Mixing
Linear operator statement
$begingroup$
I'm stuck with a practice problem question:
'' Is it true that a linear operator, T : V → V, on one-dimensional vector space V over field F has the form T(v) = av, for all v ∈ V and a a scalar from the field?''
I think this statement is true? From what i've learned, linear operations keep the addition and scalar multiplication. If the linear operation goes from V to V then we call it a linear operator? T(v) = av in the question means we're applying a linear transformation (av) on V and we obtain V. Am i right?
linear-algebra
asked Jan 20 at 23:03
ximxim
516
$endgroup$
add a comment |
$begingroup$
I'm stuck with a practice problem question:
'' Is it true that a linear operator, T : V → V, on one-dimensional vector space V over field F has the form T(v) = av, for all v ∈ V and a a scalar from the field?''
I think this statement is true? From what i've learned, linear operations keep the addition and scalar multiplication. If the linear operation goes from V to V then we call it a linear operator? T(v) = av in the question means we're applying a linear transformation (av) on V and we obtain V. Am i right?
linear-algebra
asked Jan 20 at 23:03
ximxim
516
$endgroup$
$begingroup$
Do you think that what you wrote is a proof of the statement?
$endgroup$
– José Carlos Santos
Jan 20 at 23:09
$begingroup$
Your answer (yes) is correct but your reasoning is not. There are many ways to prove the statement. Try starting by describing a 1-dimensional vector space. Then look up the definition of a linear operator. You should see what to do after that.
$endgroup$
– Ben W
Jan 20 at 23:12
$begingroup$
I thought that when it's false, you give a counter example and if it's true you state the 'theorem' or the 'theoretical statement'. From what i'm seeing, T : V → V is keeping the same dimension after the linear transformation so it's true that it's one dimensional. And if i would take any scalar from the field and use another vector space in that same field, the linear transformation still keeps the dimension from the preimage.
$endgroup$
– xim
Jan 20 at 23:14
$begingroup$
You need to be more careful. What does "keeping the same dimension" mean? It sounds like you think $V$ and $TV$ must have the same dimension but this is not so. And what do you mean, "another vector space"? Unless you want to try to use the zero subspace (which you shouldn't), you are only working with one vector space $V$.
$endgroup$
– Ben W
Jan 20 at 23:18
$begingroup$
If i would take a matrice 2 by 2 and apply a linear transformation to it, for example reflexion, the end result would still be a matrice 2 by 2 so we stay in the same dimension.
$endgroup$
– xim
Jan 20 at 23:29
add a comment |
$begingroup$
I'm stuck with a practice problem question:
'' Is it true that a linear operator, T : V → V, on one-dimensional vector space V over field F has the form T(v) = av, for all v ∈ V and a a scalar from the field?''
I think this statement is true? From what i've learned, linear operations keep the addition and scalar multiplication. If the linear operation goes from V to V then we call it a linear operator? T(v) = av in the question means we're applying a linear transformation (av) on V and we obtain V. Am i right?
linear-algebra
asked Jan 20 at 23:03
ximxim
516
$endgroup$
I'm stuck with a practice problem question:
'' Is it true that a linear operator, T : V → V, on one-dimensional vector space V over field F has the form T(v) = av, for all v ∈ V and a a scalar from the field?''
I think this statement is true? From what i've learned, linear operations keep the addition and scalar multiplication. If the linear operation goes from V to V then we call it a linear operator? T(v) = av in the question means we're applying a linear transformation (av) on V and we obtain V. Am i right?
linear-algebra
linear-algebra
asked Jan 20 at 23:03
ximxim
516
asked Jan 20 at 23:03
ximxim
516
asked Jan 20 at 23:03
ximxim
516
asked Jan 20 at 23:03
ximxim
516
asked Jan 20 at 23:03
ximxim
516
516
$begingroup$
Do you think that what you wrote is a proof of the statement?
$endgroup$
– José Carlos Santos
Jan 20 at 23:09
$begingroup$
Your answer (yes) is correct but your reasoning is not. There are many ways to prove the statement. Try starting by describing a 1-dimensional vector space. Then look up the definition of a linear operator. You should see what to do after that.
$endgroup$
– Ben W
Jan 20 at 23:12
$begingroup$
I thought that when it's false, you give a counter example and if it's true you state the 'theorem' or the 'theoretical statement'. From what i'm seeing, T : V → V is keeping the same dimension after the linear transformation so it's true that it's one dimensional. And if i would take any scalar from the field and use another vector space in that same field, the linear transformation still keeps the dimension from the preimage.
