the smallest (or biggest) eigenvalue, by an iterative method.
answered Jan 9 at 2:09
leonbloyleonbloy
40.7k645107
$endgroup$
add a comment |
$begingroup$
Your "definition" of minimum eigenvalue has several problems. For one thing, it would imply that every matrix has a real and positive eigenvalue, which is not true in general.
Also, dont' forget a simple sanity check: your matrix property should apply for a $1times 1$ matrix... and it clearly doesn't.
What you actually mean, I guess, is : if $M$ is hermitian and positive definite (which implies that its eigenvalues are real and positive), then letting $lambda_1$ be its smallest eigenvalue, we have
$$lambda_1 = min_{x ne 0} frac{x^t M x }{x^t x} $$
Further, if we decompose $M=A^t A$ the above gives
$$lambda_1 = min_{x ne 0} frac{x^t A^t A x }{x^t x} =min_{,,,x\xneq 0} frac{|A x|^2}{|x|^2}$$
But, bear in mind, here $lambda_1$ is an eigenvalue of $M$, not of $A$.
To answer the question in the title: there cannot be a general analytical procedure for computing the smallest eingenvalue of a matrix, because the problem maps to finding the smallest root of a polinomial, and for degrees five and above there's no analytical procedure (in the usual sense of the expression) for doing that (Abel-Ruffini theorem)
On the other side, if the matrix is $2times 2$, then it's trivial. But you would't use the Rayleigh quotient for that. This is only useful for computing numerically the smallest (or biggest) eigenvalue, by an iterative method.
answered Jan 9 at 2:09
leonbloyleonbloy
40.7k645107
$endgroup$
add a comment |
$begingroup$
Your "definition" of minimum eigenvalue has several problems. For one thing, it would imply that every matrix has a real and positive eigenvalue, which is not true in general.
Also, dont' forget a simple sanity check: your matrix property should apply for a $1times 1$ matrix... and it clearly doesn't.
What you actually mean, I guess, is : if $M$ is hermitian and positive definite (which implies that its eigenvalues are real and positive), then letting $lambda_1$ be its smallest eigenvalue, we have
$$lambda_1 = min_{x ne 0} frac{x^t M x }{x^t x} $$
Further, if we decompose $M=A^t A$ the above gives
$$lambda_1 = min_{x ne 0} frac{x^t A^t A x }{x^t x} =min_{,,,x\xneq 0} frac{|A x|^2}{|x|^2}$$
But, bear in mind, here $lambda_1$ is an eigenvalue of $M$, not of $A$.
To answer the question in the title: there cannot be a general analytical procedure for computing the smallest eingenvalue of a matrix, because the problem maps to finding the smallest root of a polinomial, and for degrees five and above there's no analytical procedure (in the usual sense of the expression) for doing that (Abel-Ruffini theorem)
On the other side, if the matrix is $2times 2$, then it's trivial. But you would't use the Rayleigh quotient for that. This is only useful for computing numerically the smallest (or biggest) eigenvalue, by an iterative method.
answered Jan 9 at 2:09
leonbloyleonbloy
40.7k645107
$endgroup$
Your "definition" of minimum eigenvalue has several problems. For one thing, it would imply that every matrix has a real and positive eigenvalue, which is not true in general.
Also, dont' forget a simple sanity check: your matrix property should apply for a $1times 1$ matrix... and it clearly doesn't.
What you actually mean, I guess, is : if $M$ is hermitian and positive definite (which implies that its eigenvalues are real and positive), then letting $lambda_1$ be its smallest eigenvalue, we have
$$lambda_1 = min_{x ne 0} frac{x^t M x }{x^t x} $$
Further, if we decompose $M=A^t A$ the above gives
$$lambda_1 = min_{x ne 0} frac{x^t A^t A x }{x^t x} =min_{,,,x\xneq 0} frac{|A x|^2}{|x|^2}$$
But, bear in mind, here $lambda_1$ is an eigenvalue of $M$, not of $A$.
To answer the question in the title: there cannot be a general analytical procedure for computing the smallest eingenvalue of a matrix, because the problem maps to finding the smallest root of a polinomial, and for degrees five and above there's no analytical procedure (in the usual sense of the expression) for doing that (Abel-Ruffini theorem)
On the other side, if the matrix is $2times 2$, then it's trivial. But you would't use the Rayleigh quotient for that. This is only useful for computing numerically the smallest (or biggest) eigenvalue, by an iterative method.
answered Jan 9 at 2:09
leonbloyleonbloy
40.7k645107
edited Jan 9 at 2:14
answered Jan 9 at 2:09
leonbloyleonbloy
40.7k645107
answered Jan 9 at 2:09
leonbloyleonbloy
40.7k645107
answered Jan 9 at 2:09
leonbloyleonbloy
40.7k645107
40.7k645107
add a comment |
add a comment |
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$begingroup$
For an eigenvector $x$ with eigenvalue $lambda$, $x^TAx=lambdalVert xrVert^2$. So it would seem that you need to divide that expression by $lVert xrVert^2$ before taking the min in order to find $lambda$.
$endgroup$
– alex.jordan
Jan 9 at 0:54
$begingroup$
You are right. I added the normalization but the system does not have a solution either. I will update my question.
$endgroup$
– divB
Jan 9 at 1:03
$begingroup$
In the most recent edit you replaced $x^TAx$ with $lVert AxrVert^2$. These are not the same. $lVert AxrVert^2$ is the same as $x^TA^TAx$.
$endgroup$
– alex.jordan
Jan 9 at 1:47