A Matrix Inequality for positive definite matrices
9
$begingroup$
Let $X$ and $Y$ be positive semi-definite self-adjoint complex matrices of same finite order. The, is it true that $|X-Y|leq X+Y$ where for any matrix $A$, $|A|$ is defined to be $|A|:=(A^*A)^{frac{1}{2}}$ ?
PS. I think the answer is No. But I could not find any counterexample!
linear-algebra matrices inequalities operator-theory linear-matrix-inequalities
edited Apr 4 at 20:10
Ali Taghavi
20852085
asked Apr 2 at 9:31
Samya Kumar RaySamya Kumar Ray
614
$endgroup$
add a comment |
9
$begingroup$
Let $X$ and $Y$ be positive semi-definite self-adjoint complex matrices of same finite order. The, is it true that $|X-Y|leq X+Y$ where for any matrix $A$, $|A|$ is defined to be $|A|:=(A^*A)^{frac{1}{2}}$ ?
PS. I think the answer is No. But I could not find any counterexample!
linear-algebra matrices inequalities operator-theory linear-matrix-inequalities
edited Apr 4 at 20:10
Ali Taghavi
20852085
asked Apr 2 at 9:31
Samya Kumar RaySamya Kumar Ray
614
$endgroup$
1
$begingroup$
Consider $X,Y$ to be of trace 1 and their difference to be block diagonal $sigma^x,sigma^x$. Now $|X-Y|=$ identity matrix. Take the trace of both sides, the LHS gives 4 the RHS is 2.
$endgroup$
– lcv
Apr 2 at 9:56
$begingroup$
@lcv. can you please elaborate your notations?
$endgroup$
– Samya Kumar Ray
Apr 2 at 11:16
add a comment |
9
9
9
$begingroup$
Let $X$ and $Y$ be positive semi-definite self-adjoint complex matrices of same finite order. The, is it true that $|X-Y|leq X+Y$ where for any matrix $A$, $|A|$ is defined to be $|A|:=(A^*A)^{frac{1}{2}}$ ?
PS. I think the answer is No. But I could not find any counterexample!
linear-algebra matrices inequalities operator-theory linear-matrix-inequalities
edited Apr 4 at 20:10
Ali Taghavi
20852085
asked Apr 2 at 9:31
Samya Kumar RaySamya Kumar Ray
614
$endgroup$
Let $X$ and $Y$ be positive semi-definite self-adjoint complex matrices of same finite order. The, is it true that $|X-Y|leq X+Y$ where for any matrix $A$, $|A|$ is defined to be $|A|:=(A^*A)^{frac{1}{2}}$ ?
PS. I think the answer is No. But I could not find any counterexample!
linear-algebra matrices inequalities operator-theory linear-matrix-inequalities
linear-algebra matrices inequalities operator-theory linear-matrix-inequalities
edited Apr 4 at 20:10
Ali Taghavi
20852085
asked Apr 2 at 9:31
Samya Kumar RaySamya Kumar Ray
614
edited Apr 4 at 20:10
Ali Taghavi
20852085
asked Apr 2 at 9:31
Samya Kumar RaySamya Kumar Ray
614
edited Apr 4 at 20:10
Ali Taghavi
20852085
edited Apr 4 at 20:10
Ali Taghavi
20852085
edited Apr 4 at 20:10
Ali Taghavi
20852085
20852085
asked Apr 2 at 9:31
Samya Kumar RaySamya Kumar Ray
614
asked Apr 2 at 9:31
Samya Kumar RaySamya Kumar Ray
614
asked Apr 2 at 9:31
Samya Kumar RaySamya Kumar Ray
614
614
1
$begingroup$
Consider $X,Y$ to be of trace 1 and their difference to be block diagonal $sigma^x,sigma^x$. Now $|X-Y|=$ identity matrix. Take the trace of both sides, the LHS gives 4 the RHS is 2.
$endgroup$
– lcv
Apr 2 at 9:56
$begingroup$
@lcv. can you please elaborate your notations?
