Inner product on V
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In book say that , let W be an inner product space and let T:V --> W be an isomorphism Then an inner product on V can be defined as
(u ,v) = (Tu ,Tv) , u,v ∈V.
My question,
Is there required T is an isomorphism for Inner product on V. I found that there is also inner product on V when T is an Linear transformation. Then why they wrote T is isomorphism.
linear-algebra
asked Sep 8 at 9:08
user499117
427
add a comment |Â
up vote
1
down vote
favorite
In book say that , let W be an inner product space and let T:V --> W be an isomorphism Then an inner product on V can be defined as
(u ,v) = (Tu ,Tv) , u,v ∈V.
My question,
Is there required T is an isomorphism for Inner product on V. I found that there is also inner product on V when T is an Linear transformation. Then why they wrote T is isomorphism.
linear-algebra
asked Sep 8 at 9:08
user499117
427
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In book say that , let W be an inner product space and let T:V --> W be an isomorphism Then an inner product on V can be defined as
(u ,v) = (Tu ,Tv) , u,v ∈V.
My question,
Is there required T is an isomorphism for Inner product on V. I found that there is also inner product on V when T is an Linear transformation. Then why they wrote T is isomorphism.
linear-algebra
asked Sep 8 at 9:08
user499117
427
In book say that , let W be an inner product space and let T:V --> W be an isomorphism Then an inner product on V can be defined as
(u ,v) = (Tu ,Tv) , u,v ∈V.
My question,
Is there required T is an isomorphism for Inner product on V. I found that there is also inner product on V when T is an Linear transformation. Then why they wrote T is isomorphism.
linear-algebra
linear-algebra
asked Sep 8 at 9:08
user499117
427
asked Sep 8 at 9:08
user499117
427
asked Sep 8 at 9:08
user499117
427
asked Sep 8 at 9:08
user499117
427
asked Sep 8 at 9:08
user499117
427
427
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You'll certainly need $T$ to be injective, but being surjective is optional. The reason you need $T$ to be injective is because of the definiteness part of the positive-definite axiom. If we define
$$(v, w)_V = (Tv, Tw)_W,$$
where $( cdot, cdot)_W$ is the inner product of $W$, then certainly,
$$(v, v)_V = (Tv, Tv)_W ge 0,$$
by the positivity of $( cdot, cdot)_W$. We also have,
$$(v, v)_V = 0 implies (Tv, Tv)_W = 0 implies Tv = 0,$$
but we really need $v = 0$ in order to satisfy definiteness. To get that final step from $Tv = 0$ to $v = 0$, we need injectivity.
answered Sep 8 at 9:40
Theo Bendit
13.7k12045
But in positive definiteness only required that (v,v) is strictly greater than zero for non-zero v in V. Is i am right or wrong.?
– user499117
Sep 8 at 9:55
1
You're right. But this is the contrapositive of $(v, v)_V implies v = 0$, which is equivalent. If $T$ is not injective, then there will be non-zero $v$ such that $Tv = 0$. For such non-zero vectors, $$(v, v)_V = (Tv, Tv)_W = (0, 0)_W = 0.$$
– Theo Bendit
Sep 8 at 10:18
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You'll certainly need $T$ to be injective, but being surjective is optional. The reason you need $T$ to be injective is because of the definiteness part of the positive-definite axiom. If we define
$$(v, w)_V = (Tv, Tw)_W,$$
where $( cdot, cdot)_W$ is the inner product of $W$, then certainly,
$$(v, v)_V = (Tv, Tv)_W ge 0,$$
by the positivity of $( cdot, cdot)_W$. We also have,
$$(v, v)_V = 0 implies (Tv, Tv)_W = 0 implies Tv = 0,$$
but we really need $v = 0$ in order to satisfy definiteness. To get that final step from $Tv = 0$ to $v = 0$, we need injectivity.
answered Sep 8 at 9:40
Theo Bendit
13.7k12045
But in positive definiteness only required that (v,v) is strictly greater than zero for non-zero v in V. Is i am right or wrong.?
– user499117
Sep 8 at 9:55
1
You're right. But this is the contrapositive of $(v, v)_V implies v = 0$, which is equivalent. If $T$ is not injective, then there will be non-zero $v$ such that $Tv = 0$. For such non-zero vectors, $$(v, v)_V = (Tv, Tv)_W = (0, 0)_W = 0.$$
– Theo Bendit
Sep 8 at 10:18
add a comment |Â
up vote
1
down vote
accepted
You'll certainly need $T$ to be injective, but being surjective is optional. The reason you need $T$ to be injective is because of the definiteness part of the positive-definite axiom. If we define
$$(v, w)_V = (Tv, Tw)_W,$$
where $( cdot, cdot)_W$ is the inner product of $W$, then certainly,
$$(v, v)_V = (Tv, Tv)_W ge 0,$$
by the positivity of $( cdot, cdot)_W$. We also have,
$$(v, v)_V = 0 implies (Tv, Tv)_W = 0 implies Tv = 0,$$
but we really need $v = 0$ in order to satisfy definiteness. To get that final step from $Tv = 0$ to $v = 0$, we need injectivity.
answered Sep 8 at 9:40
Theo Bendit
13.7k12045
But in positive definiteness only required that (v,v) is strictly greater than zero for non-zero v in V. Is i am right or wrong.?
