is $det(A^2 + I)$ always non negative?
Obviously $det(A^2)$ is (casework), but is the above matrix non-negative? $det(A)det(A) geq 0$ as $det(A) > 0$ or $det(A) < 0$ yields positive when squared. However, I am not sure that when adding the identity matrix that it is also positive.
linear-algebra determinant
edited Dec 11 '18 at 11:14
Widawensen
4,42121445
add a comment |
Obviously $det(A^2)$ is (casework), but is the above matrix non-negative? $det(A)det(A) geq 0$ as $det(A) > 0$ or $det(A) < 0$ yields positive when squared. However, I am not sure that when adding the identity matrix that it is also positive.
linear-algebra determinant
edited Dec 11 '18 at 11:14
Widawensen
4,42121445
add a comment |
Obviously $det(A^2)$ is (casework), but is the above matrix non-negative? $det(A)det(A) geq 0$ as $det(A) > 0$ or $det(A) < 0$ yields positive when squared. However, I am not sure that when adding the identity matrix that it is also positive.
linear-algebra determinant
edited Dec 11 '18 at 11:14
Widawensen
4,42121445
Obviously $det(A^2)$ is (casework), but is the above matrix non-negative? $det(A)det(A) geq 0$ as $det(A) > 0$ or $det(A) < 0$ yields positive when squared. However, I am not sure that when adding the identity matrix that it is also positive.
linear-algebra determinant
linear-algebra determinant
edited Dec 11 '18 at 11:14
Widawensen
4,42121445
edited Dec 11 '18 at 11:14
Widawensen
4,42121445
edited Dec 11 '18 at 11:14
Widawensen
4,42121445
edited Dec 11 '18 at 11:14
Widawensen
4,42121445
edited Dec 11 '18 at 11:14
Widawensen
4,42121445
4,42121445
asked Dec 9 '18 at 22:25
user624697
add a comment |
add a comment |
2 Answers
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I assume $A$ is a real matrix; then $iA$ is purely imaginary, and so
$overline{iA} = -iA; tag 1$
then
$det(I + A^2) = det ((I + iA)(I - iA)) = det(I + iA) det(I - iA)$
$= det(I + iA)det(overline{I + iA)} = det(I + iA) overline{det(I + iA)} ge 0. tag 2$
answered Dec 9 '18 at 22:42
Robert Lewis
43.7k22963
1
Excellent solution and very compact. Especially step $det(overline{M})=overline{det(M)}$ is very fast.
– Widawensen
Dec 11 '18 at 15:16
@Widawensen: thank you sir!
– Robert Lewis
Dec 11 '18 at 16:50
add a comment |
It is possible also to use eigenvalues to prove the claim.
Determinant is the product of eigenvalues and eigenvalues of the polynomial are polynomials of eigenvalues.
Denote $B=A^2+I$ and eigenvalues of $A$ as $r_{Aj}$ when they are real, and $c_{Ak}=a_k+b_ki$ when they are complex with non-zero imaginary part.
As we know for real matrices complex eigenvalues come in conjugate pairs $c_{Ak}=a_k+b_ki,c_{Ak*}=a_k-b_ki$.
Then eigenvalues of $B$ are $r_{Bj}^2+1$ ( which are obviously positive, even $ge 1$) for real eigenvalues, and $c_{Bk}=(a_k+b_ki)^2+1,c_{Bk*}=(a_k-b_ki)^2+1$ for complex ones.
$$c_{Bk}=a_k^2-b_k^2+1 + 2a_kb_ki$$
$$c_{Bk*}=a_k^2-b_k^2+1 - 2a_kb_ki$$
We have received once again a pair of conjugated complex numbers.
The product of conjugated complex numbers is non-negative number ( $cc^*=vert cvert^2$) and the whole product is non-negative.
