How to prove this matrix inequality $det(A+B)ge 2^nsqrt{det(A)det(B)}$
$begingroup$
Question:
Let $A_{ntimes n}$ and $B_{ntimes n}$ be positive Hermitian matrices.
Show that
$$det(A+B)ge 2^nsqrt{det(A)det(B)}.$$
I know that
$$det(A+B)ge det(A)+det(B)$$
But my problem is that I can't,(maybe this is an old reslut,and also I can't find it),
Thank you very much!
linear-algebra matrices inequality
edited Dec 20 '18 at 9:05
dmtri
1,5042521
asked Dec 5 '13 at 10:44
china mathchina math
10.2k631117
$endgroup$
add a comment |
$begingroup$
Question:
Let $A_{ntimes n}$ and $B_{ntimes n}$ be positive Hermitian matrices.
Show that
$$det(A+B)ge 2^nsqrt{det(A)det(B)}.$$
I know that
$$det(A+B)ge det(A)+det(B)$$
But my problem is that I can't,(maybe this is an old reslut,and also I can't find it),
Thank you very much!
linear-algebra matrices inequality
edited Dec 20 '18 at 9:05
dmtri
1,5042521
asked Dec 5 '13 at 10:44
china mathchina math
10.2k631117
$endgroup$
add a comment |
$begingroup$
Question:
Let $A_{ntimes n}$ and $B_{ntimes n}$ be positive Hermitian matrices.
Show that
$$det(A+B)ge 2^nsqrt{det(A)det(B)}.$$
I know that
$$det(A+B)ge det(A)+det(B)$$
But my problem is that I can't,(maybe this is an old reslut,and also I can't find it),
Thank you very much!
linear-algebra matrices inequality
edited Dec 20 '18 at 9:05
dmtri
1,5042521
asked Dec 5 '13 at 10:44
china mathchina math
10.2k631117
$endgroup$
Question:
Let $A_{ntimes n}$ and $B_{ntimes n}$ be positive Hermitian matrices.
Show that
$$det(A+B)ge 2^nsqrt{det(A)det(B)}.$$
I know that
$$det(A+B)ge det(A)+det(B)$$
But my problem is that I can't,(maybe this is an old reslut,and also I can't find it),
Thank you very much!
linear-algebra matrices inequality
linear-algebra matrices inequality
edited Dec 20 '18 at 9:05
dmtri
1,5042521
asked Dec 5 '13 at 10:44
china mathchina math
10.2k631117
edited Dec 20 '18 at 9:05
dmtri
1,5042521
asked Dec 5 '13 at 10:44
china mathchina math
10.2k631117
edited Dec 20 '18 at 9:05
dmtri
1,5042521
edited Dec 20 '18 at 9:05
dmtri
1,5042521
edited Dec 20 '18 at 9:05
dmtri
1,5042521
1,5042521
asked Dec 5 '13 at 10:44
china mathchina math
10.2k631117
asked Dec 5 '13 at 10:44
china mathchina math
10.2k631117
asked Dec 5 '13 at 10:44
china mathchina math
10.2k631117
10.2k631117
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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This is a corollary of Minkowski's Determinant Theorem: $det(A+B)^frac{1}{n}geq det(A)^frac{1}{n}+det(B)^frac{1}{n}.$ Apply AM-GM inequality to the right-hand side.
answered Dec 5 '13 at 10:59
CasteelsCasteels
9,99742234
$endgroup$
$begingroup$
this can AM-GM inequality? Thank you
$endgroup$
– china math
Dec 5 '13 at 11:02
$begingroup$
Sorry, I am not sure what you are asking?
$endgroup$
– Casteels
Dec 5 '13 at 11:03
add a comment |
$begingroup$
This seems to be a bit late for an answer, but the following theorem is proven in "Linear Algebra" by Lax in chapter 10.
Let $A$ and $B$ be self-adjoint, positive, $n times n$ matrices. Then for all $0<t<1,$
begin{align}
det(tA + (1-t)B) geq (det A)^{t}(det B)^{1-t}.
end{align}
Your answer follows with $t = frac{1}{2}$.
answered Apr 15 '14 at 0:48
DRichDRich
45438
$endgroup$
$begingroup$
It's never too late for a good answer!
$endgroup$
– Casteels
Apr 15 '14 at 7:42
$begingroup$
Can you send me pdf document or links ? Thanks you very much!
$endgroup$
– Road Human
Apr 6 '15 at 16:07
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a corollary of Minkowski's Determinant Theorem: $det(A+B)^frac{1}{n}geq det(A)^frac{1}{n}+det(B)^frac{1}{n}.$ Apply AM-GM inequality to the right-hand side.
answered Dec 5 '13 at 10:59
CasteelsCasteels
9,99742234
$endgroup$
$begingroup$
this can AM-GM inequality? Thank you
$endgroup$
– china math
Dec 5 '13 at 11:02
$begingroup$
Sorry, I am not sure what you are asking?
$endgroup$
– Casteels
Dec 5 '13 at 11:03
add a comment |
$begingroup$
This is a corollary of Minkowski's Determinant Theorem: $det(A+B)^frac{1}{n}geq det(A)^frac{1}{n}+det(B)^frac{1}{n}.$ Apply AM-GM inequality to the right-hand side.
answered Dec 5 '13 at 10:59
CasteelsCasteels
9,99742234
$endgroup$
$begingroup$
this can AM-GM inequality? Thank you
$endgroup$
– china math
Dec 5 '13 at 11:02
$begingroup$
Sorry, I am not sure what you are asking?
