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Calculate $det'(A)$ using a function $f: mathcalM_2 times 2(mathbbR) to mathcalM_2 times 2(mathbbR)$
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Problem. Let $mathcalM_2 times 2(mathbbR)$ the space of square matrix of order $2$ and consider the fuction $f: mathcalM_2 times 2(mathbbR) to mathcalM_2 times 2(mathbbR)$ defined by:
$$A=beginpmatrix a&b\c&d\ endpmatrix longmapsto f(A) =beginpmatrix d&-b\-c&a\ endpmatrix.$$
Show that
$$det(A + H) = det(A) + operatornametr(f(A)H) + det(H),$$
and conclude that $det$ is differentiable, with $det'(A) = f(A)^t$. Note that if $det(A) neq 0$, then $f(A) = det(A)A^-1$.
Notation. $f(A)^t$ is the transposed.
I proved that $det(A + H) = det(A) + operatornametr(f(A)H) + det(H)$, but I don't know to use it for show that $det'(A) = f(A)^t$. I tried to write $det(A+H) - det(A) = f(A)^t cdot H + o(H)$ using the equality, and to show that $displaystyle lim_H to 0fraco(H)H = 0$, but I couldn't. I appreciate any hints!
real-analysis derivatives
asked Jul 28 at 3:25
Lucas Corrêa
1,087319
add a comment |Â
up vote
1
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Problem. Let $mathcalM_2 times 2(mathbbR)$ the space of square matrix of order $2$ and consider the fuction $f: mathcalM_2 times 2(mathbbR) to mathcalM_2 times 2(mathbbR)$ defined by:
$$A=beginpmatrix a&b\c&d\ endpmatrix longmapsto f(A) =beginpmatrix d&-b\-c&a\ endpmatrix.$$
Show that
$$det(A + H) = det(A) + operatornametr(f(A)H) + det(H),$$
and conclude that $det$ is differentiable, with $det'(A) = f(A)^t$. Note that if $det(A) neq 0$, then $f(A) = det(A)A^-1$.
Notation. $f(A)^t$ is the transposed.
I proved that $det(A + H) = det(A) + operatornametr(f(A)H) + det(H)$, but I don't know to use it for show that $det'(A) = f(A)^t$. I tried to write $det(A+H) - det(A) = f(A)^t cdot H + o(H)$ using the equality, and to show that $displaystyle lim_H to 0fraco(H)H = 0$, but I couldn't. I appreciate any hints!
real-analysis derivatives
asked Jul 28 at 3:25
Lucas Corrêa
1,087319
try expanding $H$ into a linear combination of elementary matrices
– Max
Jul 28 at 3:37
$|det H | le |H_11 H_22| + |H_12H_21| le 2 |H|^2$. (Cf. Hadamard's inequality.)
– copper.hat
Jul 28 at 5:47
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Problem. Let $mathcalM_2 times 2(mathbbR)$ the space of square matrix of order $2$ and consider the fuction $f: mathcalM_2 times 2(mathbbR) to mathcalM_2 times 2(mathbbR)$ defined by:
$$A=beginpmatrix a&b\c&d\ endpmatrix longmapsto f(A) =beginpmatrix d&-b\-c&a\ endpmatrix.$$
Show that
$$det(A + H) = det(A) + operatornametr(f(A)H) + det(H),$$
and conclude that $det$ is differentiable, with $det'(A) = f(A)^t$. Note that if $det(A) neq 0$, then $f(A) = det(A)A^-1$.
Notation. $f(A)^t$ is the transposed.
