Functions over a $C$ vector space with geometric importance. (How to find the basis?)
$begingroup$
Searching through our suggested exercises of linear and abstract algebra for solving, I found the following exercise. The reason I am posting this, is that because we haven't went through complex spaces and theory but the exercise is still in the suggested ones, which means I am missing something :
Let $V$ be the $C$-vector space of the complex functions over the set of the vertexes of a square. We define a linear operator $T$ over $V$ to be :
$Tf(x) = f(r(x)) $
where for a vertex $x$, $r(x)$ is the exactly next vertex (going clockwise).
->Describe a basis of $V$, write the matrix of the operator $T$ over the particular base you found, find the eigenvalues and describe the eigenvectors of $T$ as functions over the vertex of the square.
What I searched and tried was :
First of all, going clockwise and since we are speaking for a complex vector space, from what I know I think that $r(x) = -ix$ (correct me if that's wrong).
Also, $f$ is a functional defined as : $f : mathbb C^m -> mathbb C$ and $T$ is an operator defined as : $T: V->mathbb C$. From the definition of the operator and the functional (if I did it correctly), how will I procced to finding the basis? I believe that once I define the basis, I will not have an issue solving the rest of the exercise because it goes as a chain, since you need the basis to find the matrix, then you need the matrix to find the eigenvalues and the eigenvalues for the eigenvectors. Given that I know perfectly the theory for bases, matrices, eigenvalues and eigenvectors for the space $mathbb R$ of the real numbers, I think that by scrolling through complex-algebra parts of the books I won't have a problem solving it. But I couldn't find such a type of exercise so I can understand how to proceed with the basis in the $mathbb C$-complex space. I would really appreciate if you could explain how to find the basis, the rest I will finish them, but that's where I am stuck at (the basis).
linear-algebra abstract-algebra functional-analysis operator-theory complex-geometry
asked May 3 '16 at 19:53
RebellosRebellos
14.5k31246
$endgroup$
|
show 3 more comments
$begingroup$
Searching through our suggested exercises of linear and abstract algebra for solving, I found the following exercise. The reason I am posting this, is that because we haven't went through complex spaces and theory but the exercise is still in the suggested ones, which means I am missing something :
Let $V$ be the $C$-vector space of the complex functions over the set of the vertexes of a square. We define a linear operator $T$ over $V$ to be :
$Tf(x) = f(r(x)) $
where for a vertex $x$, $r(x)$ is the exactly next vertex (going clockwise).
->Describe a basis of $V$, write the matrix of the operator $T$ over the particular base you found, find the eigenvalues and describe the eigenvectors of $T$ as functions over the vertex of the square.
What I searched and tried was :
First of all, going clockwise and since we are speaking for a complex vector space, from what I know I think that $r(x) = -ix$ (correct me if that's wrong).
Also, $f$ is a functional defined as : $f : mathbb C^m -> mathbb C$ and $T$ is an operator defined as : $T: V->mathbb C$. From the definition of the operator and the functional (if I did it correctly), how will I procced to finding the basis? I believe that once I define the basis, I will not have an issue solving the rest of the exercise because it goes as a chain, since you need the basis to find the matrix, then you need the matrix to find the eigenvalues and the eigenvalues for the eigenvectors. Given that I know perfectly the theory for bases, matrices, eigenvalues and eigenvectors for the space $mathbb R$ of the real numbers, I think that by scrolling through complex-algebra parts of the books I won't have a problem solving it. But I couldn't find such a type of exercise so I can understand how to proceed with the basis in the $mathbb C$-complex space. I would really appreciate if you could explain how to find the basis, the rest I will finish them, but that's where I am stuck at (the basis).
linear-algebra abstract-algebra functional-analysis operator-theory complex-geometry
asked May 3 '16 at 19:53
RebellosRebellos
14.5k31246
$endgroup$
1
$begingroup$
A square has four vertices. Your space $V$ contains functions that assign a (complex) number to each of these four vertices. So an element of $V$ is just a tuple of four numbers! Does this sound familiar? Now can you give me a basis for $V$? Don't let the complex part bother you too much. It will not affect anything until the finding of the eigenvalues.