Generally we can say even more: the determinant is mostly positive, except the case when matrix $A$ has $pm i $ eigenvalues (in that case determinant is $0$, $B=A^2+I$ is singular).
answered Dec 11 '18 at 11:10
Widawensen
4,42121445
With an argumentation similar to the presented one we could extend the claim even to the form that determinant of $A^{2m}+alpha I$, where $alphage 0$, is non-negative. It's worth to notice that even in the case of negative $alpha$ the sign of determinant depends solely on real eigenvalues of $A$ not the complex ones as they always come in conjugated pairs.
– Widawensen
Dec 11 '18 at 11:50
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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I assume $A$ is a real matrix; then $iA$ is purely imaginary, and so
$overline{iA} = -iA; tag 1$
then
$det(I + A^2) = det ((I + iA)(I - iA)) = det(I + iA) det(I - iA)$
$= det(I + iA)det(overline{I + iA)} = det(I + iA) overline{det(I + iA)} ge 0. tag 2$
answered Dec 9 '18 at 22:42
Robert Lewis
43.7k22963
1
Excellent solution and very compact. Especially step $det(overline{M})=overline{det(M)}$ is very fast.
– Widawensen
Dec 11 '18 at 15:16
@Widawensen: thank you sir!
– Robert Lewis
Dec 11 '18 at 16:50
add a comment |
I assume $A$ is a real matrix; then $iA$ is purely imaginary, and so
$overline{iA} = -iA; tag 1$
then
$det(I + A^2) = det ((I + iA)(I - iA)) = det(I + iA) det(I - iA)$
$= det(I + iA)det(overline{I + iA)} = det(I + iA) overline{det(I + iA)} ge 0. tag 2$
answered Dec 9 '18 at 22:42
Robert Lewis
43.7k22963
1
Excellent solution and very compact. Especially step $det(overline{M})=overline{det(M)}$ is very fast.
– Widawensen
Dec 11 '18 at 15:16
@Widawensen: thank you sir!
– Robert Lewis
Dec 11 '18 at 16:50
add a comment |
I assume $A$ is a real matrix; then $iA$ is purely imaginary, and so
$overline{iA} = -iA; tag 1$
then
$det(I + A^2) = det ((I + iA)(I - iA)) = det(I + iA) det(I - iA)$
$= det(I + iA)det(overline{I + iA)} = det(I + iA) overline{det(I + iA)} ge 0. tag 2$
answered Dec 9 '18 at 22:42
Robert Lewis
43.7k22963
I assume $A$ is a real matrix; then $iA$ is purely imaginary, and so
$overline{iA} = -iA; tag 1$
then
$det(I + A^2) = det ((I + iA)(I - iA)) = det(I + iA) det(I - iA)$
$= det(I + iA)det(overline{I + iA)} = det(I + iA) overline{det(I + iA)} ge 0. tag 2$
answered Dec 9 '18 at 22:42
Robert Lewis
43.7k22963
answered Dec 9 '18 at 22:42
Robert Lewis
43.7k22963
answered Dec 9 '18 at 22:42
Robert Lewis
43.7k22963
answered Dec 9 '18 at 22:42
Robert Lewis
43.7k22963
43.7k22963
1
Excellent solution and very compact. Especially step $det(overline{M})=overline{det(M)}$ is very fast.
– Widawensen
Dec 11 '18 at 15:16
@Widawensen: thank you sir!
– Robert Lewis
Dec 11 '18 at 16:50
add a comment |
1
Excellent solution and very compact. Especially step $det(overline{M})=overline{det(M)}$ is very fast.
– Widawensen
Dec 11 '18 at 15:16
@Widawensen: thank you sir!
– Robert Lewis
Dec 11 '18 at 16:50
1
1
Excellent solution and very compact. Especially step $det(overline{M})=overline{det(M)}$ is very fast.
– Widawensen
Dec 11 '18 at 15:16
Excellent solution and very compact. Especially step $det(overline{M})=overline{det(M)}$ is very fast.
– Widawensen
Dec 11 '18 at 15:16
@Widawensen: thank you sir!
– Robert Lewis
Dec 11 '18 at 16:50
@Widawensen: thank you sir!
– Robert Lewis
Dec 11 '18 at 16:50
add a comment |
It is possible also to use eigenvalues to prove the claim.
