Mixing
determinant of the sum of two matrices det(A+B)
$begingroup$
I have met with a formula, that
$$det(lambda^2A + lambda B +C) = lambda^{2n} det(A) + text{ lower order terms}.$$
Here, $lambda$ is a scalar and $A,B,C$ are $n times n$-matrices.
Can you help me prove it?
matrices determinant
edited Jan 21 at 3:42
JavaMan
11.1k12755
asked Jan 21 at 3:38
chen xi chen xi
82
$endgroup$
add a comment |
$begingroup$
I have met with a formula, that
$$det(lambda^2A + lambda B +C) = lambda^{2n} det(A) + text{ lower order terms}.$$
Here, $lambda$ is a scalar and $A,B,C$ are $n times n$-matrices.
Can you help me prove it?
matrices determinant
edited Jan 21 at 3:42
JavaMan
11.1k12755
asked Jan 21 at 3:38
chen xi chen xi
82
$endgroup$
$begingroup$
Do you mean lower order terms of $lambda$?
$endgroup$
– JavaMan
Jan 21 at 3:42
$begingroup$
Try induction on n
$endgroup$
– Dante Grevino
Jan 21 at 3:50
$begingroup$
Yes, lower order terms of $lambda$.
$endgroup$
– chen xi
Jan 21 at 3:51
add a comment |
$begingroup$
I have met with a formula, that
$$det(lambda^2A + lambda B +C) = lambda^{2n} det(A) + text{ lower order terms}.$$
Here, $lambda$ is a scalar and $A,B,C$ are $n times n$-matrices.
Can you help me prove it?
matrices determinant
edited Jan 21 at 3:42
JavaMan
11.1k12755
asked Jan 21 at 3:38
chen xi chen xi
82
$endgroup$
I have met with a formula, that
$$det(lambda^2A + lambda B +C) = lambda^{2n} det(A) + text{ lower order terms}.$$
Here, $lambda$ is a scalar and $A,B,C$ are $n times n$-matrices.
Can you help me prove it?
matrices determinant
matrices determinant
edited Jan 21 at 3:42
JavaMan
11.1k12755
asked Jan 21 at 3:38
chen xi chen xi
82
edited Jan 21 at 3:42
JavaMan
11.1k12755
asked Jan 21 at 3:38
chen xi chen xi
82
edited Jan 21 at 3:42
JavaMan
11.1k12755
edited Jan 21 at 3:42
JavaMan
11.1k12755
edited Jan 21 at 3:42
JavaMan
11.1k12755
11.1k12755
asked Jan 21 at 3:38
chen xi chen xi
82
asked Jan 21 at 3:38
chen xi chen xi
82
asked Jan 21 at 3:38
chen xi chen xi
82
82
$begingroup$
Do you mean lower order terms of $lambda$?
$endgroup$
– JavaMan
Jan 21 at 3:42
$begingroup$
Try induction on n
$endgroup$
– Dante Grevino
Jan 21 at 3:50
$begingroup$
Yes, lower order terms of $lambda$.
$endgroup$
– chen xi
Jan 21 at 3:51
add a comment |
$begingroup$
Do you mean lower order terms of $lambda$?
$endgroup$
– JavaMan
Jan 21 at 3:42
$begingroup$
Try induction on n
$endgroup$
– Dante Grevino
Jan 21 at 3:50
$begingroup$
Yes, lower order terms of $lambda$.
$endgroup$
– chen xi
Jan 21 at 3:51
$begingroup$
Do you mean lower order terms of $lambda$?
$endgroup$
– JavaMan
Jan 21 at 3:42
$begingroup$
Do you mean lower order terms of $lambda$?
$endgroup$
– JavaMan
Jan 21 at 3:42
$begingroup$
Try induction on n
$endgroup$
– Dante Grevino
Jan 21 at 3:50
$begingroup$
Try induction on n
$endgroup$
– Dante Grevino
Jan 21 at 3:50
$begingroup$
Yes, lower order terms of $lambda$.
$endgroup$
– chen xi
Jan 21 at 3:51
$begingroup$
Yes, lower order terms of $lambda$.
$endgroup$
– chen xi
Jan 21 at 3:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No better a time than now to learn about the exterior algebra perspective on determinants! Unless you prefer inductive proofs...
Let's view these matrices as linear maps $alpha, beta, gamma$ instead. In terms of the exterior algebra approach, we wish to show that $Lambda^n ( lambda^2 alpha + lambda beta + gamma)$ is $lambda^{2n} Lambda alpha$ up to lower order terms in $lambda$.