$endgroup$
– Samya Kumar Ray
Apr 2 at 11:16
add a comment |
1
$begingroup$
Consider $X,Y$ to be of trace 1 and their difference to be block diagonal $sigma^x,sigma^x$. Now $|X-Y|=$ identity matrix. Take the trace of both sides, the LHS gives 4 the RHS is 2.
$endgroup$
– lcv
Apr 2 at 9:56
$begingroup$
@lcv. can you please elaborate your notations?
$endgroup$
– Samya Kumar Ray
Apr 2 at 11:16
1
1
$begingroup$
Consider $X,Y$ to be of trace 1 and their difference to be block diagonal $sigma^x,sigma^x$. Now $|X-Y|=$ identity matrix. Take the trace of both sides, the LHS gives 4 the RHS is 2.
$endgroup$
– lcv
Apr 2 at 9:56
$begingroup$
Consider $X,Y$ to be of trace 1 and their difference to be block diagonal $sigma^x,sigma^x$. Now $|X-Y|=$ identity matrix. Take the trace of both sides, the LHS gives 4 the RHS is 2.
$endgroup$
– lcv
Apr 2 at 9:56
$begingroup$
@lcv. can you please elaborate your notations?
$endgroup$
– Samya Kumar Ray
Apr 2 at 11:16
$begingroup$
@lcv. can you please elaborate your notations?
$endgroup$
– Samya Kumar Ray
Apr 2 at 11:16
add a comment |
2 Answers
2
active
oldest
votes
16
$begingroup$
The answer is No. Here is a counter-example:
$$X=begin{pmatrix} 9 & 3 \ 3 & 1 end{pmatrix},qquad Y=begin{pmatrix} 1 & 3 \ 3 & 9 end{pmatrix}.$$
answered Apr 2 at 12:04
Denis SerreDenis Serre
29.9k795199
$endgroup$
add a comment |
1
$begingroup$
Partial solution. When $X$ and $Y$ commute, the answer is "Yes". Indeed, in this case, the equality in the question is equivalent to $XY+YXgeq0$, and sure this is true.
answered Apr 2 at 15:57
Meisam Soleimani MalekanMeisam Soleimani Malekan
1,1181512
$endgroup$
4
$begingroup$
I went also through this inequality. However your claim is incorrect: $XY+YXge0$ is only equivalent to $|X+Y|^2le(X+Y)^2$. But $A^2le B^2$ with $A,B$ positive definite is not equivalent to $Ale B$. We only have $(A^2le B^2)Longrightarrow(Ale B)$. In other words, the map $tmapstosqrt t$ is operator monotone over $(0,+infty)$, but the reciprocal $xmapsto x^2$ is not.
$endgroup$
– Denis Serre
Apr 2 at 16:48
$begingroup$
To complete the previous comment, I confirm that $XY+YX$ is not always true for $X,Y$ positive definite when $nge2$. Besides the fact that it would imply that $xmapsto x^2$ be operator monotone, a false fact, it is also related to the theory of so-called Jordan algebra. See the book by Faraut and Koranyi.
$endgroup$
– Denis Serre
Apr 2 at 16:51
$begingroup$
@DenisSerre. The inverse is true if $A$ and $B$ commute.
$endgroup$
– Meisam Soleimani Malekan
Apr 2 at 16:52
$begingroup$
If matrices commutte, then also the positivity of $X$ and $Y$ implies that of $XY+YX$.
$endgroup$
– Denis Serre
Apr 2 at 17:22
2
$begingroup$
@Meisan. Of course they don't.