$endgroup$
– Ravi
May 3 '16 at 19:58
$begingroup$
Oh, let me give it a hand that way then !
$endgroup$
– Rebellos
May 3 '16 at 20:10
$begingroup$
By the way, what does "tuple" mean ? Because even the greek dictionary word is something that we don't use anymore :P
$endgroup$
– Rebellos
May 3 '16 at 20:11
$begingroup$
An $n$-tuple is a list of $n$ numbers. I.e. $(a_1,cdots ,a_n)$.
$endgroup$
– Ravi
May 3 '16 at 20:13
$begingroup$
Oh I get it now. So we have a vector over that space which is $(a_1,a_2,a_3,a_4) $ . Now, the complex number which is assigned, is assigned by the way I thought it did ? $ r(x) = -ix$ ? Then going on like : $Tf(x) = f(-ix) $
$endgroup$
– Rebellos
May 3 '16 at 20:17
|
show 3 more comments
$begingroup$
Searching through our suggested exercises of linear and abstract algebra for solving, I found the following exercise. The reason I am posting this, is that because we haven't went through complex spaces and theory but the exercise is still in the suggested ones, which means I am missing something :
Let $V$ be the $C$-vector space of the complex functions over the set of the vertexes of a square. We define a linear operator $T$ over $V$ to be :
$Tf(x) = f(r(x)) $
where for a vertex $x$, $r(x)$ is the exactly next vertex (going clockwise).
->Describe a basis of $V$, write the matrix of the operator $T$ over the particular base you found, find the eigenvalues and describe the eigenvectors of $T$ as functions over the vertex of the square.
What I searched and tried was :
First of all, going clockwise and since we are speaking for a complex vector space, from what I know I think that $r(x) = -ix$ (correct me if that's wrong).
Also, $f$ is a functional defined as : $f : mathbb C^m -> mathbb C$ and $T$ is an operator defined as : $T: V->mathbb C$. From the definition of the operator and the functional (if I did it correctly), how will I procced to finding the basis? I believe that once I define the basis, I will not have an issue solving the rest of the exercise because it goes as a chain, since you need the basis to find the matrix, then you need the matrix to find the eigenvalues and the eigenvalues for the eigenvectors. Given that I know perfectly the theory for bases, matrices, eigenvalues and eigenvectors for the space $mathbb R$ of the real numbers, I think that by scrolling through complex-algebra parts of the books I won't have a problem solving it. But I couldn't find such a type of exercise so I can understand how to proceed with the basis in the $mathbb C$-complex space. I would really appreciate if you could explain how to find the basis, the rest I will finish them, but that's where I am stuck at (the basis).
linear-algebra abstract-algebra functional-analysis operator-theory complex-geometry
asked May 3 '16 at 19:53
RebellosRebellos
14.5k31246
$endgroup$
Searching through our suggested exercises of linear and abstract algebra for solving, I found the following exercise. The reason I am posting this, is that because we haven't went through complex spaces and theory but the exercise is still in the suggested ones, which means I am missing something :
Let $V$ be the $C$-vector space of the complex functions over the set of the vertexes of a square. We define a linear operator $T$ over $V$ to be :
$Tf(x) = f(r(x)) $
where for a vertex $x$, $r(x)$ is the exactly next vertex (going clockwise).
->Describe a basis of $V$, write the matrix of the operator $T$ over the particular base you found, find the eigenvalues and describe the eigenvectors of $T$ as functions over the vertex of the square.
What I searched and tried was :
First of all, going clockwise and since we are speaking for a complex vector space, from what I know I think that $r(x) = -ix$ (correct me if that's wrong).
Also, $f$ is a functional defined as : $f : mathbb C^m -> mathbb C$ and $T$ is an operator defined as : $T: V->mathbb C$. From the definition of the operator and the functional (if I did it correctly), how will I procced to finding the basis? I believe that once I define the basis, I will not have an issue solving the rest of the exercise because it goes as a chain, since you need the basis to find the matrix, then you need the matrix to find the eigenvalues and the eigenvalues for the eigenvectors. Given that I know perfectly the theory for bases, matrices, eigenvalues and eigenvectors for the space $mathbb R$ of the real numbers, I think that by scrolling through complex-algebra parts of the books I won't have a problem solving it. But I couldn't find such a type of exercise so I can understand how to proceed with the basis in the $mathbb C$-complex space. I would really appreciate if you could explain how to find the basis, the rest I will finish them, but that's where I am stuck at (the basis).