I proved that $det(A + H) = det(A) + operatornametr(f(A)H) + det(H)$, but I don't know to use it for show that $det'(A) = f(A)^t$. I tried to write $det(A+H) - det(A) = f(A)^t cdot H + o(H)$ using the equality, and to show that $displaystyle lim_H to 0fraco(H)H = 0$, but I couldn't. I appreciate any hints!
real-analysis derivatives
asked Jul 28 at 3:25
Lucas Corrêa
1,087319
Problem. Let $mathcalM_2 times 2(mathbbR)$ the space of square matrix of order $2$ and consider the fuction $f: mathcalM_2 times 2(mathbbR) to mathcalM_2 times 2(mathbbR)$ defined by:
$$A=beginpmatrix a&b\c&d\ endpmatrix longmapsto f(A) =beginpmatrix d&-b\-c&a\ endpmatrix.$$
Show that
$$det(A + H) = det(A) + operatornametr(f(A)H) + det(H),$$
and conclude that $det$ is differentiable, with $det'(A) = f(A)^t$. Note that if $det(A) neq 0$, then $f(A) = det(A)A^-1$.
Notation. $f(A)^t$ is the transposed.
I proved that $det(A + H) = det(A) + operatornametr(f(A)H) + det(H)$, but I don't know to use it for show that $det'(A) = f(A)^t$. I tried to write $det(A+H) - det(A) = f(A)^t cdot H + o(H)$ using the equality, and to show that $displaystyle lim_H to 0fraco(H)H = 0$, but I couldn't. I appreciate any hints!
real-analysis derivatives
asked Jul 28 at 3:25
Lucas Corrêa
1,087319
asked Jul 28 at 3:25
Lucas Corrêa
1,087319
asked Jul 28 at 3:25
Lucas Corrêa
1,087319
asked Jul 28 at 3:25
Lucas Corrêa
1,087319
1,087319
try expanding $H$ into a linear combination of elementary matrices
– Max
Jul 28 at 3:37
$|det H | le |H_11 H_22| + |H_12H_21| le 2 |H|^2$. (Cf. Hadamard's inequality.)
– copper.hat
Jul 28 at 5:47
add a comment |Â
try expanding $H$ into a linear combination of elementary matrices
– Max
Jul 28 at 3:37
$|det H | le |H_11 H_22| + |H_12H_21| le 2 |H|^2$. (Cf. Hadamard's inequality.)
– copper.hat
Jul 28 at 5:47
try expanding $H$ into a linear combination of elementary matrices
– Max
Jul 28 at 3:37
try expanding $H$ into a linear combination of elementary matrices
– Max
Jul 28 at 3:37
$|det H | le |H_11 H_22| + |H_12H_21| le 2 |H|^2$. (Cf. Hadamard's inequality.)
– copper.hat
Jul 28 at 5:47
$|det H | le |H_11 H_22| + |H_12H_21| le 2 |H|^2$. (Cf. Hadamard's inequality.)
– copper.hat
Jul 28 at 5:47
add a comment |Â
1 Answer
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Not exactly an answer, but may illuminate slightly.
I guess I would note that the functions $h mapsto det(A+h E_ij)$ are all
polynomials in $h$ and hence smooth. Hence the partial derivatives exist and are
continuous and so $det$ is differentiable.
To find a formula, we can start with special cases.
Note that $det(I+ t E_ij) = 1 + t delta_ij$ and so $D det(I) (E_ij) = delta_ij$.
Hence $D det (I)(H) = sum_ij H_ij det(I)(E_ij) = operatornametr H$.
Now suppose $A$ is invertible. Then $det(A+H) = det A det (I+ A^-1 H)$,
and it follows from this that
$D det(A)(H) = (det A )D det(I)(A^-1H) = det A operatornametr (A^-1H)$. Since
$A operatornameadj(A) = (det A )I$, we have
$D det(A)(H) = operatornametr(operatornameadj(A) H)$.
Since both sides of the latter formula are continuous and the invertible
matrices are dense in the space of matrices, this formula is true for all $A$.
To obtain the other representation of the derivative, we use the inner product
$langle A , B rangle = operatornametr(A^TB)$.
Then $D det(A)(H) = langle operatornameadj(A)^T , H rangle$.
answered Jul 28 at 6:48
copper.hat
122k557156
Thank you! I understood your idea. So, using $langle A, B rangle = operatornametr(A^TB)$ and write $det(A+H) - det(H) = operatornametr(f(A)H) + det(H)$ and take $o(H) = det(H)$, I can also conclude the same thing, right?