Determinant is the product of eigenvalues and eigenvalues of the polynomial are polynomials of eigenvalues.
Denote $B=A^2+I$ and eigenvalues of $A$ as $r_{Aj}$ when they are real, and $c_{Ak}=a_k+b_ki$ when they are complex with non-zero imaginary part.
As we know for real matrices complex eigenvalues come in conjugate pairs $c_{Ak}=a_k+b_ki,c_{Ak*}=a_k-b_ki$.
Then eigenvalues of $B$ are $r_{Bj}^2+1$ ( which are obviously positive, even $ge 1$) for real eigenvalues, and $c_{Bk}=(a_k+b_ki)^2+1,c_{Bk*}=(a_k-b_ki)^2+1$ for complex ones.
$$c_{Bk}=a_k^2-b_k^2+1 + 2a_kb_ki$$
$$c_{Bk*}=a_k^2-b_k^2+1 - 2a_kb_ki$$
We have received once again a pair of conjugated complex numbers.
The product of conjugated complex numbers is non-negative number ( $cc^*=vert cvert^2$) and the whole product is non-negative.
Generally we can say even more: the determinant is mostly positive, except the case when matrix $A$ has $pm i $ eigenvalues (in that case determinant is $0$, $B=A^2+I$ is singular).
answered Dec 11 '18 at 11:10
Widawensen
4,42121445
With an argumentation similar to the presented one we could extend the claim even to the form that determinant of $A^{2m}+alpha I$, where $alphage 0$, is non-negative. It's worth to notice that even in the case of negative $alpha$ the sign of determinant depends solely on real eigenvalues of $A$ not the complex ones as they always come in conjugated pairs.
– Widawensen
Dec 11 '18 at 11:50
add a comment |
It is possible also to use eigenvalues to prove the claim.
Determinant is the product of eigenvalues and eigenvalues of the polynomial are polynomials of eigenvalues.
Denote $B=A^2+I$ and eigenvalues of $A$ as $r_{Aj}$ when they are real, and $c_{Ak}=a_k+b_ki$ when they are complex with non-zero imaginary part.
As we know for real matrices complex eigenvalues come in conjugate pairs $c_{Ak}=a_k+b_ki,c_{Ak*}=a_k-b_ki$.
Then eigenvalues of $B$ are $r_{Bj}^2+1$ ( which are obviously positive, even $ge 1$) for real eigenvalues, and $c_{Bk}=(a_k+b_ki)^2+1,c_{Bk*}=(a_k-b_ki)^2+1$ for complex ones.
$$c_{Bk}=a_k^2-b_k^2+1 + 2a_kb_ki$$
$$c_{Bk*}=a_k^2-b_k^2+1 - 2a_kb_ki$$
We have received once again a pair of conjugated complex numbers.
The product of conjugated complex numbers is non-negative number ( $cc^*=vert cvert^2$) and the whole product is non-negative.
Generally we can say even more: the determinant is mostly positive, except the case when matrix $A$ has $pm i $ eigenvalues (in that case determinant is $0$, $B=A^2+I$ is singular).
answered Dec 11 '18 at 11:10
Widawensen
4,42121445
With an argumentation similar to the presented one we could extend the claim even to the form that determinant of $A^{2m}+alpha I$, where $alphage 0$, is non-negative. It's worth to notice that even in the case of negative $alpha$ the sign of determinant depends solely on real eigenvalues of $A$ not the complex ones as they always come in conjugated pairs.
– Widawensen
Dec 11 '18 at 11:50
add a comment |
It is possible also to use eigenvalues to prove the claim.
Determinant is the product of eigenvalues and eigenvalues of the polynomial are polynomials of eigenvalues.
Denote $B=A^2+I$ and eigenvalues of $A$ as $r_{Aj}$ when they are real, and $c_{Ak}=a_k+b_ki$ when they are complex with non-zero imaginary part.
As we know for real matrices complex eigenvalues come in conjugate pairs $c_{Ak}=a_k+b_ki,c_{Ak*}=a_k-b_ki$.