$endgroup$
– Denis Serre
Apr 2 at 19:31
|
show 2 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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oldest
votes
16
$begingroup$
The answer is No. Here is a counter-example:
$$X=begin{pmatrix} 9 & 3 \ 3 & 1 end{pmatrix},qquad Y=begin{pmatrix} 1 & 3 \ 3 & 9 end{pmatrix}.$$
answered Apr 2 at 12:04
Denis SerreDenis Serre
29.9k795199
$endgroup$
add a comment |
16
$begingroup$
The answer is No. Here is a counter-example:
$$X=begin{pmatrix} 9 & 3 \ 3 & 1 end{pmatrix},qquad Y=begin{pmatrix} 1 & 3 \ 3 & 9 end{pmatrix}.$$
answered Apr 2 at 12:04
Denis SerreDenis Serre
29.9k795199
$endgroup$
add a comment |
16
16
16
$begingroup$
The answer is No. Here is a counter-example:
$$X=begin{pmatrix} 9 & 3 \ 3 & 1 end{pmatrix},qquad Y=begin{pmatrix} 1 & 3 \ 3 & 9 end{pmatrix}.$$
answered Apr 2 at 12:04
Denis SerreDenis Serre
29.9k795199
$endgroup$
The answer is No. Here is a counter-example:
$$X=begin{pmatrix} 9 & 3 \ 3 & 1 end{pmatrix},qquad Y=begin{pmatrix} 1 & 3 \ 3 & 9 end{pmatrix}.$$
answered Apr 2 at 12:04
Denis SerreDenis Serre
29.9k795199
edited Apr 2 at 13:33
answered Apr 2 at 12:04
Denis SerreDenis Serre
29.9k795199
answered Apr 2 at 12:04
Denis SerreDenis Serre
29.9k795199
answered Apr 2 at 12:04
Denis SerreDenis Serre
29.9k795199
29.9k795199
add a comment |
add a comment |
1
$begingroup$
Partial solution. When $X$ and $Y$ commute, the answer is "Yes". Indeed, in this case, the equality in the question is equivalent to $XY+YXgeq0$, and sure this is true.
answered Apr 2 at 15:57
Meisam Soleimani MalekanMeisam Soleimani Malekan
1,1181512
$endgroup$
4
$begingroup$
I went also through this inequality. However your claim is incorrect: $XY+YXge0$ is only equivalent to $|X+Y|^2le(X+Y)^2$. But $A^2le B^2$ with $A,B$ positive definite is not equivalent to $Ale B$. We only have $(A^2le B^2)Longrightarrow(Ale B)$. In other words, the map $tmapstosqrt t$ is operator monotone over $(0,+infty)$, but the reciprocal $xmapsto x^2$ is not.
$endgroup$
– Denis Serre
Apr 2 at 16:48
$begingroup$
To complete the previous comment, I confirm that $XY+YX$ is not always true for $X,Y$ positive definite when $nge2$. Besides the fact that it would imply that $xmapsto x^2$ be operator monotone, a false fact, it is also related to the theory of so-called Jordan algebra. See the book by Faraut and Koranyi.
$endgroup$
– Denis Serre
Apr 2 at 16:51
$begingroup$
@DenisSerre. The inverse is true if $A$ and $B$ commute.
$endgroup$
– Meisam Soleimani Malekan
Apr 2 at 16:52
$begingroup$
If matrices commutte, then also the positivity of $X$ and $Y$ implies that of $XY+YX$.
$endgroup$
– Denis Serre
Apr 2 at 17:22
2
$begingroup$
@Meisan. Of course they don't.
$endgroup$
– Denis Serre
Apr 2 at 19:31
|
show 2 more comments
1
$begingroup$
Partial solution. When $X$ and $Y$ commute, the answer is "Yes". Indeed, in this case, the equality in the question is equivalent to $XY+YXgeq0$, and sure this is true.
answered Apr 2 at 15:57
Meisam Soleimani MalekanMeisam Soleimani Malekan
1,1181512
$endgroup$
4
$begingroup$
I went also through this inequality. However your claim is incorrect: $XY+YXge0$ is only equivalent to $|X+Y|^2le(X+Y)^2$. But $A^2le B^2$ with $A,B$ positive definite is not equivalent to $Ale B$. We only have $(A^2le B^2)Longrightarrow(Ale B)$. In other words, the map $tmapstosqrt t$ is operator monotone over $(0,+infty)$, but the reciprocal $xmapsto x^2$ is not.