– user499117
Sep 8 at 9:55
1
You're right. But this is the contrapositive of $(v, v)_V implies v = 0$, which is equivalent. If $T$ is not injective, then there will be non-zero $v$ such that $Tv = 0$. For such non-zero vectors, $$(v, v)_V = (Tv, Tv)_W = (0, 0)_W = 0.$$
– Theo Bendit
Sep 8 at 10:18
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You'll certainly need $T$ to be injective, but being surjective is optional. The reason you need $T$ to be injective is because of the definiteness part of the positive-definite axiom. If we define
$$(v, w)_V = (Tv, Tw)_W,$$
where $( cdot, cdot)_W$ is the inner product of $W$, then certainly,
$$(v, v)_V = (Tv, Tv)_W ge 0,$$
by the positivity of $( cdot, cdot)_W$. We also have,
$$(v, v)_V = 0 implies (Tv, Tv)_W = 0 implies Tv = 0,$$
but we really need $v = 0$ in order to satisfy definiteness. To get that final step from $Tv = 0$ to $v = 0$, we need injectivity.
answered Sep 8 at 9:40
Theo Bendit
13.7k12045
You'll certainly need $T$ to be injective, but being surjective is optional. The reason you need $T$ to be injective is because of the definiteness part of the positive-definite axiom. If we define
$$(v, w)_V = (Tv, Tw)_W,$$
where $( cdot, cdot)_W$ is the inner product of $W$, then certainly,
$$(v, v)_V = (Tv, Tv)_W ge 0,$$
by the positivity of $( cdot, cdot)_W$. We also have,
$$(v, v)_V = 0 implies (Tv, Tv)_W = 0 implies Tv = 0,$$
but we really need $v = 0$ in order to satisfy definiteness. To get that final step from $Tv = 0$ to $v = 0$, we need injectivity.
answered Sep 8 at 9:40
Theo Bendit
13.7k12045
answered Sep 8 at 9:40
Theo Bendit
13.7k12045
answered Sep 8 at 9:40
Theo Bendit
13.7k12045
answered Sep 8 at 9:40
Theo Bendit
13.7k12045
13.7k12045
But in positive definiteness only required that (v,v) is strictly greater than zero for non-zero v in V. Is i am right or wrong.?
– user499117
Sep 8 at 9:55
1
You're right. But this is the contrapositive of $(v, v)_V implies v = 0$, which is equivalent. If $T$ is not injective, then there will be non-zero $v$ such that $Tv = 0$. For such non-zero vectors, $$(v, v)_V = (Tv, Tv)_W = (0, 0)_W = 0.$$
– Theo Bendit
Sep 8 at 10:18
add a comment |Â
But in positive definiteness only required that (v,v) is strictly greater than zero for non-zero v in V. Is i am right or wrong.?
– user499117
Sep 8 at 9:55
1
You're right. But this is the contrapositive of $(v, v)_V implies v = 0$, which is equivalent. If $T$ is not injective, then there will be non-zero $v$ such that $Tv = 0$. For such non-zero vectors, $$(v, v)_V = (Tv, Tv)_W = (0, 0)_W = 0.$$
– Theo Bendit
Sep 8 at 10:18
But in positive definiteness only required that (v,v) is strictly greater than zero for non-zero v in V. Is i am right or wrong.?
– user499117
Sep 8 at 9:55
But in positive definiteness only required that (v,v) is strictly greater than zero for non-zero v in V. Is i am right or wrong.?
– user499117
Sep 8 at 9:55
1
1
You're right. But this is the contrapositive of $(v, v)_V implies v = 0$, which is equivalent. If $T$ is not injective, then there will be non-zero $v$ such that $Tv = 0$. For such non-zero vectors, $$(v, v)_V = (Tv, Tv)_W = (0, 0)_W = 0.$$
– Theo Bendit
Sep 8 at 10:18
You're right. But this is the contrapositive of $(v, v)_V implies v = 0$, which is equivalent. If $T$ is not injective, then there will be non-zero $v$ such that $Tv = 0$. For such non-zero vectors, $$(v, v)_V = (Tv, Tv)_W = (0, 0)_W = 0.$$
– Theo Bendit
Sep 8 at 10:18
add a comment |Â
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