– Lucas Corrêa
Jul 28 at 15:12
1
@LucasCorrêa: Well, there are two different approaches. The one in the answer is a general approach that works for all dimensions. The one in the question is a nice formula for the $2 times 2$ case, and you need to show that $det H$ is $o(|H|)$. If you look at my comment to the question you can see that in fact, $det H$ is $O(|H|^2)$ ('big' O). The formula in the question doesn't easily generalise to higher dimensions.
– copper.hat
Jul 28 at 15:16
1
Yes, I just asked you why I wanted to use the back of the exercise. But your response was very good! In fact, you gave me a much better proof than mine to $det'(A)H = det A operatornametr(A^-1H)$.
– Lucas Corrêa
Jul 28 at 15:21
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Not exactly an answer, but may illuminate slightly.
I guess I would note that the functions $h mapsto det(A+h E_ij)$ are all
polynomials in $h$ and hence smooth. Hence the partial derivatives exist and are
continuous and so $det$ is differentiable.
To find a formula, we can start with special cases.
Note that $det(I+ t E_ij) = 1 + t delta_ij$ and so $D det(I) (E_ij) = delta_ij$.
Hence $D det (I)(H) = sum_ij H_ij det(I)(E_ij) = operatornametr H$.
Now suppose $A$ is invertible. Then $det(A+H) = det A det (I+ A^-1 H)$,
and it follows from this that
$D det(A)(H) = (det A )D det(I)(A^-1H) = det A operatornametr (A^-1H)$. Since
$A operatornameadj(A) = (det A )I$, we have
$D det(A)(H) = operatornametr(operatornameadj(A) H)$.
Since both sides of the latter formula are continuous and the invertible
matrices are dense in the space of matrices, this formula is true for all $A$.
To obtain the other representation of the derivative, we use the inner product
$langle A , B rangle = operatornametr(A^TB)$.
Then $D det(A)(H) = langle operatornameadj(A)^T , H rangle$.
answered Jul 28 at 6:48
copper.hat
122k557156
Thank you! I understood your idea. So, using $langle A, B rangle = operatornametr(A^TB)$ and write $det(A+H) - det(H) = operatornametr(f(A)H) + det(H)$ and take $o(H) = det(H)$, I can also conclude the same thing, right?
– Lucas Corrêa
Jul 28 at 15:12
1
@LucasCorrêa: Well, there are two different approaches. The one in the answer is a general approach that works for all dimensions. The one in the question is a nice formula for the $2 times 2$ case, and you need to show that $det H$ is $o(|H|)$. If you look at my comment to the question you can see that in fact, $det H$ is $O(|H|^2)$ ('big' O). The formula in the question doesn't easily generalise to higher dimensions.
– copper.hat
Jul 28 at 15:16
1
Yes, I just asked you why I wanted to use the back of the exercise. But your response was very good! In fact, you gave me a much better proof than mine to $det'(A)H = det A operatornametr(A^-1H)$.
– Lucas Corrêa
Jul 28 at 15:21
add a comment |Â
up vote
1
down vote
accepted
Not exactly an answer, but may illuminate slightly.
I guess I would note that the functions $h mapsto det(A+h E_ij)$ are all
polynomials in $h$ and hence smooth. Hence the partial derivatives exist and are
continuous and so $det$ is differentiable.
To find a formula, we can start with special cases.
Note that $det(I+ t E_ij) = 1 + t delta_ij$ and so $D det(I) (E_ij) = delta_ij$.
Hence $D det (I)(H) = sum_ij H_ij det(I)(E_ij) = operatornametr H$.
Now suppose $A$ is invertible. Then $det(A+H) = det A det (I+ A^-1 H)$,
and it follows from this that
$D det(A)(H) = (det A )D det(I)(A^-1H) = det A operatornametr (A^-1H)$. Since
$A operatornameadj(A) = (det A )I$, we have
$D det(A)(H) = operatornametr(operatornameadj(A) H)$.
Since both sides of the latter formula are continuous and the invertible
matrices are dense in the space of matrices, this formula is true for all $A$.
To obtain the other representation of the derivative, we use the inner product
$langle A , B rangle = operatornametr(A^TB)$.
