A one-to-one map between $M_{ntimes n}$ and the $mathbb{R}^{n^{2}}$?
up vote
-1
down vote
favorite
I've been trying to think in something that can make this happen, but i'm not get anywhere. Plus, i have to show something through this map that can make this space($M_{ntimes n}$) a metric space, so a think to use the trace of a matrix. Is that correct?
linear-algebra general-topology matrices metric-spaces
edited Dec 4 at 20:02
J.G.
21.1k21933
asked Dec 4 at 19:51
gbcosta
64
add a comment |
up vote
-1
down vote
favorite
I've been trying to think in something that can make this happen, but i'm not get anywhere. Plus, i have to show something through this map that can make this space($M_{ntimes n}$) a metric space, so a think to use the trace of a matrix. Is that correct?
linear-algebra general-topology matrices metric-spaces
edited Dec 4 at 20:02
J.G.
21.1k21933
asked Dec 4 at 19:51
gbcosta
64
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I've been trying to think in something that can make this happen, but i'm not get anywhere. Plus, i have to show something through this map that can make this space($M_{ntimes n}$) a metric space, so a think to use the trace of a matrix. Is that correct?
linear-algebra general-topology matrices metric-spaces
edited Dec 4 at 20:02
J.G.
21.1k21933
asked Dec 4 at 19:51
gbcosta
64
I've been trying to think in something that can make this happen, but i'm not get anywhere. Plus, i have to show something through this map that can make this space($M_{ntimes n}$) a metric space, so a think to use the trace of a matrix. Is that correct?
linear-algebra general-topology matrices metric-spaces
linear-algebra general-topology matrices metric-spaces
edited Dec 4 at 20:02
J.G.
21.1k21933
asked Dec 4 at 19:51
gbcosta
64
edited Dec 4 at 20:02
J.G.
21.1k21933
asked Dec 4 at 19:51
gbcosta
64
edited Dec 4 at 20:02
J.G.
21.1k21933
edited Dec 4 at 20:02
J.G.
21.1k21933
edited Dec 4 at 20:02
J.G.
21.1k21933
21.1k21933
asked Dec 4 at 19:51
gbcosta
64
asked Dec 4 at 19:51
gbcosta
64
asked Dec 4 at 19:51
gbcosta
64
64
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The trace of matrix is invalid since it reduces the dimension from $n^2$ to $n$. You can define the set of $ntimes n$ matrices with Frobenius norm and reorder the rows of the matrices back to back and define the Euclidean $2$-norm on $Bbb R^{n^2}$. In this manner the two metric spaces would become homeomorphic.
answered Dec 4 at 19:57
Mostafa Ayaz
13.6k3836
Actually, $mathrm{tr}:M_{ntimes n}tomathbb R$ reduces the dimension to just $1$.
– Federico
Dec 4 at 20:05
1
I mean the degree of freedom reduces to $n$ since only diagonal entries are important.
– Mostafa Ayaz
Dec 4 at 20:05
Yeah, i saw my mistake about the trace. I'm gonna look the sugestion of Frobenius norm here, thanks a lot.
– gbcosta
Dec 4 at 20:33
You're welcome. I also can help you further if you needed....
– Mostafa Ayaz
Dec 4 at 20:40
@Federico actually, any metric $m: X times X to mathbb{R}$ has dimension 1 codomain, so I don't think that affects anything...
– Monstrous Moonshiner
Dec 4 at 21:10
|
show 3 more comments
up vote
1
down vote
Just take
$$
f:Ato (a_{1,1},a_{1,2},dots,a_{1,n},a_{2,1},dots a_{2,n},a_{3,1}dots,a_{n,1},a_{n,2},dots, a_{n,n})
$$
i.e just rearrange the entries of $A$ into a vector in $mathbb{R}^{n^2}$.