$endgroup$
– Denis Serre
Apr 2 at 16:48
$begingroup$
To complete the previous comment, I confirm that $XY+YX$ is not always true for $X,Y$ positive definite when $nge2$. Besides the fact that it would imply that $xmapsto x^2$ be operator monotone, a false fact, it is also related to the theory of so-called Jordan algebra. See the book by Faraut and Koranyi.
$endgroup$
– Denis Serre
Apr 2 at 16:51
$begingroup$
@DenisSerre. The inverse is true if $A$ and $B$ commute.
$endgroup$
– Meisam Soleimani Malekan
Apr 2 at 16:52
$begingroup$
If matrices commutte, then also the positivity of $X$ and $Y$ implies that of $XY+YX$.
$endgroup$
– Denis Serre
Apr 2 at 17:22
2
$begingroup$
@Meisan. Of course they don't.
$endgroup$
– Denis Serre
Apr 2 at 19:31
|
show 2 more comments
1
1
1
$begingroup$
Partial solution. When $X$ and $Y$ commute, the answer is "Yes". Indeed, in this case, the equality in the question is equivalent to $XY+YXgeq0$, and sure this is true.
answered Apr 2 at 15:57
Meisam Soleimani MalekanMeisam Soleimani Malekan
1,1181512
$endgroup$
Partial solution. When $X$ and $Y$ commute, the answer is "Yes". Indeed, in this case, the equality in the question is equivalent to $XY+YXgeq0$, and sure this is true.
answered Apr 2 at 15:57
Meisam Soleimani MalekanMeisam Soleimani Malekan
1,1181512
edited Apr 3 at 8:48
answered Apr 2 at 15:57
Meisam Soleimani MalekanMeisam Soleimani Malekan
1,1181512
answered Apr 2 at 15:57
Meisam Soleimani MalekanMeisam Soleimani Malekan
1,1181512
answered Apr 2 at 15:57
Meisam Soleimani MalekanMeisam Soleimani Malekan
1,1181512
1,1181512
4
$begingroup$
I went also through this inequality. However your claim is incorrect: $XY+YXge0$ is only equivalent to $|X+Y|^2le(X+Y)^2$. But $A^2le B^2$ with $A,B$ positive definite is not equivalent to $Ale B$. We only have $(A^2le B^2)Longrightarrow(Ale B)$. In other words, the map $tmapstosqrt t$ is operator monotone over $(0,+infty)$, but the reciprocal $xmapsto x^2$ is not.
$endgroup$
– Denis Serre
Apr 2 at 16:48
$begingroup$
To complete the previous comment, I confirm that $XY+YX$ is not always true for $X,Y$ positive definite when $nge2$. Besides the fact that it would imply that $xmapsto x^2$ be operator monotone, a false fact, it is also related to the theory of so-called Jordan algebra. See the book by Faraut and Koranyi.
$endgroup$
– Denis Serre
Apr 2 at 16:51
$begingroup$
@DenisSerre. The inverse is true if $A$ and $B$ commute.
$endgroup$
– Meisam Soleimani Malekan
Apr 2 at 16:52
$begingroup$
If matrices commutte, then also the positivity of $X$ and $Y$ implies that of $XY+YX$.
$endgroup$
– Denis Serre
Apr 2 at 17:22
2
$begingroup$
@Meisan. Of course they don't.
$endgroup$
– Denis Serre
Apr 2 at 19:31
|
show 2 more comments
4
$begingroup$
I went also through this inequality. However your claim is incorrect: $XY+YXge0$ is only equivalent to $|X+Y|^2le(X+Y)^2$. But $A^2le B^2$ with $A,B$ positive definite is not equivalent to $Ale B$. We only have $(A^2le B^2)Longrightarrow(Ale B)$. In other words, the map $tmapstosqrt t$ is operator monotone over $(0,+infty)$, but the reciprocal $xmapsto x^2$ is not.