Then $D det(A)(H) = langle operatornameadj(A)^T , H rangle$.
answered Jul 28 at 6:48
copper.hat
122k557156
Thank you! I understood your idea. So, using $langle A, B rangle = operatornametr(A^TB)$ and write $det(A+H) - det(H) = operatornametr(f(A)H) + det(H)$ and take $o(H) = det(H)$, I can also conclude the same thing, right?
– Lucas Corrêa
Jul 28 at 15:12
1
@LucasCorrêa: Well, there are two different approaches. The one in the answer is a general approach that works for all dimensions. The one in the question is a nice formula for the $2 times 2$ case, and you need to show that $det H$ is $o(|H|)$. If you look at my comment to the question you can see that in fact, $det H$ is $O(|H|^2)$ ('big' O). The formula in the question doesn't easily generalise to higher dimensions.
– copper.hat
Jul 28 at 15:16
1
Yes, I just asked you why I wanted to use the back of the exercise. But your response was very good! In fact, you gave me a much better proof than mine to $det'(A)H = det A operatornametr(A^-1H)$.
– Lucas Corrêa
Jul 28 at 15:21
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Not exactly an answer, but may illuminate slightly.
I guess I would note that the functions $h mapsto det(A+h E_ij)$ are all
polynomials in $h$ and hence smooth. Hence the partial derivatives exist and are
continuous and so $det$ is differentiable.
To find a formula, we can start with special cases.
Note that $det(I+ t E_ij) = 1 + t delta_ij$ and so $D det(I) (E_ij) = delta_ij$.
Hence $D det (I)(H) = sum_ij H_ij det(I)(E_ij) = operatornametr H$.
Now suppose $A$ is invertible. Then $det(A+H) = det A det (I+ A^-1 H)$,
and it follows from this that
$D det(A)(H) = (det A )D det(I)(A^-1H) = det A operatornametr (A^-1H)$. Since
$A operatornameadj(A) = (det A )I$, we have
$D det(A)(H) = operatornametr(operatornameadj(A) H)$.
Since both sides of the latter formula are continuous and the invertible
matrices are dense in the space of matrices, this formula is true for all $A$.
To obtain the other representation of the derivative, we use the inner product
$langle A , B rangle = operatornametr(A^TB)$.
Then $D det(A)(H) = langle operatornameadj(A)^T , H rangle$.
answered Jul 28 at 6:48
copper.hat
122k557156
Not exactly an answer, but may illuminate slightly.
I guess I would note that the functions $h mapsto det(A+h E_ij)$ are all
polynomials in $h$ and hence smooth. Hence the partial derivatives exist and are
continuous and so $det$ is differentiable.
To find a formula, we can start with special cases.
Note that $det(I+ t E_ij) = 1 + t delta_ij$ and so $D det(I) (E_ij) = delta_ij$.
Hence $D det (I)(H) = sum_ij H_ij det(I)(E_ij) = operatornametr H$.
Now suppose $A$ is invertible. Then $det(A+H) = det A det (I+ A^-1 H)$,
and it follows from this that
$D det(A)(H) = (det A )D det(I)(A^-1H) = det A operatornametr (A^-1H)$. Since
$A operatornameadj(A) = (det A )I$, we have
$D det(A)(H) = operatornametr(operatornameadj(A) H)$.
Since both sides of the latter formula are continuous and the invertible
matrices are dense in the space of matrices, this formula is true for all $A$.
To obtain the other representation of the derivative, we use the inner product
$langle A , B rangle = operatornametr(A^TB)$.
Then $D det(A)(H) = langle operatornameadj(A)^T , H rangle$.
answered Jul 28 at 6:48
copper.hat
122k557156
answered Jul 28 at 6:48
copper.hat
122k557156
answered Jul 28 at 6:48
copper.hat
122k557156
answered Jul 28 at 6:48
copper.hat
122k557156
122k557156
Thank you! I understood your idea. So, using $langle A, B rangle = operatornametr(A^TB)$ and write $det(A+H) - det(H) = operatornametr(f(A)H) + det(H)$ and take $o(H) = det(H)$, I can also conclude the same thing, right?