Then define $d(A,B):=||f(A)-f(B)||$.
answered Dec 4 at 20:08
qbert
22k32459
But this map is one-to-one? I think that $$f:Arightarrow (a_{1,1},..., a_{1,n}, a_{2,1}...,a{2,n},...,a_{n,1},...,a_{n,n})$$ answer the question, but i'm not sure of that.
– gbcosta
Dec 4 at 20:40
Take n=3. If $A=begin{array}{lcr} a & b & c \ d & e & f \ g & h & j end{array}$ and $B = begin{array}{lcr} a & b & c \ k & l & m \ g & h & j end{array}$. So $Aneq B$ but f(A)=f(B)
– gbcosta
Dec 4 at 20:57
No problem, anyway you helped me a lot, thank you so much.
– gbcosta
Dec 4 at 21:04
@gbcosta no problem
– qbert
Dec 4 at 21:04
2
Your objection : "Take n=3..." is not valid : we order the entries f(A)=(a, b, c, d, e, f, g, h, i) is different from f(B)=(a, b, c, k, l, m, g, h, i) !
– Jean Marie
Dec 4 at 21:12
|
show 1 more comment
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The trace of matrix is invalid since it reduces the dimension from $n^2$ to $n$. You can define the set of $ntimes n$ matrices with Frobenius norm and reorder the rows of the matrices back to back and define the Euclidean $2$-norm on $Bbb R^{n^2}$. In this manner the two metric spaces would become homeomorphic.
answered Dec 4 at 19:57
Mostafa Ayaz
13.6k3836
Actually, $mathrm{tr}:M_{ntimes n}tomathbb R$ reduces the dimension to just $1$.
– Federico
Dec 4 at 20:05
1
I mean the degree of freedom reduces to $n$ since only diagonal entries are important.
– Mostafa Ayaz
Dec 4 at 20:05
Yeah, i saw my mistake about the trace. I'm gonna look the sugestion of Frobenius norm here, thanks a lot.
– gbcosta
Dec 4 at 20:33
You're welcome. I also can help you further if you needed....
– Mostafa Ayaz
Dec 4 at 20:40
@Federico actually, any metric $m: X times X to mathbb{R}$ has dimension 1 codomain, so I don't think that affects anything...
– Monstrous Moonshiner
Dec 4 at 21:10
|
show 3 more comments
up vote
1
down vote
accepted
The trace of matrix is invalid since it reduces the dimension from $n^2$ to $n$. You can define the set of $ntimes n$ matrices with Frobenius norm and reorder the rows of the matrices back to back and define the Euclidean $2$-norm on $Bbb R^{n^2}$. In this manner the two metric spaces would become homeomorphic.
answered Dec 4 at 19:57
Mostafa Ayaz
13.6k3836
Actually, $mathrm{tr}:M_{ntimes n}tomathbb R$ reduces the dimension to just $1$.
– Federico
Dec 4 at 20:05
1
I mean the degree of freedom reduces to $n$ since only diagonal entries are important.
– Mostafa Ayaz
Dec 4 at 20:05
Yeah, i saw my mistake about the trace. I'm gonna look the sugestion of Frobenius norm here, thanks a lot.
– gbcosta
Dec 4 at 20:33
You're welcome. I also can help you further if you needed....
– Mostafa Ayaz
Dec 4 at 20:40
@Federico actually, any metric $m: X times X to mathbb{R}$ has dimension 1 codomain, so I don't think that affects anything...
– Monstrous Moonshiner
Dec 4 at 21:10
|
show 3 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The trace of matrix is invalid since it reduces the dimension from $n^2$ to $n$. You can define the set of $ntimes n$ matrices with Frobenius norm and reorder the rows of the matrices back to back and define the Euclidean $2$-norm on $Bbb R^{n^2}$. In this manner the two metric spaces would become homeomorphic.
answered Dec 4 at 19:57
Mostafa Ayaz
13.6k3836
The trace of matrix is invalid since it reduces the dimension from $n^2$ to $n$. You can define the set of $ntimes n$ matrices with Frobenius norm and reorder the rows of the matrices back to back and define the Euclidean $2$-norm on $Bbb R^{n^2}$. In this manner the two metric spaces would become homeomorphic.
answered Dec 4 at 19:57
Mostafa Ayaz
13.6k3836
answered Dec 4 at 19:57
Mostafa Ayaz
13.6k3836
answered Dec 4 at 19:57
Mostafa Ayaz
13.6k3836
answered Dec 4 at 19:57
Mostafa Ayaz
13.6k3836
13.6k3836
Actually, $mathrm{tr}:M_{ntimes n}tomathbb R$ reduces the dimension to just $1$.