$endgroup$
– Denis Serre
Apr 2 at 16:48
$begingroup$
To complete the previous comment, I confirm that $XY+YX$ is not always true for $X,Y$ positive definite when $nge2$. Besides the fact that it would imply that $xmapsto x^2$ be operator monotone, a false fact, it is also related to the theory of so-called Jordan algebra. See the book by Faraut and Koranyi.
$endgroup$
– Denis Serre
Apr 2 at 16:51
$begingroup$
@DenisSerre. The inverse is true if $A$ and $B$ commute.
$endgroup$
– Meisam Soleimani Malekan
Apr 2 at 16:52
$begingroup$
If matrices commutte, then also the positivity of $X$ and $Y$ implies that of $XY+YX$.
$endgroup$
– Denis Serre
Apr 2 at 17:22
2
$begingroup$
@Meisan. Of course they don't.
$endgroup$
– Denis Serre
Apr 2 at 19:31
4
4
$begingroup$
I went also through this inequality. However your claim is incorrect: $XY+YXge0$ is only equivalent to $|X+Y|^2le(X+Y)^2$. But $A^2le B^2$ with $A,B$ positive definite is not equivalent to $Ale B$. We only have $(A^2le B^2)Longrightarrow(Ale B)$. In other words, the map $tmapstosqrt t$ is operator monotone over $(0,+infty)$, but the reciprocal $xmapsto x^2$ is not.
$endgroup$
– Denis Serre
Apr 2 at 16:48
$begingroup$
I went also through this inequality. However your claim is incorrect: $XY+YXge0$ is only equivalent to $|X+Y|^2le(X+Y)^2$. But $A^2le B^2$ with $A,B$ positive definite is not equivalent to $Ale B$. We only have $(A^2le B^2)Longrightarrow(Ale B)$. In other words, the map $tmapstosqrt t$ is operator monotone over $(0,+infty)$, but the reciprocal $xmapsto x^2$ is not.
$endgroup$
– Denis Serre
Apr 2 at 16:48
$begingroup$
To complete the previous comment, I confirm that $XY+YX$ is not always true for $X,Y$ positive definite when $nge2$. Besides the fact that it would imply that $xmapsto x^2$ be operator monotone, a false fact, it is also related to the theory of so-called Jordan algebra. See the book by Faraut and Koranyi.
$endgroup$
– Denis Serre
Apr 2 at 16:51
$begingroup$
To complete the previous comment, I confirm that $XY+YX$ is not always true for $X,Y$ positive definite when $nge2$. Besides the fact that it would imply that $xmapsto x^2$ be operator monotone, a false fact, it is also related to the theory of so-called Jordan algebra. See the book by Faraut and Koranyi.
$endgroup$
– Denis Serre
Apr 2 at 16:51
$begingroup$
@DenisSerre. The inverse is true if $A$ and $B$ commute.
$endgroup$
– Meisam Soleimani Malekan
Apr 2 at 16:52
$begingroup$
@DenisSerre. The inverse is true if $A$ and $B$ commute.
$endgroup$
– Meisam Soleimani Malekan
Apr 2 at 16:52
$begingroup$
If matrices commutte, then also the positivity of $X$ and $Y$ implies that of $XY+YX$.
$endgroup$
– Denis Serre
Apr 2 at 17:22
$begingroup$
If matrices commutte, then also the positivity of $X$ and $Y$ implies that of $XY+YX$.
$endgroup$
– Denis Serre
Apr 2 at 17:22
2
2
$begingroup$
@Meisan. Of course they don't.
$endgroup$
– Denis Serre
Apr 2 at 19:31
$begingroup$
@Meisan. Of course they don't.
$endgroup$
– Denis Serre
Apr 2 at 19:31
|
show 2 more comments
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$begingroup$
Consider $X,Y$ to be of trace 1 and their difference to be block diagonal $sigma^x,sigma^x$. Now $|X-Y|=$ identity matrix. Take the trace of both sides, the LHS gives 4 the RHS is 2.
$endgroup$
– lcv
Apr 2 at 9:56
$begingroup$
@lcv. can you please elaborate your notations?
$endgroup$
– Samya Kumar Ray
Apr 2 at 11:16