Inverse of the adjugate operation
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In projective geometry, the map between a primal and dual quadric is the adjugate: $adj(Q) = Q^*$. The map from dual to primal is then the inverse adjugate, $Q = adj^{-1}(Q^*)$, as in this paper.
How is this calculated? I know the adjugate is the inverse multiplied by the determinant, but I don't see how to invert this operation.
Help is appreciated!
linear-algebra projective-geometry
asked Dec 9 '18 at 18:11
JoshuaFJoshuaF
1012
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add a comment |
$begingroup$
In projective geometry, the map between a primal and dual quadric is the adjugate: $adj(Q) = Q^*$. The map from dual to primal is then the inverse adjugate, $Q = adj^{-1}(Q^*)$, as in this paper.
How is this calculated? I know the adjugate is the inverse multiplied by the determinant, but I don't see how to invert this operation.
Help is appreciated!
linear-algebra projective-geometry
asked Dec 9 '18 at 18:11
JoshuaFJoshuaF
1012
$endgroup$
add a comment |
$begingroup$
In projective geometry, the map between a primal and dual quadric is the adjugate: $adj(Q) = Q^*$. The map from dual to primal is then the inverse adjugate, $Q = adj^{-1}(Q^*)$, as in this paper.
How is this calculated? I know the adjugate is the inverse multiplied by the determinant, but I don't see how to invert this operation.
Help is appreciated!
linear-algebra projective-geometry
asked Dec 9 '18 at 18:11
JoshuaFJoshuaF
1012
$endgroup$
In projective geometry, the map between a primal and dual quadric is the adjugate: $adj(Q) = Q^*$. The map from dual to primal is then the inverse adjugate, $Q = adj^{-1}(Q^*)$, as in this paper.
How is this calculated? I know the adjugate is the inverse multiplied by the determinant, but I don't see how to invert this operation.
Help is appreciated!
linear-algebra projective-geometry
linear-algebra projective-geometry
asked Dec 9 '18 at 18:11
JoshuaFJoshuaF
1012
asked Dec 9 '18 at 18:11
JoshuaFJoshuaF
1012
edited Dec 9 '18 at 18:28
JoshuaF
asked Dec 9 '18 at 18:11
JoshuaFJoshuaF
1012
asked Dec 9 '18 at 18:11
JoshuaFJoshuaF
1012
asked Dec 9 '18 at 18:11
JoshuaFJoshuaF
1012
1012
add a comment |
add a comment |
1 Answer
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After some math with determinants, it looks like $adj^{-1}(A) = A^{-1} * |A|^{frac{1}{1 - dim(A)}}$ (based on The determinant of adjugate matrix).
answered Dec 10 '18 at 4:16
JoshuaFJoshuaF
1012
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add a comment |
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$begingroup$
After some math with determinants, it looks like $adj^{-1}(A) = A^{-1} * |A|^{frac{1}{1 - dim(A)}}$ (based on The determinant of adjugate matrix).
answered Dec 10 '18 at 4:16
JoshuaFJoshuaF
1012
$endgroup$
add a comment |
$begingroup$
After some math with determinants, it looks like $adj^{-1}(A) = A^{-1} * |A|^{frac{1}{1 - dim(A)}}$ (based on The determinant of adjugate matrix).
answered Dec 10 '18 at 4:16
JoshuaFJoshuaF
1012
$endgroup$
add a comment |
$begingroup$
After some math with determinants, it looks like $adj^{-1}(A) = A^{-1} * |A|^{frac{1}{1 - dim(A)}}$ (based on The determinant of adjugate matrix).
answered Dec 10 '18 at 4:16
JoshuaFJoshuaF
1012
$endgroup$
After some math with determinants, it looks like $adj^{-1}(A) = A^{-1} * |A|^{frac{1}{1 - dim(A)}}$ (based on The determinant of adjugate matrix).
answered Dec 10 '18 at 4:16
JoshuaFJoshuaF
1012
answered Dec 10 '18 at 4:16
JoshuaFJoshuaF
1012
answered Dec 10 '18 at 4:16
JoshuaFJoshuaF
1012
answered Dec 10 '18 at 4:16
JoshuaFJoshuaF
1012
1012
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