– Federico
Dec 4 at 20:05
1
I mean the degree of freedom reduces to $n$ since only diagonal entries are important.
– Mostafa Ayaz
Dec 4 at 20:05
Yeah, i saw my mistake about the trace. I'm gonna look the sugestion of Frobenius norm here, thanks a lot.
– gbcosta
Dec 4 at 20:33
You're welcome. I also can help you further if you needed....
– Mostafa Ayaz
Dec 4 at 20:40
@Federico actually, any metric $m: X times X to mathbb{R}$ has dimension 1 codomain, so I don't think that affects anything...
– Monstrous Moonshiner
Dec 4 at 21:10
|
show 3 more comments
Actually, $mathrm{tr}:M_{ntimes n}tomathbb R$ reduces the dimension to just $1$.
– Federico
Dec 4 at 20:05
1
I mean the degree of freedom reduces to $n$ since only diagonal entries are important.
– Mostafa Ayaz
Dec 4 at 20:05
Yeah, i saw my mistake about the trace. I'm gonna look the sugestion of Frobenius norm here, thanks a lot.
– gbcosta
Dec 4 at 20:33
You're welcome. I also can help you further if you needed....
– Mostafa Ayaz
Dec 4 at 20:40
@Federico actually, any metric $m: X times X to mathbb{R}$ has dimension 1 codomain, so I don't think that affects anything...
– Monstrous Moonshiner
Dec 4 at 21:10
Actually, $mathrm{tr}:M_{ntimes n}tomathbb R$ reduces the dimension to just $1$.
– Federico
Dec 4 at 20:05
Actually, $mathrm{tr}:M_{ntimes n}tomathbb R$ reduces the dimension to just $1$.
– Federico
Dec 4 at 20:05
1
1
I mean the degree of freedom reduces to $n$ since only diagonal entries are important.
– Mostafa Ayaz
Dec 4 at 20:05
I mean the degree of freedom reduces to $n$ since only diagonal entries are important.
– Mostafa Ayaz
Dec 4 at 20:05
Yeah, i saw my mistake about the trace. I'm gonna look the sugestion of Frobenius norm here, thanks a lot.
– gbcosta
Dec 4 at 20:33
Yeah, i saw my mistake about the trace. I'm gonna look the sugestion of Frobenius norm here, thanks a lot.
– gbcosta
Dec 4 at 20:33
You're welcome. I also can help you further if you needed....
– Mostafa Ayaz
Dec 4 at 20:40
You're welcome. I also can help you further if you needed....
– Mostafa Ayaz
Dec 4 at 20:40
@Federico actually, any metric $m: X times X to mathbb{R}$ has dimension 1 codomain, so I don't think that affects anything...
– Monstrous Moonshiner
Dec 4 at 21:10
@Federico actually, any metric $m: X times X to mathbb{R}$ has dimension 1 codomain, so I don't think that affects anything...
– Monstrous Moonshiner
Dec 4 at 21:10
|
show 3 more comments
up vote
1
down vote
Just take
$$
f:Ato (a_{1,1},a_{1,2},dots,a_{1,n},a_{2,1},dots a_{2,n},a_{3,1}dots,a_{n,1},a_{n,2},dots, a_{n,n})
$$
i.e just rearrange the entries of $A$ into a vector in $mathbb{R}^{n^2}$.