$endgroup$
– Casteels
Dec 5 '13 at 11:03
add a comment |
$begingroup$
This is a corollary of Minkowski's Determinant Theorem: $det(A+B)^frac{1}{n}geq det(A)^frac{1}{n}+det(B)^frac{1}{n}.$ Apply AM-GM inequality to the right-hand side.
answered Dec 5 '13 at 10:59
CasteelsCasteels
9,99742234
$endgroup$
This is a corollary of Minkowski's Determinant Theorem: $det(A+B)^frac{1}{n}geq det(A)^frac{1}{n}+det(B)^frac{1}{n}.$ Apply AM-GM inequality to the right-hand side.
answered Dec 5 '13 at 10:59
CasteelsCasteels
9,99742234
answered Dec 5 '13 at 10:59
CasteelsCasteels
9,99742234
answered Dec 5 '13 at 10:59
CasteelsCasteels
9,99742234
answered Dec 5 '13 at 10:59
CasteelsCasteels
9,99742234
9,99742234
$begingroup$
this can AM-GM inequality? Thank you
$endgroup$
– china math
Dec 5 '13 at 11:02
$begingroup$
Sorry, I am not sure what you are asking?
$endgroup$
– Casteels
Dec 5 '13 at 11:03
add a comment |
$begingroup$
this can AM-GM inequality? Thank you
$endgroup$
– china math
Dec 5 '13 at 11:02
$begingroup$
Sorry, I am not sure what you are asking?
$endgroup$
– Casteels
Dec 5 '13 at 11:03
$begingroup$
this can AM-GM inequality? Thank you
$endgroup$
– china math
Dec 5 '13 at 11:02
$begingroup$
this can AM-GM inequality? Thank you
$endgroup$
– china math
Dec 5 '13 at 11:02
$begingroup$
Sorry, I am not sure what you are asking?
$endgroup$
– Casteels
Dec 5 '13 at 11:03
$begingroup$
Sorry, I am not sure what you are asking?
$endgroup$
– Casteels
Dec 5 '13 at 11:03
add a comment |
$begingroup$
This seems to be a bit late for an answer, but the following theorem is proven in "Linear Algebra" by Lax in chapter 10.
Let $A$ and $B$ be self-adjoint, positive, $n times n$ matrices. Then for all $0<t<1,$
begin{align}
det(tA + (1-t)B) geq (det A)^{t}(det B)^{1-t}.
end{align}
Your answer follows with $t = frac{1}{2}$.
answered Apr 15 '14 at 0:48
DRichDRich
45438
$endgroup$
$begingroup$
It's never too late for a good answer!
$endgroup$
– Casteels
Apr 15 '14 at 7:42
$begingroup$
Can you send me pdf document or links ? Thanks you very much!
$endgroup$
– Road Human
Apr 6 '15 at 16:07
add a comment |
$begingroup$
This seems to be a bit late for an answer, but the following theorem is proven in "Linear Algebra" by Lax in chapter 10.
Let $A$ and $B$ be self-adjoint, positive, $n times n$ matrices. Then for all $0<t<1,$
begin{align}
det(tA + (1-t)B) geq (det A)^{t}(det B)^{1-t}.
end{align}
Your answer follows with $t = frac{1}{2}$.
answered Apr 15 '14 at 0:48
DRichDRich
45438
$endgroup$
$begingroup$
It's never too late for a good answer!
$endgroup$
– Casteels
Apr 15 '14 at 7:42
$begingroup$
Can you send me pdf document or links ? Thanks you very much!
$endgroup$
– Road Human
Apr 6 '15 at 16:07
add a comment |
$begingroup$
This seems to be a bit late for an answer, but the following theorem is proven in "Linear Algebra" by Lax in chapter 10.
Let $A$ and $B$ be self-adjoint, positive, $n times n$ matrices. Then for all $0<t<1,$
begin{align}
det(tA + (1-t)B) geq (det A)^{t}(det B)^{1-t}.
end{align}
Your answer follows with $t = frac{1}{2}$.
answered Apr 15 '14 at 0:48
DRichDRich
45438
$endgroup$
This seems to be a bit late for an answer, but the following theorem is proven in "Linear Algebra" by Lax in chapter 10.
Let $A$ and $B$ be self-adjoint, positive, $n times n$ matrices. Then for all $0<t<1,$
begin{align}
det(tA + (1-t)B) geq (det A)^{t}(det B)^{1-t}.
end{align}
Your answer follows with $t = frac{1}{2}$.
answered Apr 15 '14 at 0:48
DRichDRich
45438
answered Apr 15 '14 at 0:48
DRichDRich
45438
answered Apr 15 '14 at 0:48
DRichDRich
45438
answered Apr 15 '14 at 0:48
DRichDRich
45438
45438
$begingroup$
It's never too late for a good answer!
$endgroup$
– Casteels
Apr 15 '14 at 7:42
$begingroup$
Can you send me pdf document or links ? Thanks you very much!
$endgroup$
– Road Human
Apr 6 '15 at 16:07
add a comment |
$begingroup$
It's never too late for a good answer!
$endgroup$
– Casteels
Apr 15 '14 at 7:42
$begingroup$
Can you send me pdf document or links ? Thanks you very much!
$endgroup$
– Road Human
Apr 6 '15 at 16:07
$begingroup$
It's never too late for a good answer!
$endgroup$
– Casteels
Apr 15 '14 at 7:42
$begingroup$
It's never too late for a good answer!
$endgroup$
– Casteels
Apr 15 '14 at 7:42
$begingroup$
Can you send me pdf document or links ? Thanks you very much!
$endgroup$
– Road Human
Apr 6 '15 at 16:07
$begingroup$
Can you send me pdf document or links ? Thanks you very much!
$endgroup$
– Road Human
Apr 6 '15 at 16:07
add a comment |
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