Every inner product space is a metric space.
$begingroup$
Show that every inner product space is a metric space.
To show this should I set the distance metric as $d(x,y) = <x-y,x-y>$, then show properties of being metric space such as d(x,y) = d(y,x) etc.?
If so the point I do not understand is why we set metric as $d(x,y) = <x-y,x-y>$ (this metric is mentioned in wolfram)
real-analysis metric-spaces inner-product-space
asked Dec 6 '18 at 19:43
PumpkinPumpkin
5021418
$endgroup$
add a comment |
$begingroup$
Show that every inner product space is a metric space.
To show this should I set the distance metric as $d(x,y) = <x-y,x-y>$, then show properties of being metric space such as d(x,y) = d(y,x) etc.?
If so the point I do not understand is why we set metric as $d(x,y) = <x-y,x-y>$ (this metric is mentioned in wolfram)
real-analysis metric-spaces inner-product-space
asked Dec 6 '18 at 19:43
PumpkinPumpkin
5021418
$endgroup$
$begingroup$
you should use $d(x,y) = sqrt{langle x-y, x-yrangle}$
$endgroup$
– Thomas
Dec 6 '18 at 19:46
add a comment |
$begingroup$
Show that every inner product space is a metric space.
To show this should I set the distance metric as $d(x,y) = <x-y,x-y>$, then show properties of being metric space such as d(x,y) = d(y,x) etc.?
If so the point I do not understand is why we set metric as $d(x,y) = <x-y,x-y>$ (this metric is mentioned in wolfram)
real-analysis metric-spaces inner-product-space
asked Dec 6 '18 at 19:43
PumpkinPumpkin
5021418
$endgroup$
Show that every inner product space is a metric space.
To show this should I set the distance metric as $d(x,y) = <x-y,x-y>$, then show properties of being metric space such as d(x,y) = d(y,x) etc.?
If so the point I do not understand is why we set metric as $d(x,y) = <x-y,x-y>$ (this metric is mentioned in wolfram)
real-analysis metric-spaces inner-product-space
real-analysis metric-spaces inner-product-space
asked Dec 6 '18 at 19:43
PumpkinPumpkin
5021418
asked Dec 6 '18 at 19:43
PumpkinPumpkin
5021418
asked Dec 6 '18 at 19:43
PumpkinPumpkin
5021418
asked Dec 6 '18 at 19:43
PumpkinPumpkin
5021418
asked Dec 6 '18 at 19:43
PumpkinPumpkin
5021418
5021418
$begingroup$
you should use $d(x,y) = sqrt{langle x-y, x-yrangle}$
$endgroup$
– Thomas
Dec 6 '18 at 19:46
add a comment |
$begingroup$
you should use $d(x,y) = sqrt{langle x-y, x-yrangle}$
$endgroup$
– Thomas
Dec 6 '18 at 19:46
$begingroup$
you should use $d(x,y) = sqrt{langle x-y, x-yrangle}$
$endgroup$
– Thomas
Dec 6 '18 at 19:46
$begingroup$
you should use $d(x,y) = sqrt{langle x-y, x-yrangle}$
$endgroup$
– Thomas
Dec 6 '18 at 19:46
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
That is wrong. It should be $d(x,y)=sqrt{langle x-y,x-yrangle}$ because the map $xmapstosqrt{langle x,xrangle}$ is a norm. And, whenever you have a norm $lVertcdotrVert$, the map $(x,y)mapstolVert x-yrVert$ is a distance.
answered Dec 6 '18 at 19:50
José Carlos SantosJosé Carlos Santos
159k22126229
$endgroup$
add a comment |
$begingroup$
The inner product induces a norm via
$$||x||:=sqrt{langle x,xrangle}$$
and then, the norm induces a metric via
$$d(x,y):=||x-y||$$
which is precisely what is mentioned in the comment above. In general, one has the relation $text{Inner Product Spaces}subsettext{Normed Vector Spaces}subsettext{Metric Spaces}$.
answered Dec 6 '18 at 19:50
JWP_HTXJWP_HTX
410314
$endgroup$
add a comment |
$begingroup$
You are setting $d(x,y)=||x-y||$. So your metric here comes from the norm. You can check that it satisfies the required metric conditions.
We are not sating that this is the only way to choose a metric on the space, but it is one way.
answered Dec 6 '18 at 19:53
AnyADAnyAD
2,108812
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That is wrong. It should be $d(x,y)=sqrt{langle x-y,x-yrangle}$ because the map $xmapstosqrt{langle x,xrangle}$ is a norm. And, whenever you have a norm $lVertcdotrVert$, the map $(x,y)mapstolVert x-yrVert$ is a distance.
answered Dec 6 '18 at 19:50
José Carlos SantosJosé Carlos Santos
159k22126229
$endgroup$
add a comment |
$begingroup$
That is wrong. It should be $d(x,y)=sqrt{langle x-y,x-yrangle}$ because the map $xmapstosqrt{langle x,xrangle}$ is a norm. And, whenever you have a norm $lVertcdotrVert$, the map $(x,y)mapstolVert x-yrVert$ is a distance.