Then define $d(A,B):=||f(A)-f(B)||$.
answered Dec 4 at 20:08
qbert
22k32459
But this map is one-to-one? I think that $$f:Arightarrow (a_{1,1},..., a_{1,n}, a_{2,1}...,a{2,n},...,a_{n,1},...,a_{n,n})$$ answer the question, but i'm not sure of that.
– gbcosta
Dec 4 at 20:40
Take n=3. If $A=begin{array}{lcr} a & b & c \ d & e & f \ g & h & j end{array}$ and $B = begin{array}{lcr} a & b & c \ k & l & m \ g & h & j end{array}$. So $Aneq B$ but f(A)=f(B)
– gbcosta
Dec 4 at 20:57
No problem, anyway you helped me a lot, thank you so much.
– gbcosta
Dec 4 at 21:04
@gbcosta no problem
– qbert
Dec 4 at 21:04
2
Your objection : "Take n=3..." is not valid : we order the entries f(A)=(a, b, c, d, e, f, g, h, i) is different from f(B)=(a, b, c, k, l, m, g, h, i) !
– Jean Marie
Dec 4 at 21:12
|
show 1 more comment
up vote
1
down vote
Just take
$$
f:Ato (a_{1,1},a_{1,2},dots,a_{1,n},a_{2,1},dots a_{2,n},a_{3,1}dots,a_{n,1},a_{n,2},dots, a_{n,n})
$$
i.e just rearrange the entries of $A$ into a vector in $mathbb{R}^{n^2}$.
Then define $d(A,B):=||f(A)-f(B)||$.
answered Dec 4 at 20:08
qbert
22k32459
But this map is one-to-one? I think that $$f:Arightarrow (a_{1,1},..., a_{1,n}, a_{2,1}...,a{2,n},...,a_{n,1},...,a_{n,n})$$ answer the question, but i'm not sure of that.
– gbcosta
Dec 4 at 20:40
Take n=3. If $A=begin{array}{lcr} a & b & c \ d & e & f \ g & h & j end{array}$ and $B = begin{array}{lcr} a & b & c \ k & l & m \ g & h & j end{array}$. So $Aneq B$ but f(A)=f(B)
– gbcosta
Dec 4 at 20:57
No problem, anyway you helped me a lot, thank you so much.
– gbcosta
Dec 4 at 21:04
@gbcosta no problem
– qbert
Dec 4 at 21:04
2
Your objection : "Take n=3..." is not valid : we order the entries f(A)=(a, b, c, d, e, f, g, h, i) is different from f(B)=(a, b, c, k, l, m, g, h, i) !
– Jean Marie
Dec 4 at 21:12
|
show 1 more comment
up vote
1
down vote
up vote
1
down vote
Just take
$$
f:Ato (a_{1,1},a_{1,2},dots,a_{1,n},a_{2,1},dots a_{2,n},a_{3,1}dots,a_{n,1},a_{n,2},dots, a_{n,n})
$$
i.e just rearrange the entries of $A$ into a vector in $mathbb{R}^{n^2}$.
Then define $d(A,B):=||f(A)-f(B)||$.
answered Dec 4 at 20:08
qbert
22k32459
Just take
$$
f:Ato (a_{1,1},a_{1,2},dots,a_{1,n},a_{2,1},dots a_{2,n},a_{3,1}dots,a_{n,1},a_{n,2},dots, a_{n,n})
$$
i.e just rearrange the entries of $A$ into a vector in $mathbb{R}^{n^2}$.
Then define $d(A,B):=||f(A)-f(B)||$.
answered Dec 4 at 20:08
qbert
22k32459
edited Dec 4 at 21:01
answered Dec 4 at 20:08
qbert
22k32459
answered Dec 4 at 20:08
qbert
22k32459
answered Dec 4 at 20:08
qbert
22k32459
22k32459
But this map is one-to-one? I think that $$f:Arightarrow (a_{1,1},..., a_{1,n}, a_{2,1}...,a{2,n},...,a_{n,1},...,a_{n,n})$$ answer the question, but i'm not sure of that.
