PSD vector inner product with positive vectors
Suppose $A in mathbb{R}^{n times n}$ be a PSD matrix. Let $x,y in mathbb{R}^n$ such that $x = [x_1,...,x_n]^T$ and $y = [y_1,...,y_n]^T$. We require that $x,y$ are element-wise positive, that is $x_i >0$ and $y_i > 0$ for all $i in {1,2,...,n}$. In that case can it be concluded that $x^T A y geq 0$ in general?
linear-algebra
asked Dec 19 at 20:50
rajatsen91
33918
add a comment |
Suppose $A in mathbb{R}^{n times n}$ be a PSD matrix. Let $x,y in mathbb{R}^n$ such that $x = [x_1,...,x_n]^T$ and $y = [y_1,...,y_n]^T$. We require that $x,y$ are element-wise positive, that is $x_i >0$ and $y_i > 0$ for all $i in {1,2,...,n}$. In that case can it be concluded that $x^T A y geq 0$ in general?
linear-algebra
asked Dec 19 at 20:50
rajatsen91
33918
add a comment |
Suppose $A in mathbb{R}^{n times n}$ be a PSD matrix. Let $x,y in mathbb{R}^n$ such that $x = [x_1,...,x_n]^T$ and $y = [y_1,...,y_n]^T$. We require that $x,y$ are element-wise positive, that is $x_i >0$ and $y_i > 0$ for all $i in {1,2,...,n}$. In that case can it be concluded that $x^T A y geq 0$ in general?
linear-algebra
asked Dec 19 at 20:50
rajatsen91
33918
Suppose $A in mathbb{R}^{n times n}$ be a PSD matrix. Let $x,y in mathbb{R}^n$ such that $x = [x_1,...,x_n]^T$ and $y = [y_1,...,y_n]^T$. We require that $x,y$ are element-wise positive, that is $x_i >0$ and $y_i > 0$ for all $i in {1,2,...,n}$. In that case can it be concluded that $x^T A y geq 0$ in general?
linear-algebra
linear-algebra
asked Dec 19 at 20:50
rajatsen91
33918
asked Dec 19 at 20:50
rajatsen91
33918
asked Dec 19 at 20:50
rajatsen91
33918
asked Dec 19 at 20:50
rajatsen91
33918
asked Dec 19 at 20:50
rajatsen91
33918
33918
add a comment |
add a comment |
3 Answers
3
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oldest
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No. Let
$$M=newcommandbmat{begin{pmatrix}}newcommandemat{end{pmatrix}}bmat 0 & 1 \ -1 & 0 emat,$$
then $$bmat x & y emat bmat 0 & 1 \ -1 & 0 emat bmat a \ bemat = bx-ay,$$
so in particular when $x=a$, $y=b$, we have that the quadratic form corresponding to $M$ is always zero on any vector, so $M$ is PSD.
However if $a=y=1$, $x=b=frac{1}{2}$, then the product of the vectors with the matrix will be $frac{1}{4}-1=frac{-3}{4}$.
The Idea:
The idea is that rotation by 90 degrees is PSD, since the dot product of a vector and its 90 degree rotation will always be zero.
However, if we choose the vector to be rotated by 90 degrees to already be closer to the $y$-axis, and the vector to compare it to to be closer to the $x$-axis, then the final product will be a dot product of vectors with an obtuse angle between them, which will be negative.
The same idea allows us to replace $M$ with a matrix which is e.g. rotation by 45 degrees to get a positive definite matrix which also doesn't have $v^TMu>0$ for positive vectors $v$ and $u$.
answered Dec 19 at 21:02
jgon
12.3k21940
add a comment |
No. Take any large positive $n$ and consider
$$
pmatrix{1&n}pmatrix{5&-2\ -2&1}pmatrix{1\ 1}=3-n.
$$
answered Dec 19 at 21:01
user1551
71.2k566125
add a comment |
I believe the matrix $A=pmatrix{1&-1\0&1}$ is PSD, however $pmatrix{1&0}^TApmatrix{0&1}=-1$. You can replace $0$ by small enough $epsilon>0$ and it will not change the fact that this results in a negative number.
answered Dec 19 at 20:58
SmileyCraft
2,114313
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
No. Let
$$M=newcommandbmat{begin{pmatrix}}newcommandemat{end{pmatrix}}bmat 0 & 1 \ -1 & 0 emat,$$
then $$bmat x & y emat bmat 0 & 1 \ -1 & 0 emat bmat a \ bemat = bx-ay,$$
so in particular when $x=a$, $y=b$, we have that the quadratic form corresponding to $M$ is always zero on any vector, so $M$ is PSD.