linear-algebra abstract-algebra functional-analysis operator-theory complex-geometry
linear-algebra abstract-algebra functional-analysis operator-theory complex-geometry
asked May 3 '16 at 19:53
RebellosRebellos
14.5k31246
asked May 3 '16 at 19:53
RebellosRebellos
14.5k31246
asked May 3 '16 at 19:53
RebellosRebellos
14.5k31246
asked May 3 '16 at 19:53
RebellosRebellos
14.5k31246
asked May 3 '16 at 19:53
RebellosRebellos
14.5k31246
14.5k31246
1
$begingroup$
A square has four vertices. Your space $V$ contains functions that assign a (complex) number to each of these four vertices. So an element of $V$ is just a tuple of four numbers! Does this sound familiar? Now can you give me a basis for $V$? Don't let the complex part bother you too much. It will not affect anything until the finding of the eigenvalues.
$endgroup$
– Ravi
May 3 '16 at 19:58
$begingroup$
Oh, let me give it a hand that way then !
$endgroup$
– Rebellos
May 3 '16 at 20:10
$begingroup$
By the way, what does "tuple" mean ? Because even the greek dictionary word is something that we don't use anymore :P
$endgroup$
– Rebellos
May 3 '16 at 20:11
$begingroup$
An $n$-tuple is a list of $n$ numbers. I.e. $(a_1,cdots ,a_n)$.
$endgroup$
– Ravi
May 3 '16 at 20:13
$begingroup$
Oh I get it now. So we have a vector over that space which is $(a_1,a_2,a_3,a_4) $ . Now, the complex number which is assigned, is assigned by the way I thought it did ? $ r(x) = -ix$ ? Then going on like : $Tf(x) = f(-ix) $
$endgroup$
– Rebellos
May 3 '16 at 20:17
|
show 3 more comments
1
$begingroup$
A square has four vertices. Your space $V$ contains functions that assign a (complex) number to each of these four vertices. So an element of $V$ is just a tuple of four numbers! Does this sound familiar? Now can you give me a basis for $V$? Don't let the complex part bother you too much. It will not affect anything until the finding of the eigenvalues.
$endgroup$
– Ravi
May 3 '16 at 19:58
$begingroup$
Oh, let me give it a hand that way then !
$endgroup$
– Rebellos
May 3 '16 at 20:10
$begingroup$
By the way, what does "tuple" mean ? Because even the greek dictionary word is something that we don't use anymore :P
$endgroup$
– Rebellos
May 3 '16 at 20:11
$begingroup$
An $n$-tuple is a list of $n$ numbers. I.e. $(a_1,cdots ,a_n)$.
$endgroup$
– Ravi
May 3 '16 at 20:13
$begingroup$
Oh I get it now. So we have a vector over that space which is $(a_1,a_2,a_3,a_4) $ . Now, the complex number which is assigned, is assigned by the way I thought it did ? $ r(x) = -ix$ ? Then going on like : $Tf(x) = f(-ix) $
$endgroup$
– Rebellos
May 3 '16 at 20:17
1
1
$begingroup$
A square has four vertices. Your space $V$ contains functions that assign a (complex) number to each of these four vertices. So an element of $V$ is just a tuple of four numbers! Does this sound familiar? Now can you give me a basis for $V$? Don't let the complex part bother you too much. It will not affect anything until the finding of the eigenvalues.
$endgroup$
– Ravi
May 3 '16 at 19:58
$begingroup$
A square has four vertices. Your space $V$ contains functions that assign a (complex) number to each of these four vertices. So an element of $V$ is just a tuple of four numbers! Does this sound familiar? Now can you give me a basis for $V$? Don't let the complex part bother you too much. It will not affect anything until the finding of the eigenvalues.
$endgroup$
– Ravi
May 3 '16 at 19:58
$begingroup$
Oh, let me give it a hand that way then !
$endgroup$
– Rebellos
May 3 '16 at 20:10
$begingroup$
Oh, let me give it a hand that way then !