– Lucas Corrêa
Jul 28 at 15:12
1
@LucasCorrêa: Well, there are two different approaches. The one in the answer is a general approach that works for all dimensions. The one in the question is a nice formula for the $2 times 2$ case, and you need to show that $det H$ is $o(|H|)$. If you look at my comment to the question you can see that in fact, $det H$ is $O(|H|^2)$ ('big' O). The formula in the question doesn't easily generalise to higher dimensions.
– copper.hat
Jul 28 at 15:16
1
Yes, I just asked you why I wanted to use the back of the exercise. But your response was very good! In fact, you gave me a much better proof than mine to $det'(A)H = det A operatornametr(A^-1H)$.
– Lucas Corrêa
Jul 28 at 15:21
add a comment |Â
Thank you! I understood your idea. So, using $langle A, B rangle = operatornametr(A^TB)$ and write $det(A+H) - det(H) = operatornametr(f(A)H) + det(H)$ and take $o(H) = det(H)$, I can also conclude the same thing, right?
– Lucas Corrêa
Jul 28 at 15:12
1
@LucasCorrêa: Well, there are two different approaches. The one in the answer is a general approach that works for all dimensions. The one in the question is a nice formula for the $2 times 2$ case, and you need to show that $det H$ is $o(|H|)$. If you look at my comment to the question you can see that in fact, $det H$ is $O(|H|^2)$ ('big' O). The formula in the question doesn't easily generalise to higher dimensions.
– copper.hat
Jul 28 at 15:16
1
Yes, I just asked you why I wanted to use the back of the exercise. But your response was very good! In fact, you gave me a much better proof than mine to $det'(A)H = det A operatornametr(A^-1H)$.
– Lucas Corrêa
Jul 28 at 15:21
Thank you! I understood your idea. So, using $langle A, B rangle = operatornametr(A^TB)$ and write $det(A+H) - det(H) = operatornametr(f(A)H) + det(H)$ and take $o(H) = det(H)$, I can also conclude the same thing, right?
– Lucas Corrêa
Jul 28 at 15:12
Thank you! I understood your idea. So, using $langle A, B rangle = operatornametr(A^TB)$ and write $det(A+H) - det(H) = operatornametr(f(A)H) + det(H)$ and take $o(H) = det(H)$, I can also conclude the same thing, right?
– Lucas Corrêa
Jul 28 at 15:12
1
1
@LucasCorrêa: Well, there are two different approaches. The one in the answer is a general approach that works for all dimensions. The one in the question is a nice formula for the $2 times 2$ case, and you need to show that $det H$ is $o(|H|)$. If you look at my comment to the question you can see that in fact, $det H$ is $O(|H|^2)$ ('big' O). The formula in the question doesn't easily generalise to higher dimensions.
– copper.hat
Jul 28 at 15:16
@LucasCorrêa: Well, there are two different approaches. The one in the answer is a general approach that works for all dimensions. The one in the question is a nice formula for the $2 times 2$ case, and you need to show that $det H$ is $o(|H|)$. If you look at my comment to the question you can see that in fact, $det H$ is $O(|H|^2)$ ('big' O). The formula in the question doesn't easily generalise to higher dimensions.
– copper.hat
Jul 28 at 15:16
1
1
Yes, I just asked you why I wanted to use the back of the exercise. But your response was very good! In fact, you gave me a much better proof than mine to $det'(A)H = det A operatornametr(A^-1H)$.
– Lucas Corrêa
Jul 28 at 15:21
Yes, I just asked you why I wanted to use the back of the exercise. But your response was very good! In fact, you gave me a much better proof than mine to $det'(A)H = det A operatornametr(A^-1H)$.
– Lucas Corrêa
Jul 28 at 15:21
add a comment |Â
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try expanding $H$ into a linear combination of elementary matrices
– Max
Jul 28 at 3:37
$|det H | le |H_11 H_22| + |H_12H_21| le 2 |H|^2$. (Cf. Hadamard's inequality.)
– copper.hat
Jul 28 at 5:47