Then eigenvalues of $B$ are $r_{Bj}^2+1$ ( which are obviously positive, even $ge 1$) for real eigenvalues, and $c_{Bk}=(a_k+b_ki)^2+1,c_{Bk*}=(a_k-b_ki)^2+1$ for complex ones.
$$c_{Bk}=a_k^2-b_k^2+1 + 2a_kb_ki$$
$$c_{Bk*}=a_k^2-b_k^2+1 - 2a_kb_ki$$
We have received once again a pair of conjugated complex numbers.
The product of conjugated complex numbers is non-negative number ( $cc^*=vert cvert^2$) and the whole product is non-negative.
Generally we can say even more: the determinant is mostly positive, except the case when matrix $A$ has $pm i $ eigenvalues (in that case determinant is $0$, $B=A^2+I$ is singular).
answered Dec 11 '18 at 11:10
Widawensen
4,42121445
It is possible also to use eigenvalues to prove the claim.
Determinant is the product of eigenvalues and eigenvalues of the polynomial are polynomials of eigenvalues.
Denote $B=A^2+I$ and eigenvalues of $A$ as $r_{Aj}$ when they are real, and $c_{Ak}=a_k+b_ki$ when they are complex with non-zero imaginary part.
As we know for real matrices complex eigenvalues come in conjugate pairs $c_{Ak}=a_k+b_ki,c_{Ak*}=a_k-b_ki$.
Then eigenvalues of $B$ are $r_{Bj}^2+1$ ( which are obviously positive, even $ge 1$) for real eigenvalues, and $c_{Bk}=(a_k+b_ki)^2+1,c_{Bk*}=(a_k-b_ki)^2+1$ for complex ones.
$$c_{Bk}=a_k^2-b_k^2+1 + 2a_kb_ki$$
$$c_{Bk*}=a_k^2-b_k^2+1 - 2a_kb_ki$$
We have received once again a pair of conjugated complex numbers.
The product of conjugated complex numbers is non-negative number ( $cc^*=vert cvert^2$) and the whole product is non-negative.
Generally we can say even more: the determinant is mostly positive, except the case when matrix $A$ has $pm i $ eigenvalues (in that case determinant is $0$, $B=A^2+I$ is singular).
answered Dec 11 '18 at 11:10
Widawensen
4,42121445
edited Dec 11 '18 at 14:28
answered Dec 11 '18 at 11:10
Widawensen
4,42121445
answered Dec 11 '18 at 11:10
Widawensen
4,42121445
answered Dec 11 '18 at 11:10
Widawensen
4,42121445
4,42121445
With an argumentation similar to the presented one we could extend the claim even to the form that determinant of $A^{2m}+alpha I$, where $alphage 0$, is non-negative. It's worth to notice that even in the case of negative $alpha$ the sign of determinant depends solely on real eigenvalues of $A$ not the complex ones as they always come in conjugated pairs.
– Widawensen
Dec 11 '18 at 11:50
add a comment |
With an argumentation similar to the presented one we could extend the claim even to the form that determinant of $A^{2m}+alpha I$, where $alphage 0$, is non-negative. It's worth to notice that even in the case of negative $alpha$ the sign of determinant depends solely on real eigenvalues of $A$ not the complex ones as they always come in conjugated pairs.
– Widawensen
Dec 11 '18 at 11:50
With an argumentation similar to the presented one we could extend the claim even to the form that determinant of $A^{2m}+alpha I$, where $alphage 0$, is non-negative. It's worth to notice that even in the case of negative $alpha$ the sign of determinant depends solely on real eigenvalues of $A$ not the complex ones as they always come in conjugated pairs.
– Widawensen
Dec 11 '18 at 11:50
With an argumentation similar to the presented one we could extend the claim even to the form that determinant of $A^{2m}+alpha I$, where $alphage 0$, is non-negative. It's worth to notice that even in the case of negative $alpha$ the sign of determinant depends solely on real eigenvalues of $A$ not the complex ones as they always come in conjugated pairs.
– Widawensen
Dec 11 '18 at 11:50
add a comment |
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