Why not different diagonal elements in orthogonal reduction of quadratic form? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)diagonalize quadratic formIs there iterative eigen decomposition method?How to put $2x^2 + 4xy + 6y^2 + 6x + 2y = 6$ in canonical formConic matrix and diagonalizationDiagonalisation of a quadratic form.Matrix of the quadratic formWhy use decoupling diagonalization method to solve system of ODE's instead of just the eigen vectors?How to find generalized Eigen vectors of a matrix with Eigen vectors already on diagonal?Understanding repeated Eigen valuesReducing the quadratic form
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Why not different diagonal elements in orthogonal reduction of quadratic form?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)diagonalize quadratic formIs there iterative eigen decomposition method?How to put $2x^2 + 4xy + 6y^2 + 6x + 2y = 6$ in canonical formConic matrix and diagonalizationDiagonalisation of a quadratic form.Matrix of the quadratic formWhy use decoupling diagonalization method to solve system of ODE's instead of just the eigen vectors?How to find generalized Eigen vectors of a matrix with Eigen vectors already on diagonal?Understanding repeated Eigen valuesReducing the quadratic form
$begingroup$
Suppose we have quadratic form
$(3x)^2 -(2y)^2-z^2-(4xy)+(8xz)+(12yz)$
and we are asked to reduce it to cannonical form. The steps would be:
- Find eigen values (3,6,-9 in this case) of the matrix of quadratic form (A)
- Modal matrix(P) from eigen vectors
- Normalisation of P.(N)
- Finding $D=N^T×A×N$
D would look like
$$D =beginbmatrix
3 & 0 & 0 \
0 & 6 & 0 \
0 & 0& -9
endbmatrix.$$
Now the diagonal elements are exactly same as eigen values!
My query is till now, every time, eigen values come out as the diagonal elements, then why we go about finding the matrix in the first place?
eigenvalues-eigenvectors diagonalization
asked Apr 1 at 7:56
VimathVimath
788
$endgroup$
add a comment |
$begingroup$
Suppose we have quadratic form
$(3x)^2 -(2y)^2-z^2-(4xy)+(8xz)+(12yz)$
and we are asked to reduce it to cannonical form. The steps would be:
- Find eigen values (3,6,-9 in this case) of the matrix of quadratic form (A)
- Modal matrix(P) from eigen vectors
- Normalisation of P.(N)
- Finding $D=N^T×A×N$
D would look like
$$D =beginbmatrix
3 & 0 & 0 \
0 & 6 & 0 \
0 & 0& -9
endbmatrix.$$
Now the diagonal elements are exactly same as eigen values!
My query is till now, every time, eigen values come out as the diagonal elements, then why we go about finding the matrix in the first place?
eigenvalues-eigenvectors diagonalization
asked Apr 1 at 7:56
VimathVimath
788
$endgroup$
$begingroup$
1) For practice. 2) To check your own work. Suppose that you ended up with a $D$ that didn’t have eigenvalues along its diagonal. That could mean that you computed them incorrectly in the first place.
$endgroup$
– amd
Apr 1 at 19:24
$begingroup$
Ok but even then, why it's so?
$endgroup$
– Vimath
Apr 2 at 8:05
add a comment |
$begingroup$
Suppose we have quadratic form
$(3x)^2 -(2y)^2-z^2-(4xy)+(8xz)+(12yz)$
and we are asked to reduce it to cannonical form. The steps would be:
- Find eigen values (3,6,-9 in this case) of the matrix of quadratic form (A)
- Modal matrix(P) from eigen vectors
- Normalisation of P.(N)
- Finding $D=N^T×A×N$
D would look like
$$D =beginbmatrix
3 & 0 & 0 \
0 & 6 & 0 \
0 & 0& -9
endbmatrix.$$
Now the diagonal elements are exactly same as eigen values!
My query is till now, every time, eigen values come out as the diagonal elements, then why we go about finding the matrix in the first place?
eigenvalues-eigenvectors diagonalization
asked Apr 1 at 7:56
VimathVimath
788
$endgroup$
Suppose we have quadratic form
$(3x)^2 -(2y)^2-z^2-(4xy)+(8xz)+(12yz)$
and we are asked to reduce it to cannonical form. The steps would be:
- Find eigen values (3,6,-9 in this case) of the matrix of quadratic form (A)
- Modal matrix(P) from eigen vectors
- Normalisation of P.(N)
- Finding $D=N^T×A×N$
D would look like
$$D =beginbmatrix
3 & 0 & 0 \
0 & 6 & 0 \
0 & 0& -9
endbmatrix.$$
Now the diagonal elements are exactly same as eigen values!
My query is till now, every time, eigen values come out as the diagonal elements, then why we go about finding the matrix in the first place?
eigenvalues-eigenvectors diagonalization
eigenvalues-eigenvectors diagonalization
asked Apr 1 at 7:56
VimathVimath
788
asked Apr 1 at 7:56
VimathVimath
788
asked Apr 1 at 7:56
VimathVimath
788
asked Apr 1 at 7:56
VimathVimath
788
asked Apr 1 at 7:56
VimathVimath
788
788
$begingroup$
1) For practice. 2) To check your own work. Suppose that you ended up with a $D$ that didn’t have eigenvalues along its diagonal. That could mean that you computed them incorrectly in the first place.
$endgroup$
– amd
Apr 1 at 19:24
$begingroup$
Ok but even then, why it's so?
$endgroup$
– Vimath
Apr 2 at 8:05
add a comment |
$begingroup$
1) For practice. 2) To check your own work. Suppose that you ended up with a $D$ that didn’t have eigenvalues along its diagonal. That could mean that you computed them incorrectly in the first place.
$endgroup$
– amd
Apr 1 at 19:24
$begingroup$
Ok but even then, why it's so?
$endgroup$
– Vimath
Apr 2 at 8:05
$begingroup$
1) For practice. 2) To check your own work. Suppose that you ended up with a $D$ that didn’t have eigenvalues along its diagonal. That could mean that you computed them incorrectly in the first place.
$endgroup$
– amd
Apr 1 at 19:24
$begingroup$
1) For practice. 2) To check your own work. Suppose that you ended up with a $D$ that didn’t have eigenvalues along its diagonal. That could mean that you computed them incorrectly in the first place.
$endgroup$
– amd
Apr 1 at 19:24
$begingroup$
Ok but even then, why it's so?
$endgroup$
– Vimath
Apr 2 at 8:05
$begingroup$
Ok but even then, why it's so?
$endgroup$
– Vimath
Apr 2 at 8:05
add a comment |
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$begingroup$
1) For practice. 2) To check your own work. Suppose that you ended up with a $D$ that didn’t have eigenvalues along its diagonal. That could mean that you computed them incorrectly in the first place.
$endgroup$
– amd
Apr 1 at 19:24
$begingroup$
Ok but even then, why it's so?
$endgroup$
– Vimath
Apr 2 at 8:05