However if $a=y=1$, $x=b=frac{1}{2}$, then the product of the vectors with the matrix will be $frac{1}{4}-1=frac{-3}{4}$.
The Idea:
The idea is that rotation by 90 degrees is PSD, since the dot product of a vector and its 90 degree rotation will always be zero.
However, if we choose the vector to be rotated by 90 degrees to already be closer to the $y$-axis, and the vector to compare it to to be closer to the $x$-axis, then the final product will be a dot product of vectors with an obtuse angle between them, which will be negative.
The same idea allows us to replace $M$ with a matrix which is e.g. rotation by 45 degrees to get a positive definite matrix which also doesn't have $v^TMu>0$ for positive vectors $v$ and $u$.
answered Dec 19 at 21:02
jgon
12.3k21940
add a comment |
No. Let
$$M=newcommandbmat{begin{pmatrix}}newcommandemat{end{pmatrix}}bmat 0 & 1 \ -1 & 0 emat,$$
then $$bmat x & y emat bmat 0 & 1 \ -1 & 0 emat bmat a \ bemat = bx-ay,$$
so in particular when $x=a$, $y=b$, we have that the quadratic form corresponding to $M$ is always zero on any vector, so $M$ is PSD.
However if $a=y=1$, $x=b=frac{1}{2}$, then the product of the vectors with the matrix will be $frac{1}{4}-1=frac{-3}{4}$.
The Idea:
The idea is that rotation by 90 degrees is PSD, since the dot product of a vector and its 90 degree rotation will always be zero.
However, if we choose the vector to be rotated by 90 degrees to already be closer to the $y$-axis, and the vector to compare it to to be closer to the $x$-axis, then the final product will be a dot product of vectors with an obtuse angle between them, which will be negative.
The same idea allows us to replace $M$ with a matrix which is e.g. rotation by 45 degrees to get a positive definite matrix which also doesn't have $v^TMu>0$ for positive vectors $v$ and $u$.
answered Dec 19 at 21:02
jgon
12.3k21940
add a comment |
No. Let
$$M=newcommandbmat{begin{pmatrix}}newcommandemat{end{pmatrix}}bmat 0 & 1 \ -1 & 0 emat,$$
then $$bmat x & y emat bmat 0 & 1 \ -1 & 0 emat bmat a \ bemat = bx-ay,$$
so in particular when $x=a$, $y=b$, we have that the quadratic form corresponding to $M$ is always zero on any vector, so $M$ is PSD.
However if $a=y=1$, $x=b=frac{1}{2}$, then the product of the vectors with the matrix will be $frac{1}{4}-1=frac{-3}{4}$.
The Idea:
The idea is that rotation by 90 degrees is PSD, since the dot product of a vector and its 90 degree rotation will always be zero.
However, if we choose the vector to be rotated by 90 degrees to already be closer to the $y$-axis, and the vector to compare it to to be closer to the $x$-axis, then the final product will be a dot product of vectors with an obtuse angle between them, which will be negative.
The same idea allows us to replace $M$ with a matrix which is e.g. rotation by 45 degrees to get a positive definite matrix which also doesn't have $v^TMu>0$ for positive vectors $v$ and $u$.
answered Dec 19 at 21:02
jgon
12.3k21940
No. Let
$$M=newcommandbmat{begin{pmatrix}}newcommandemat{end{pmatrix}}bmat 0 & 1 \ -1 & 0 emat,$$
then $$bmat x & y emat bmat 0 & 1 \ -1 & 0 emat bmat a \ bemat = bx-ay,$$
so in particular when $x=a$, $y=b$, we have that the quadratic form corresponding to $M$ is always zero on any vector, so $M$ is PSD.
