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Which inner products preserve positive correlation?
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Suppose we have a symmetric PD or PSD matrix M which induces an inner product $langle cdot, cdot rangle_M$. If we have that $langle x, y rangle > 0$ for two unit vectors $x$, $y$, are there any sufficient conditions on $x, y$, and/or $M$ that ensure that $langlex, yrangle_M > 0$ (other than the obvious $x=y$ or $M = I$)?.
linear-algebra matrices inner-product
edited 46 mins ago
Federico Poloni
12.3k25380
asked 5 hours ago
B Merlot
183
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up vote
2
down vote
favorite
Suppose we have a symmetric PD or PSD matrix M which induces an inner product $langle cdot, cdot rangle_M$. If we have that $langle x, y rangle > 0$ for two unit vectors $x$, $y$, are there any sufficient conditions on $x, y$, and/or $M$ that ensure that $langlex, yrangle_M > 0$ (other than the obvious $x=y$ or $M = I$)?.
linear-algebra matrices inner-product
edited 46 mins ago
Federico Poloni
12.3k25380
asked 5 hours ago
B Merlot
183
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose we have a symmetric PD or PSD matrix M which induces an inner product $langle cdot, cdot rangle_M$. If we have that $langle x, y rangle > 0$ for two unit vectors $x$, $y$, are there any sufficient conditions on $x, y$, and/or $M$ that ensure that $langlex, yrangle_M > 0$ (other than the obvious $x=y$ or $M = I$)?.
linear-algebra matrices inner-product
edited 46 mins ago
Federico Poloni
12.3k25380
asked 5 hours ago
B Merlot
183
Suppose we have a symmetric PD or PSD matrix M which induces an inner product $langle cdot, cdot rangle_M$. If we have that $langle x, y rangle > 0$ for two unit vectors $x$, $y$, are there any sufficient conditions on $x, y$, and/or $M$ that ensure that $langlex, yrangle_M > 0$ (other than the obvious $x=y$ or $M = I$)?.
linear-algebra matrices inner-product
linear-algebra matrices inner-product
edited 46 mins ago
Federico Poloni
12.3k25380
asked 5 hours ago
B Merlot
183
edited 46 mins ago
Federico Poloni
12.3k25380
asked 5 hours ago
B Merlot
183
edited 46 mins ago
Federico Poloni
12.3k25380
edited 46 mins ago
Federico Poloni
12.3k25380
edited 46 mins ago
Federico Poloni
12.3k25380
12.3k25380
asked 5 hours ago
B Merlot
183
asked 5 hours ago
B Merlot
183
asked 5 hours ago
B Merlot
183
183
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
Since $M$ is PSD, it can be decomposed as $M=A^T A$, where $A$ is another matrix; see
https://math.stackexchange.com/questions/1801403/decomposition-of-a-positive-semidefinite-matrix
Now $langle x,yrangle=x^T M y=x^T A^T Ay=langle Ax,Ayrangle$ and so your question is equivalent to: Which matrices $A$ preserve positive correlations between vectors? It is known that the matrices that preserve orthogonality are precisely those that are scalar multiples of orthogonal transformations:
https://math.stackexchange.com/questions/2355551/linear-transformations-that-preserve-orthogonality
Certainly if a matrix does not preserve orthogonality, it does not preserve positive correlations (it can fix $x$ and map $y$ to a vector orthogonal to $x$). Conversely, scalar multiples of orthogonal matrices do preserve positive correlations -- more generally, they preserve dot products:
https://math.stackexchange.com/questions/2161729/how-orthogonal-matrices-preserve-dot-product-and-volume-proof
We conclude that $A$ must be a scalar multiple of an orthogonal matrix, and hence $M$ must be a positive multiple of the identity $I$.
answered 1 hour ago
Aryeh Kontorovich
2,1741124
Minor edit: $M$ must be a positive multiple of $I$ (previously I had "scalar").
