Numerically stable determinant of matrix product
What is the best way in general to numerically calculate the determinant of a product of two matrices?
Calculate the matrix product first and then calculate the determinant of the result or calculate the determinants of the two matrices and subsequently multiply them?
The determinant calculation is done using tested routines (python-numpy i.e. LAPACK).
numerical-linear-algebra
asked Apr 21 '17 at 9:39
TheCat
11
add a comment |
What is the best way in general to numerically calculate the determinant of a product of two matrices?
Calculate the matrix product first and then calculate the determinant of the result or calculate the determinants of the two matrices and subsequently multiply them?
The determinant calculation is done using tested routines (python-numpy i.e. LAPACK).
numerical-linear-algebra
asked Apr 21 '17 at 9:39
TheCat
11
"calculate the determinants of the two matrices and subsequently multiply them" - this is the more stable route. Generally, if you can avoid multiplying matrices before computing anything else (determinants, eigenvalues, etc.), do so.
– J. M. is not a mathematician
Apr 21 '17 at 10:13
It's probably more stable, but determinants in general are rather expensive to calculate.
– mathreadler
Apr 21 '17 at 10:29
add a comment |
What is the best way in general to numerically calculate the determinant of a product of two matrices?
Calculate the matrix product first and then calculate the determinant of the result or calculate the determinants of the two matrices and subsequently multiply them?
The determinant calculation is done using tested routines (python-numpy i.e. LAPACK).
numerical-linear-algebra
asked Apr 21 '17 at 9:39
TheCat
11
What is the best way in general to numerically calculate the determinant of a product of two matrices?
Calculate the matrix product first and then calculate the determinant of the result or calculate the determinants of the two matrices and subsequently multiply them?
The determinant calculation is done using tested routines (python-numpy i.e. LAPACK).
numerical-linear-algebra
numerical-linear-algebra
asked Apr 21 '17 at 9:39
TheCat
11
asked Apr 21 '17 at 9:39
TheCat
11
asked Apr 21 '17 at 9:39
TheCat
11
asked Apr 21 '17 at 9:39
TheCat
11
asked Apr 21 '17 at 9:39
TheCat
11
11
"calculate the determinants of the two matrices and subsequently multiply them" - this is the more stable route. Generally, if you can avoid multiplying matrices before computing anything else (determinants, eigenvalues, etc.), do so.
– J. M. is not a mathematician
Apr 21 '17 at 10:13
It's probably more stable, but determinants in general are rather expensive to calculate.
– mathreadler
Apr 21 '17 at 10:29
add a comment |
"calculate the determinants of the two matrices and subsequently multiply them" - this is the more stable route. Generally, if you can avoid multiplying matrices before computing anything else (determinants, eigenvalues, etc.), do so.
– J. M. is not a mathematician
Apr 21 '17 at 10:13
It's probably more stable, but determinants in general are rather expensive to calculate.
– mathreadler
Apr 21 '17 at 10:29
"calculate the determinants of the two matrices and subsequently multiply them" - this is the more stable route. Generally, if you can avoid multiplying matrices before computing anything else (determinants, eigenvalues, etc.), do so.
– J. M. is not a mathematician
Apr 21 '17 at 10:13
"calculate the determinants of the two matrices and subsequently multiply them" - this is the more stable route. Generally, if you can avoid multiplying matrices before computing anything else (determinants, eigenvalues, etc.), do so.
– J. M. is not a mathematician
Apr 21 '17 at 10:13
It's probably more stable, but determinants in general are rather expensive to calculate.
– mathreadler
Apr 21 '17 at 10:29
It's probably more stable, but determinants in general are rather expensive to calculate.
– mathreadler
Apr 21 '17 at 10:29
add a comment |
1 Answer
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Certainly, compute the two determinants of the factors individually. The matrix product can contain catastrophic roundoff error, destroying any evidence of the actual determinant. For example, consider the product
$$ begin{pmatrix} 1&0\10^9&1end{pmatrix}begin{pmatrix} 1&10^9\0&1end{pmatrix}
= begin{pmatrix} 1&10^9\10^9&10^{18}+1end{pmatrix}.$$
In IEEE double precision, that's
begin{pmatrix} 1&10^9\10^9&10^{18}end{pmatrix} whose determinant is $0$. Taking the product of the determinants of the original two matrics yields the correct answer $1$.
answered Nov 24 at 4:16
Don Hatch
281112
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Certainly, compute the two determinants of the factors individually. The matrix product can contain catastrophic roundoff error, destroying any evidence of the actual determinant. For example, consider the product
$$ begin{pmatrix} 1&0\10^9&1end{pmatrix}begin{pmatrix} 1&10^9\0&1end{pmatrix}
= begin{pmatrix} 1&10^9\10^9&10^{18}+1end{pmatrix}.$$
In IEEE double precision, that's
begin{pmatrix} 1&10^9\10^9&10^{18}end{pmatrix} whose determinant is $0$. Taking the product of the determinants of the original two matrics yields the correct answer $1$.
answered Nov 24 at 4:16
Don Hatch
281112
add a comment |
Certainly, compute the two determinants of the factors individually. The matrix product can contain catastrophic roundoff error, destroying any evidence of the actual determinant. For example, consider the product
$$ begin{pmatrix} 1&0\10^9&1end{pmatrix}begin{pmatrix} 1&10^9\0&1end{pmatrix}
= begin{pmatrix} 1&10^9\10^9&10^{18}+1end{pmatrix}.$$
In IEEE double precision, that's
begin{pmatrix} 1&10^9\10^9&10^{18}end{pmatrix} whose determinant is $0$. Taking the product of the determinants of the original two matrics yields the correct answer $1$.
answered Nov 24 at 4:16
Don Hatch
281112
add a comment |
Certainly, compute the two determinants of the factors individually. The matrix product can contain catastrophic roundoff error, destroying any evidence of the actual determinant. For example, consider the product
$$ begin{pmatrix} 1&0\10^9&1end{pmatrix}begin{pmatrix} 1&10^9\0&1end{pmatrix}
= begin{pmatrix} 1&10^9\10^9&10^{18}+1end{pmatrix}.$$
In IEEE double precision, that's
begin{pmatrix} 1&10^9\10^9&10^{18}end{pmatrix} whose determinant is $0$. Taking the product of the determinants of the original two matrics yields the correct answer $1$.
answered Nov 24 at 4:16
Don Hatch
281112
Certainly, compute the two determinants of the factors individually. The matrix product can contain catastrophic roundoff error, destroying any evidence of the actual determinant. For example, consider the product
$$ begin{pmatrix} 1&0\10^9&1end{pmatrix}begin{pmatrix} 1&10^9\0&1end{pmatrix}
= begin{pmatrix} 1&10^9\10^9&10^{18}+1end{pmatrix}.$$
In IEEE double precision, that's
begin{pmatrix} 1&10^9\10^9&10^{18}end{pmatrix} whose determinant is $0$. Taking the product of the determinants of the original two matrics yields the correct answer $1$.
answered Nov 24 at 4:16
Don Hatch
281112
answered Nov 24 at 4:16
Don Hatch
281112
answered Nov 24 at 4:16
Don Hatch
281112
answered Nov 24 at 4:16
Don Hatch
281112
281112
add a comment |
add a comment |
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"calculate the determinants of the two matrices and subsequently multiply them" - this is the more stable route. Generally, if you can avoid multiplying matrices before computing anything else (determinants, eigenvalues, etc.), do so.
– J. M. is not a mathematician
Apr 21 '17 at 10:13
It's probably more stable, but determinants in general are rather expensive to calculate.
– mathreadler
Apr 21 '17 at 10:29