$endgroup$
– Rebellos
May 3 '16 at 20:10
$begingroup$
By the way, what does "tuple" mean ? Because even the greek dictionary word is something that we don't use anymore :P
$endgroup$
– Rebellos
May 3 '16 at 20:11
$begingroup$
By the way, what does "tuple" mean ? Because even the greek dictionary word is something that we don't use anymore :P
$endgroup$
– Rebellos
May 3 '16 at 20:11
$begingroup$
An $n$-tuple is a list of $n$ numbers. I.e. $(a_1,cdots ,a_n)$.
$endgroup$
– Ravi
May 3 '16 at 20:13
$begingroup$
An $n$-tuple is a list of $n$ numbers. I.e. $(a_1,cdots ,a_n)$.
$endgroup$
– Ravi
May 3 '16 at 20:13
$begingroup$
Oh I get it now. So we have a vector over that space which is $(a_1,a_2,a_3,a_4) $ . Now, the complex number which is assigned, is assigned by the way I thought it did ? $ r(x) = -ix$ ? Then going on like : $Tf(x) = f(-ix) $
$endgroup$
– Rebellos
May 3 '16 at 20:17
$begingroup$
Oh I get it now. So we have a vector over that space which is $(a_1,a_2,a_3,a_4) $ . Now, the complex number which is assigned, is assigned by the way I thought it did ? $ r(x) = -ix$ ? Then going on like : $Tf(x) = f(-ix) $
$endgroup$
– Rebellos
May 3 '16 at 20:17
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Answering to my question after a great help from Jake, through comments and chat :
Let's say that $(a_1,a_2,a_3,a_4)$ are the four vertices of our square. Then, the operator that assigns them exactly to the next one will give back the vector $(a_2,a_3,a_4,a_1)$. (Note that the fourth one goes back to the first one). Then, a vector $v$ of the space $mathbb C^4$ that our vertices are assigned, is written as : $v = e_1a_2 + e_2a_3 + e_3a_4 + e_4a_1$.
From that, we get that the basis $B$ is the : $B=text{span}[(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)]$
And furthermore, the matrix of our operator T :
$ M_T =
begin{bmatrix}
0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 \
0 & 1 & 0 & 0 \
1 & 0 & 0 & 0
end{bmatrix} $
We get that exact matrix, because $(1,0,0,0)$ goes to $(0,0,0,1)$ etc.
After that, we can move on to solving the characteristic polynomial and finding the eigenvalues and eigenvectors.
answered May 3 '16 at 21:36
RebellosRebellos
14.5k31246
$endgroup$
add a comment |
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$begingroup$
Answering to my question after a great help from Jake, through comments and chat :
Let's say that $(a_1,a_2,a_3,a_4)$ are the four vertices of our square. Then, the operator that assigns them exactly to the next one will give back the vector $(a_2,a_3,a_4,a_1)$. (Note that the fourth one goes back to the first one). Then, a vector $v$ of the space $mathbb C^4$ that our vertices are assigned, is written as : $v = e_1a_2 + e_2a_3 + e_3a_4 + e_4a_1$.
From that, we get that the basis $B$ is the : $B=text{span}[(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)]$
And furthermore, the matrix of our operator T :
$ M_T =
begin{bmatrix}
0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 \
0 & 1 & 0 & 0 \
1 & 0 & 0 & 0
end{bmatrix} $
We get that exact matrix, because $(1,0,0,0)$ goes to $(0,0,0,1)$ etc.
After that, we can move on to solving the characteristic polynomial and finding the eigenvalues and eigenvectors.
answered May 3 '16 at 21:36
RebellosRebellos
14.5k31246
$endgroup$
add a comment |
$begingroup$
Answering to my question after a great help from Jake, through comments and chat :
Let's say that $(a_1,a_2,a_3,a_4)$ are the four vertices of our square. Then, the operator that assigns them exactly to the next one will give back the vector $(a_2,a_3,a_4,a_1)$. (Note that the fourth one goes back to the first one). Then, a vector $v$ of the space $mathbb C^4$ that our vertices are assigned, is written as : $v = e_1a_2 + e_2a_3 + e_3a_4 + e_4a_1$.