The determinant here will be $ (lambda^2 alpha + lambda beta + gamma )wedge cdots wedge (lambda^2 alpha + lambda beta + gamma)$. Now simply distribute, and collect into terms with the same degree of $lambda$. The term with the highest degree of $lambda$ will be the one where we have picked out all the $lambda^2 A$'s, and that's $Lambda^n lambda^2 alpha = lambda^{2n} Lambda^n alpha$, as desired.
answered Jan 21 at 4:02
Dean YoungDean Young
1,683721
$endgroup$
1
$begingroup$
Thanks a lot! I will learn about exterior algebra.
$endgroup$
– chen xi
Jan 21 at 4:53
$begingroup$
If that ends up being too much to learn for the time being, then try solving the problem using the fact that determinant is alternating multilinear in the rows. Then expand out the whole expression using multilinearlity in each row, pulling the $lambda$'s out as well. We get a large sum, which we sort into groups based on the degree of $lambda$. Then the top cofficient, the one corresponding to $lambda^{2n}$, will be $text{det}(A)$.
$endgroup$
– Dean Young
Jan 21 at 5:08
$begingroup$
This is a more comprehensive explanation. I would like to ask that $det(A+B）=det(A)+other terms$, is it right?
$endgroup$
– chen xi
Jan 21 at 5:13
$begingroup$
What would you like to ask?
$endgroup$
– Dean Young
Jan 21 at 5:14
$begingroup$
$det(A+B)=det(A)$+ other terms, is is right?
$endgroup$
– chen xi
Jan 21 at 5:18
|
show 1 more comment
$begingroup$
Hint: $ det(lambda A)=lambda^ndet (A) $ where $ Ain M_n(R) $ for some commutative ring $ R $.
answered Jan 21 at 4:01
user549397user549397
1,5061418
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No better a time than now to learn about the exterior algebra perspective on determinants! Unless you prefer inductive proofs...
Let's view these matrices as linear maps $alpha, beta, gamma$ instead. In terms of the exterior algebra approach, we wish to show that $Lambda^n ( lambda^2 alpha + lambda beta + gamma)$ is $lambda^{2n} Lambda alpha$ up to lower order terms in $lambda$.
The determinant here will be $ (lambda^2 alpha + lambda beta + gamma )wedge cdots wedge (lambda^2 alpha + lambda beta + gamma)$. Now simply distribute, and collect into terms with the same degree of $lambda$. The term with the highest degree of $lambda$ will be the one where we have picked out all the $lambda^2 A$'s, and that's $Lambda^n lambda^2 alpha = lambda^{2n} Lambda^n alpha$, as desired.
answered Jan 21 at 4:02
Dean YoungDean Young
1,683721
$endgroup$
1
$begingroup$
Thanks a lot! I will learn about exterior algebra.
$endgroup$
– chen xi
Jan 21 at 4:53
$begingroup$
If that ends up being too much to learn for the time being, then try solving the problem using the fact that determinant is alternating multilinear in the rows. Then expand out the whole expression using multilinearlity in each row, pulling the $lambda$'s out as well. We get a large sum, which we sort into groups based on the degree of $lambda$. Then the top cofficient, the one corresponding to $lambda^{2n}$, will be $text{det}(A)$.
$endgroup$
– Dean Young
Jan 21 at 5:08
$begingroup$
This is a more comprehensive explanation. I would like to ask that $det(A+B）=det(A)+other terms$, is it right?
$endgroup$
– chen xi
Jan 21 at 5:13
$begingroup$
What would you like to ask?
$endgroup$
– Dean Young
Jan 21 at 5:14
$begingroup$
$det(A+B)=det(A)$+ other terms, is is right?
$endgroup$
– chen xi
Jan 21 at 5:18
|
show 1 more comment
$begingroup$
No better a time than now to learn about the exterior algebra perspective on determinants! Unless you prefer inductive proofs...
Let's view these matrices as linear maps $alpha, beta, gamma$ instead. In terms of the exterior algebra approach, we wish to show that $Lambda^n ( lambda^2 alpha + lambda beta + gamma)$ is $lambda^{2n} Lambda alpha$ up to lower order terms in $lambda$.
The determinant here will be $ (lambda^2 alpha + lambda beta + gamma )wedge cdots wedge (lambda^2 alpha + lambda beta + gamma)$. Now simply distribute, and collect into terms with the same degree of $lambda$. The term with the highest degree of $lambda$ will be the one where we have picked out all the $lambda^2 A$'s, and that's $Lambda^n lambda^2 alpha = lambda^{2n} Lambda^n alpha$, as desired.
answered Jan 21 at 4:02
Dean YoungDean Young
1,683721
$endgroup$
1
$begingroup$
Thanks a lot! I will learn about exterior algebra.