– Aryeh Kontorovich
1 hour ago
Thank you! Are there any assumptions on $x$ and $y$ we could make instead?
– B Merlot
1 hour ago
For all positively correlated $x,y$, you can find a "bad" $M$ that will cause them to be orthogonal under the induced inner product.
– Aryeh Kontorovich
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Since $M$ is PSD, it can be decomposed as $M=A^T A$, where $A$ is another matrix; see
https://math.stackexchange.com/questions/1801403/decomposition-of-a-positive-semidefinite-matrix
Now $langle x,yrangle=x^T M y=x^T A^T Ay=langle Ax,Ayrangle$ and so your question is equivalent to: Which matrices $A$ preserve positive correlations between vectors? It is known that the matrices that preserve orthogonality are precisely those that are scalar multiples of orthogonal transformations:
https://math.stackexchange.com/questions/2355551/linear-transformations-that-preserve-orthogonality
Certainly if a matrix does not preserve orthogonality, it does not preserve positive correlations (it can fix $x$ and map $y$ to a vector orthogonal to $x$). Conversely, scalar multiples of orthogonal matrices do preserve positive correlations -- more generally, they preserve dot products:
https://math.stackexchange.com/questions/2161729/how-orthogonal-matrices-preserve-dot-product-and-volume-proof
We conclude that $A$ must be a scalar multiple of an orthogonal matrix, and hence $M$ must be a positive multiple of the identity $I$.
answered 1 hour ago
Aryeh Kontorovich
2,1741124
Minor edit: $M$ must be a positive multiple of $I$ (previously I had "scalar").
– Aryeh Kontorovich
1 hour ago
Thank you! Are there any assumptions on $x$ and $y$ we could make instead?
– B Merlot
1 hour ago
For all positively correlated $x,y$, you can find a "bad" $M$ that will cause them to be orthogonal under the induced inner product.
– Aryeh Kontorovich
1 hour ago
add a comment |Â
up vote
6
down vote
accepted
Since $M$ is PSD, it can be decomposed as $M=A^T A$, where $A$ is another matrix; see
https://math.stackexchange.com/questions/1801403/decomposition-of-a-positive-semidefinite-matrix
Now $langle x,yrangle=x^T M y=x^T A^T Ay=langle Ax,Ayrangle$ and so your question is equivalent to: Which matrices $A$ preserve positive correlations between vectors? It is known that the matrices that preserve orthogonality are precisely those that are scalar multiples of orthogonal transformations:
https://math.stackexchange.com/questions/2355551/linear-transformations-that-preserve-orthogonality
Certainly if a matrix does not preserve orthogonality, it does not preserve positive correlations (it can fix $x$ and map $y$ to a vector orthogonal to $x$). Conversely, scalar multiples of orthogonal matrices do preserve positive correlations -- more generally, they preserve dot products:
https://math.stackexchange.com/questions/2161729/how-orthogonal-matrices-preserve-dot-product-and-volume-proof
We conclude that $A$ must be a scalar multiple of an orthogonal matrix, and hence $M$ must be a positive multiple of the identity $I$.
answered 1 hour ago
Aryeh Kontorovich
2,1741124
Minor edit: $M$ must be a positive multiple of $I$ (previously I had "scalar").
– Aryeh Kontorovich
1 hour ago
Thank you! Are there any assumptions on $x$ and $y$ we could make instead?
– B Merlot
1 hour ago
For all positively correlated $x,y$, you can find a "bad" $M$ that will cause them to be orthogonal under the induced inner product.