From that, we get that the basis $B$ is the : $B=text{span}[(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)]$
And furthermore, the matrix of our operator T :
$ M_T =
begin{bmatrix}
0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 \
0 & 1 & 0 & 0 \
1 & 0 & 0 & 0
end{bmatrix} $
We get that exact matrix, because $(1,0,0,0)$ goes to $(0,0,0,1)$ etc.
After that, we can move on to solving the characteristic polynomial and finding the eigenvalues and eigenvectors.
answered May 3 '16 at 21:36
RebellosRebellos
14.5k31246
$endgroup$
add a comment |
$begingroup$
Answering to my question after a great help from Jake, through comments and chat :
Let's say that $(a_1,a_2,a_3,a_4)$ are the four vertices of our square. Then, the operator that assigns them exactly to the next one will give back the vector $(a_2,a_3,a_4,a_1)$. (Note that the fourth one goes back to the first one). Then, a vector $v$ of the space $mathbb C^4$ that our vertices are assigned, is written as : $v = e_1a_2 + e_2a_3 + e_3a_4 + e_4a_1$.
From that, we get that the basis $B$ is the : $B=text{span}[(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)]$
And furthermore, the matrix of our operator T :
$ M_T =
begin{bmatrix}
0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 \
0 & 1 & 0 & 0 \
1 & 0 & 0 & 0
end{bmatrix} $
We get that exact matrix, because $(1,0,0,0)$ goes to $(0,0,0,1)$ etc.
After that, we can move on to solving the characteristic polynomial and finding the eigenvalues and eigenvectors.
answered May 3 '16 at 21:36
RebellosRebellos
14.5k31246
$endgroup$
Answering to my question after a great help from Jake, through comments and chat :
Let's say that $(a_1,a_2,a_3,a_4)$ are the four vertices of our square. Then, the operator that assigns them exactly to the next one will give back the vector $(a_2,a_3,a_4,a_1)$. (Note that the fourth one goes back to the first one). Then, a vector $v$ of the space $mathbb C^4$ that our vertices are assigned, is written as : $v = e_1a_2 + e_2a_3 + e_3a_4 + e_4a_1$.
From that, we get that the basis $B$ is the : $B=text{span}[(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)]$
And furthermore, the matrix of our operator T :
$ M_T =
begin{bmatrix}
0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 \
0 & 1 & 0 & 0 \
1 & 0 & 0 & 0
end{bmatrix} $
We get that exact matrix, because $(1,0,0,0)$ goes to $(0,0,0,1)$ etc.
After that, we can move on to solving the characteristic polynomial and finding the eigenvalues and eigenvectors.
answered May 3 '16 at 21:36
RebellosRebellos
14.5k31246
edited Dec 16 '18 at 19:53
answered May 3 '16 at 21:36
RebellosRebellos
14.5k31246
answered May 3 '16 at 21:36
RebellosRebellos
14.5k31246
answered May 3 '16 at 21:36
RebellosRebellos
14.5k31246
14.5k31246
add a comment |
add a comment |
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1
$begingroup$
A square has four vertices. Your space $V$ contains functions that assign a (complex) number to each of these four vertices. So an element of $V$ is just a tuple of four numbers! Does this sound familiar? Now can you give me a basis for $V$? Don't let the complex part bother you too much. It will not affect anything until the finding of the eigenvalues.
$endgroup$
– Ravi
May 3 '16 at 19:58
$begingroup$
Oh, let me give it a hand that way then !
$endgroup$
– Rebellos
May 3 '16 at 20:10
$begingroup$
By the way, what does "tuple" mean ? Because even the greek dictionary word is something that we don't use anymore :P
$endgroup$
– Rebellos
May 3 '16 at 20:11
$begingroup$
An $n$-tuple is a list of $n$ numbers. I.e. $(a_1,cdots ,a_n)$.
$endgroup$
– Ravi
May 3 '16 at 20:13
$begingroup$
Oh I get it now. So we have a vector over that space which is $(a_1,a_2,a_3,a_4) $ . Now, the complex number which is assigned, is assigned by the way I thought it did ? $ r(x) = -ix$ ? Then going on like : $Tf(x) = f(-ix) $
$endgroup$
– Rebellos
May 3 '16 at 20:17