$endgroup$
– chen xi
Jan 21 at 4:53
$begingroup$
If that ends up being too much to learn for the time being, then try solving the problem using the fact that determinant is alternating multilinear in the rows. Then expand out the whole expression using multilinearlity in each row, pulling the $lambda$'s out as well. We get a large sum, which we sort into groups based on the degree of $lambda$. Then the top cofficient, the one corresponding to $lambda^{2n}$, will be $text{det}(A)$.
$endgroup$
– Dean Young
Jan 21 at 5:08
$begingroup$
This is a more comprehensive explanation. I would like to ask that $det(A+B）=det(A)+other terms$, is it right?
$endgroup$
– chen xi
Jan 21 at 5:13
$begingroup$
What would you like to ask?
$endgroup$
– Dean Young
Jan 21 at 5:14
$begingroup$
$det(A+B)=det(A)$+ other terms, is is right?
$endgroup$
– chen xi
Jan 21 at 5:18
|
show 1 more comment
$begingroup$
No better a time than now to learn about the exterior algebra perspective on determinants! Unless you prefer inductive proofs...
Let's view these matrices as linear maps $alpha, beta, gamma$ instead. In terms of the exterior algebra approach, we wish to show that $Lambda^n ( lambda^2 alpha + lambda beta + gamma)$ is $lambda^{2n} Lambda alpha$ up to lower order terms in $lambda$.
The determinant here will be $ (lambda^2 alpha + lambda beta + gamma )wedge cdots wedge (lambda^2 alpha + lambda beta + gamma)$. Now simply distribute, and collect into terms with the same degree of $lambda$. The term with the highest degree of $lambda$ will be the one where we have picked out all the $lambda^2 A$'s, and that's $Lambda^n lambda^2 alpha = lambda^{2n} Lambda^n alpha$, as desired.
answered Jan 21 at 4:02
Dean YoungDean Young
1,683721
$endgroup$
No better a time than now to learn about the exterior algebra perspective on determinants! Unless you prefer inductive proofs...
Let's view these matrices as linear maps $alpha, beta, gamma$ instead. In terms of the exterior algebra approach, we wish to show that $Lambda^n ( lambda^2 alpha + lambda beta + gamma)$ is $lambda^{2n} Lambda alpha$ up to lower order terms in $lambda$.
The determinant here will be $ (lambda^2 alpha + lambda beta + gamma )wedge cdots wedge (lambda^2 alpha + lambda beta + gamma)$. Now simply distribute, and collect into terms with the same degree of $lambda$. The term with the highest degree of $lambda$ will be the one where we have picked out all the $lambda^2 A$'s, and that's $Lambda^n lambda^2 alpha = lambda^{2n} Lambda^n alpha$, as desired.
answered Jan 21 at 4:02
Dean YoungDean Young
1,683721
answered Jan 21 at 4:02
Dean YoungDean Young
1,683721
answered Jan 21 at 4:02
Dean YoungDean Young
1,683721
answered Jan 21 at 4:02
Dean YoungDean Young
1,683721
1,683721
1
$begingroup$
Thanks a lot! I will learn about exterior algebra.
$endgroup$
– chen xi
Jan 21 at 4:53
$begingroup$
If that ends up being too much to learn for the time being, then try solving the problem using the fact that determinant is alternating multilinear in the rows. Then expand out the whole expression using multilinearlity in each row, pulling the $lambda$'s out as well. We get a large sum, which we sort into groups based on the degree of $lambda$. Then the top cofficient, the one corresponding to $lambda^{2n}$, will be $text{det}(A)$.
$endgroup$
– Dean Young
Jan 21 at 5:08
$begingroup$
This is a more comprehensive explanation. I would like to ask that $det(A+B）=det(A)+other terms$, is it right?
$endgroup$
– chen xi
Jan 21 at 5:13
$begingroup$
What would you like to ask?
$endgroup$
– Dean Young
Jan 21 at 5:14
$begingroup$
$det(A+B)=det(A)$+ other terms, is is right?
$endgroup$
– chen xi
Jan 21 at 5:18
|
show 1 more comment
1
$begingroup$
Thanks a lot! I will learn about exterior algebra.