answered Dec 6 '18 at 19:50
José Carlos SantosJosé Carlos Santos
159k22126229
$endgroup$
add a comment |
$begingroup$
That is wrong. It should be $d(x,y)=sqrt{langle x-y,x-yrangle}$ because the map $xmapstosqrt{langle x,xrangle}$ is a norm. And, whenever you have a norm $lVertcdotrVert$, the map $(x,y)mapstolVert x-yrVert$ is a distance.
answered Dec 6 '18 at 19:50
José Carlos SantosJosé Carlos Santos
159k22126229
$endgroup$
That is wrong. It should be $d(x,y)=sqrt{langle x-y,x-yrangle}$ because the map $xmapstosqrt{langle x,xrangle}$ is a norm. And, whenever you have a norm $lVertcdotrVert$, the map $(x,y)mapstolVert x-yrVert$ is a distance.
answered Dec 6 '18 at 19:50
José Carlos SantosJosé Carlos Santos
159k22126229
answered Dec 6 '18 at 19:50
José Carlos SantosJosé Carlos Santos
159k22126229
answered Dec 6 '18 at 19:50
José Carlos SantosJosé Carlos Santos
159k22126229
answered Dec 6 '18 at 19:50
José Carlos SantosJosé Carlos Santos
159k22126229
159k22126229
add a comment |
add a comment |
$begingroup$
The inner product induces a norm via
$$||x||:=sqrt{langle x,xrangle}$$
and then, the norm induces a metric via
$$d(x,y):=||x-y||$$
which is precisely what is mentioned in the comment above. In general, one has the relation $text{Inner Product Spaces}subsettext{Normed Vector Spaces}subsettext{Metric Spaces}$.
answered Dec 6 '18 at 19:50
JWP_HTXJWP_HTX
410314
$endgroup$
add a comment |
$begingroup$
The inner product induces a norm via
$$||x||:=sqrt{langle x,xrangle}$$
and then, the norm induces a metric via
$$d(x,y):=||x-y||$$
which is precisely what is mentioned in the comment above. In general, one has the relation $text{Inner Product Spaces}subsettext{Normed Vector Spaces}subsettext{Metric Spaces}$.
answered Dec 6 '18 at 19:50
JWP_HTXJWP_HTX
410314
$endgroup$
add a comment |
$begingroup$
The inner product induces a norm via
$$||x||:=sqrt{langle x,xrangle}$$
and then, the norm induces a metric via
$$d(x,y):=||x-y||$$
which is precisely what is mentioned in the comment above. In general, one has the relation $text{Inner Product Spaces}subsettext{Normed Vector Spaces}subsettext{Metric Spaces}$.
answered Dec 6 '18 at 19:50
JWP_HTXJWP_HTX
410314
$endgroup$
The inner product induces a norm via
$$||x||:=sqrt{langle x,xrangle}$$
and then, the norm induces a metric via
$$d(x,y):=||x-y||$$
which is precisely what is mentioned in the comment above. In general, one has the relation $text{Inner Product Spaces}subsettext{Normed Vector Spaces}subsettext{Metric Spaces}$.
answered Dec 6 '18 at 19:50
JWP_HTXJWP_HTX
410314
answered Dec 6 '18 at 19:50
JWP_HTXJWP_HTX
410314
answered Dec 6 '18 at 19:50
JWP_HTXJWP_HTX
410314
answered Dec 6 '18 at 19:50
JWP_HTXJWP_HTX
410314
410314
add a comment |
add a comment |
$begingroup$
You are setting $d(x,y)=||x-y||$. So your metric here comes from the norm. You can check that it satisfies the required metric conditions.
We are not sating that this is the only way to choose a metric on the space, but it is one way.
answered Dec 6 '18 at 19:53
AnyADAnyAD
2,108812
$endgroup$
add a comment |
$begingroup$
You are setting $d(x,y)=||x-y||$. So your metric here comes from the norm. You can check that it satisfies the required metric conditions.
We are not sating that this is the only way to choose a metric on the space, but it is one way.
answered Dec 6 '18 at 19:53
AnyADAnyAD
2,108812
$endgroup$
add a comment |
$begingroup$
You are setting $d(x,y)=||x-y||$. So your metric here comes from the norm. You can check that it satisfies the required metric conditions.
We are not sating that this is the only way to choose a metric on the space, but it is one way.
answered Dec 6 '18 at 19:53
AnyADAnyAD
2,108812
$endgroup$
You are setting $d(x,y)=||x-y||$. So your metric here comes from the norm. You can check that it satisfies the required metric conditions.
We are not sating that this is the only way to choose a metric on the space, but it is one way.
answered Dec 6 '18 at 19:53
AnyADAnyAD
2,108812
answered Dec 6 '18 at 19:53
AnyADAnyAD
2,108812
answered Dec 6 '18 at 19:53
AnyADAnyAD
2,108812
answered Dec 6 '18 at 19:53
AnyADAnyAD
2,108812
2,108812
add a comment |
add a comment |
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$begingroup$
you should use $d(x,y) = sqrt{langle x-y, x-yrangle}$
$endgroup$
– Thomas
Dec 6 '18 at 19:46