– gbcosta
Dec 4 at 20:40
Take n=3. If $A=begin{array}{lcr} a & b & c \ d & e & f \ g & h & j end{array}$ and $B = begin{array}{lcr} a & b & c \ k & l & m \ g & h & j end{array}$. So $Aneq B$ but f(A)=f(B)
– gbcosta
Dec 4 at 20:57
No problem, anyway you helped me a lot, thank you so much.
– gbcosta
Dec 4 at 21:04
@gbcosta no problem
– qbert
Dec 4 at 21:04
2
Your objection : "Take n=3..." is not valid : we order the entries f(A)=(a, b, c, d, e, f, g, h, i) is different from f(B)=(a, b, c, k, l, m, g, h, i) !
– Jean Marie
Dec 4 at 21:12
|
show 1 more comment
But this map is one-to-one? I think that $$f:Arightarrow (a_{1,1},..., a_{1,n}, a_{2,1}...,a{2,n},...,a_{n,1},...,a_{n,n})$$ answer the question, but i'm not sure of that.
– gbcosta
Dec 4 at 20:40
Take n=3. If $A=begin{array}{lcr} a & b & c \ d & e & f \ g & h & j end{array}$ and $B = begin{array}{lcr} a & b & c \ k & l & m \ g & h & j end{array}$. So $Aneq B$ but f(A)=f(B)
– gbcosta
Dec 4 at 20:57
No problem, anyway you helped me a lot, thank you so much.
– gbcosta
Dec 4 at 21:04
@gbcosta no problem
– qbert
Dec 4 at 21:04
2
Your objection : "Take n=3..." is not valid : we order the entries f(A)=(a, b, c, d, e, f, g, h, i) is different from f(B)=(a, b, c, k, l, m, g, h, i) !
– Jean Marie
Dec 4 at 21:12
But this map is one-to-one? I think that $$f:Arightarrow (a_{1,1},..., a_{1,n}, a_{2,1}...,a{2,n},...,a_{n,1},...,a_{n,n})$$ answer the question, but i'm not sure of that.
– gbcosta
Dec 4 at 20:40
But this map is one-to-one? I think that $$f:Arightarrow (a_{1,1},..., a_{1,n}, a_{2,1}...,a{2,n},...,a_{n,1},...,a_{n,n})$$ answer the question, but i'm not sure of that.
– gbcosta
Dec 4 at 20:40
Take n=3. If $A=begin{array}{lcr} a & b & c \ d & e & f \ g & h & j end{array}$ and $B = begin{array}{lcr} a & b & c \ k & l & m \ g & h & j end{array}$. So $Aneq B$ but f(A)=f(B)
– gbcosta
Dec 4 at 20:57
Take n=3. If $A=begin{array}{lcr} a & b & c \ d & e & f \ g & h & j end{array}$ and $B = begin{array}{lcr} a & b & c \ k & l & m \ g & h & j end{array}$. So $Aneq B$ but f(A)=f(B)
– gbcosta
Dec 4 at 20:57
No problem, anyway you helped me a lot, thank you so much.
– gbcosta
Dec 4 at 21:04
No problem, anyway you helped me a lot, thank you so much.
– gbcosta
Dec 4 at 21:04
@gbcosta no problem
– qbert
Dec 4 at 21:04
@gbcosta no problem
– qbert
Dec 4 at 21:04
2
2
Your objection : "Take n=3..." is not valid : we order the entries f(A)=(a, b, c, d, e, f, g, h, i) is different from f(B)=(a, b, c, k, l, m, g, h, i) !
– Jean Marie
Dec 4 at 21:12
Your objection : "Take n=3..." is not valid : we order the entries f(A)=(a, b, c, d, e, f, g, h, i) is different from f(B)=(a, b, c, k, l, m, g, h, i) !
– Jean Marie
Dec 4 at 21:12
|
show 1 more comment
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