$endgroup$
– chen xi
Jan 21 at 4:53
$begingroup$
If that ends up being too much to learn for the time being, then try solving the problem using the fact that determinant is alternating multilinear in the rows. Then expand out the whole expression using multilinearlity in each row, pulling the $lambda$'s out as well. We get a large sum, which we sort into groups based on the degree of $lambda$. Then the top cofficient, the one corresponding to $lambda^{2n}$, will be $text{det}(A)$.
$endgroup$
– Dean Young
Jan 21 at 5:08
$begingroup$
This is a more comprehensive explanation. I would like to ask that $det(A+B）=det(A)+other terms$, is it right?
$endgroup$
– chen xi
Jan 21 at 5:13
$begingroup$
What would you like to ask?
$endgroup$
– Dean Young
Jan 21 at 5:14
$begingroup$
$det(A+B)=det(A)$+ other terms, is is right?
$endgroup$
– chen xi
Jan 21 at 5:18
1
1
$begingroup$
Thanks a lot! I will learn about exterior algebra.
$endgroup$
– chen xi
Jan 21 at 4:53
$begingroup$
Thanks a lot! I will learn about exterior algebra.
$endgroup$
– chen xi
Jan 21 at 4:53
$begingroup$
If that ends up being too much to learn for the time being, then try solving the problem using the fact that determinant is alternating multilinear in the rows. Then expand out the whole expression using multilinearlity in each row, pulling the $lambda$'s out as well. We get a large sum, which we sort into groups based on the degree of $lambda$. Then the top cofficient, the one corresponding to $lambda^{2n}$, will be $text{det}(A)$.
$endgroup$
– Dean Young
Jan 21 at 5:08
$begingroup$
If that ends up being too much to learn for the time being, then try solving the problem using the fact that determinant is alternating multilinear in the rows. Then expand out the whole expression using multilinearlity in each row, pulling the $lambda$'s out as well. We get a large sum, which we sort into groups based on the degree of $lambda$. Then the top cofficient, the one corresponding to $lambda^{2n}$, will be $text{det}(A)$.
$endgroup$
– Dean Young
Jan 21 at 5:08
$begingroup$
This is a more comprehensive explanation. I would like to ask that $det(A+B）=det(A)+other terms$, is it right?
$endgroup$
– chen xi
Jan 21 at 5:13
$begingroup$
This is a more comprehensive explanation. I would like to ask that $det(A+B）=det(A)+other terms$, is it right?
$endgroup$
– chen xi
Jan 21 at 5:13
$begingroup$
What would you like to ask?
$endgroup$
– Dean Young
Jan 21 at 5:14
$begingroup$
What would you like to ask?
$endgroup$
– Dean Young
Jan 21 at 5:14
$begingroup$
$det(A+B)=det(A)$+ other terms, is is right?
$endgroup$
– chen xi
Jan 21 at 5:18
$begingroup$
$det(A+B)=det(A)$+ other terms, is is right?
$endgroup$
– chen xi
Jan 21 at 5:18
|
show 1 more comment
$begingroup$
Hint: $ det(lambda A)=lambda^ndet (A) $ where $ Ain M_n(R) $ for some commutative ring $ R $.
answered Jan 21 at 4:01
user549397user549397
1,5061418
$endgroup$
add a comment |
$begingroup$
Hint: $ det(lambda A)=lambda^ndet (A) $ where $ Ain M_n(R) $ for some commutative ring $ R $.
answered Jan 21 at 4:01
user549397user549397
1,5061418
$endgroup$
add a comment |
$begingroup$
Hint: $ det(lambda A)=lambda^ndet (A) $ where $ Ain M_n(R) $ for some commutative ring $ R $.
answered Jan 21 at 4:01
user549397user549397
1,5061418
$endgroup$
Hint: $ det(lambda A)=lambda^ndet (A) $ where $ Ain M_n(R) $ for some commutative ring $ R $.
answered Jan 21 at 4:01
user549397user549397
1,5061418
answered Jan 21 at 4:01
user549397user549397
1,5061418
answered Jan 21 at 4:01
user549397user549397
1,5061418
answered Jan 21 at 4:01
user549397user549397
1,5061418
1,5061418
add a comment |
add a comment |
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$begingroup$
Do you mean lower order terms of $lambda$?
$endgroup$
– JavaMan
Jan 21 at 3:42
$begingroup$
Try induction on n
$endgroup$
– Dante Grevino
Jan 21 at 3:50
$begingroup$
Yes, lower order terms of $lambda$.
$endgroup$
– chen xi
Jan 21 at 3:51