However if $a=y=1$, $x=b=frac{1}{2}$, then the product of the vectors with the matrix will be $frac{1}{4}-1=frac{-3}{4}$.
The Idea:
The idea is that rotation by 90 degrees is PSD, since the dot product of a vector and its 90 degree rotation will always be zero.
However, if we choose the vector to be rotated by 90 degrees to already be closer to the $y$-axis, and the vector to compare it to to be closer to the $x$-axis, then the final product will be a dot product of vectors with an obtuse angle between them, which will be negative.
The same idea allows us to replace $M$ with a matrix which is e.g. rotation by 45 degrees to get a positive definite matrix which also doesn't have $v^TMu>0$ for positive vectors $v$ and $u$.
answered Dec 19 at 21:02
jgon
12.3k21940
answered Dec 19 at 21:02
jgon
12.3k21940
answered Dec 19 at 21:02
jgon
12.3k21940
answered Dec 19 at 21:02
jgon
12.3k21940
12.3k21940
add a comment |
add a comment |
No. Take any large positive $n$ and consider
$$
pmatrix{1&n}pmatrix{5&-2\ -2&1}pmatrix{1\ 1}=3-n.
$$
answered Dec 19 at 21:01
user1551
71.2k566125
add a comment |
No. Take any large positive $n$ and consider
$$
pmatrix{1&n}pmatrix{5&-2\ -2&1}pmatrix{1\ 1}=3-n.
$$
answered Dec 19 at 21:01
user1551
71.2k566125
add a comment |
No. Take any large positive $n$ and consider
$$
pmatrix{1&n}pmatrix{5&-2\ -2&1}pmatrix{1\ 1}=3-n.
$$
answered Dec 19 at 21:01
user1551
71.2k566125
No. Take any large positive $n$ and consider
$$
pmatrix{1&n}pmatrix{5&-2\ -2&1}pmatrix{1\ 1}=3-n.
$$
answered Dec 19 at 21:01
user1551
71.2k566125
answered Dec 19 at 21:01
user1551
71.2k566125
answered Dec 19 at 21:01
user1551
71.2k566125
answered Dec 19 at 21:01
user1551
71.2k566125
71.2k566125
add a comment |
add a comment |
I believe the matrix $A=pmatrix{1&-1\0&1}$ is PSD, however $pmatrix{1&0}^TApmatrix{0&1}=-1$. You can replace $0$ by small enough $epsilon>0$ and it will not change the fact that this results in a negative number.
answered Dec 19 at 20:58
SmileyCraft
2,114313
add a comment |
I believe the matrix $A=pmatrix{1&-1\0&1}$ is PSD, however $pmatrix{1&0}^TApmatrix{0&1}=-1$. You can replace $0$ by small enough $epsilon>0$ and it will not change the fact that this results in a negative number.
answered Dec 19 at 20:58
SmileyCraft
2,114313
add a comment |
I believe the matrix $A=pmatrix{1&-1\0&1}$ is PSD, however $pmatrix{1&0}^TApmatrix{0&1}=-1$. You can replace $0$ by small enough $epsilon>0$ and it will not change the fact that this results in a negative number.
answered Dec 19 at 20:58
SmileyCraft
2,114313
I believe the matrix $A=pmatrix{1&-1\0&1}$ is PSD, however $pmatrix{1&0}^TApmatrix{0&1}=-1$. You can replace $0$ by small enough $epsilon>0$ and it will not change the fact that this results in a negative number.
answered Dec 19 at 20:58
SmileyCraft
2,114313
edited Dec 19 at 21:05
answered Dec 19 at 20:58
SmileyCraft
2,114313
answered Dec 19 at 20:58
SmileyCraft
2,114313
answered Dec 19 at 20:58
SmileyCraft
2,114313
2,114313
add a comment |
add a comment |
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