– Aryeh Kontorovich
1 hour ago
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Since $M$ is PSD, it can be decomposed as $M=A^T A$, where $A$ is another matrix; see
https://math.stackexchange.com/questions/1801403/decomposition-of-a-positive-semidefinite-matrix
Now $langle x,yrangle=x^T M y=x^T A^T Ay=langle Ax,Ayrangle$ and so your question is equivalent to: Which matrices $A$ preserve positive correlations between vectors? It is known that the matrices that preserve orthogonality are precisely those that are scalar multiples of orthogonal transformations:
https://math.stackexchange.com/questions/2355551/linear-transformations-that-preserve-orthogonality
Certainly if a matrix does not preserve orthogonality, it does not preserve positive correlations (it can fix $x$ and map $y$ to a vector orthogonal to $x$). Conversely, scalar multiples of orthogonal matrices do preserve positive correlations -- more generally, they preserve dot products:
https://math.stackexchange.com/questions/2161729/how-orthogonal-matrices-preserve-dot-product-and-volume-proof
We conclude that $A$ must be a scalar multiple of an orthogonal matrix, and hence $M$ must be a positive multiple of the identity $I$.
answered 1 hour ago
Aryeh Kontorovich
2,1741124
Since $M$ is PSD, it can be decomposed as $M=A^T A$, where $A$ is another matrix; see
https://math.stackexchange.com/questions/1801403/decomposition-of-a-positive-semidefinite-matrix
Now $langle x,yrangle=x^T M y=x^T A^T Ay=langle Ax,Ayrangle$ and so your question is equivalent to: Which matrices $A$ preserve positive correlations between vectors? It is known that the matrices that preserve orthogonality are precisely those that are scalar multiples of orthogonal transformations:
https://math.stackexchange.com/questions/2355551/linear-transformations-that-preserve-orthogonality
Certainly if a matrix does not preserve orthogonality, it does not preserve positive correlations (it can fix $x$ and map $y$ to a vector orthogonal to $x$). Conversely, scalar multiples of orthogonal matrices do preserve positive correlations -- more generally, they preserve dot products:
https://math.stackexchange.com/questions/2161729/how-orthogonal-matrices-preserve-dot-product-and-volume-proof
We conclude that $A$ must be a scalar multiple of an orthogonal matrix, and hence $M$ must be a positive multiple of the identity $I$.
answered 1 hour ago
Aryeh Kontorovich
2,1741124
edited 1 hour ago
answered 1 hour ago
Aryeh Kontorovich
2,1741124
answered 1 hour ago
Aryeh Kontorovich
2,1741124
answered 1 hour ago
Aryeh Kontorovich
2,1741124
2,1741124
Minor edit: $M$ must be a positive multiple of $I$ (previously I had "scalar").
– Aryeh Kontorovich
1 hour ago
Thank you! Are there any assumptions on $x$ and $y$ we could make instead?
– B Merlot
1 hour ago
For all positively correlated $x,y$, you can find a "bad" $M$ that will cause them to be orthogonal under the induced inner product.
– Aryeh Kontorovich
1 hour ago
add a comment |Â
Minor edit: $M$ must be a positive multiple of $I$ (previously I had "scalar").
– Aryeh Kontorovich
1 hour ago
Thank you! Are there any assumptions on $x$ and $y$ we could make instead?
– B Merlot
1 hour ago
For all positively correlated $x,y$, you can find a "bad" $M$ that will cause them to be orthogonal under the induced inner product.
– Aryeh Kontorovich
1 hour ago
Minor edit: $M$ must be a positive multiple of $I$ (previously I had "scalar").
– Aryeh Kontorovich
1 hour ago
Minor edit: $M$ must be a positive multiple of $I$ (previously I had "scalar").
– Aryeh Kontorovich
1 hour ago
Thank you! Are there any assumptions on $x$ and $y$ we could make instead?
– B Merlot
1 hour ago
Thank you! Are there any assumptions on $x$ and $y$ we could make instead?
– B Merlot
1 hour ago
For all positively correlated $x,y$, you can find a "bad" $M$ that will cause them to be orthogonal under the induced inner product.
– Aryeh Kontorovich
1 hour ago
For all positively correlated $x,y$, you can find a "bad" $M$ that will cause them to be orthogonal under the induced inner product.
– Aryeh Kontorovich
1 hour ago
add a comment |Â
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