$endgroup$
– xim
Jan 20 at 23:14
$begingroup$
You need to be more careful. What does "keeping the same dimension" mean? It sounds like you think $V$ and $TV$ must have the same dimension but this is not so. And what do you mean, "another vector space"? Unless you want to try to use the zero subspace (which you shouldn't), you are only working with one vector space $V$.
$endgroup$
– Ben W
Jan 20 at 23:18
$begingroup$
If i would take a matrice 2 by 2 and apply a linear transformation to it, for example reflexion, the end result would still be a matrice 2 by 2 so we stay in the same dimension.
$endgroup$
– xim
Jan 20 at 23:29
add a comment |
$begingroup$
Do you think that what you wrote is a proof of the statement?
$endgroup$
– José Carlos Santos
Jan 20 at 23:09
$begingroup$
Your answer (yes) is correct but your reasoning is not. There are many ways to prove the statement. Try starting by describing a 1-dimensional vector space. Then look up the definition of a linear operator. You should see what to do after that.
$endgroup$
– Ben W
Jan 20 at 23:12
$begingroup$
I thought that when it's false, you give a counter example and if it's true you state the 'theorem' or the 'theoretical statement'. From what i'm seeing, T : V → V is keeping the same dimension after the linear transformation so it's true that it's one dimensional. And if i would take any scalar from the field and use another vector space in that same field, the linear transformation still keeps the dimension from the preimage.
$endgroup$
– xim
Jan 20 at 23:14
$begingroup$
You need to be more careful. What does "keeping the same dimension" mean? It sounds like you think $V$ and $TV$ must have the same dimension but this is not so. And what do you mean, "another vector space"? Unless you want to try to use the zero subspace (which you shouldn't), you are only working with one vector space $V$.
$endgroup$
– Ben W
Jan 20 at 23:18
$begingroup$
If i would take a matrice 2 by 2 and apply a linear transformation to it, for example reflexion, the end result would still be a matrice 2 by 2 so we stay in the same dimension.
$endgroup$
– xim
Jan 20 at 23:29
$begingroup$
Do you think that what you wrote is a proof of the statement?
$endgroup$
– José Carlos Santos
Jan 20 at 23:09
$begingroup$
Do you think that what you wrote is a proof of the statement?
$endgroup$
– José Carlos Santos
Jan 20 at 23:09
$begingroup$
Your answer (yes) is correct but your reasoning is not. There are many ways to prove the statement. Try starting by describing a 1-dimensional vector space. Then look up the definition of a linear operator. You should see what to do after that.
$endgroup$
– Ben W
Jan 20 at 23:12
$begingroup$
Your answer (yes) is correct but your reasoning is not. There are many ways to prove the statement. Try starting by describing a 1-dimensional vector space. Then look up the definition of a linear operator. You should see what to do after that.
$endgroup$
– Ben W
Jan 20 at 23:12
$begingroup$
I thought that when it's false, you give a counter example and if it's true you state the 'theorem' or the 'theoretical statement'. From what i'm seeing, T : V → V is keeping the same dimension after the linear transformation so it's true that it's one dimensional. And if i would take any scalar from the field and use another vector space in that same field, the linear transformation still keeps the dimension from the preimage.
$endgroup$
– xim
Jan 20 at 23:14
$begingroup$
I thought that when it's false, you give a counter example and if it's true you state the 'theorem' or the 'theoretical statement'. From what i'm seeing, T : V → V is keeping the same dimension after the linear transformation so it's true that it's one dimensional. And if i would take any scalar from the field and use another vector space in that same field, the linear transformation still keeps the dimension from the preimage.
$endgroup$
– xim
Jan 20 at 23:14
$begingroup$
You need to be more careful. What does "keeping the same dimension" mean? It sounds like you think $V$ and $TV$ must have the same dimension but this is not so. And what do you mean, "another vector space"? Unless you want to try to use the zero subspace (which you shouldn't), you are only working with one vector space $V$.
$endgroup$
– Ben W
Jan 20 at 23:18
$begingroup$
You need to be more careful. What does "keeping the same dimension" mean? It sounds like you think $V$ and $TV$ must have the same dimension but this is not so. And what do you mean, "another vector space"? Unless you want to try to use the zero subspace (which you shouldn't), you are only working with one vector space $V$.
$endgroup$
– Ben W
Jan 20 at 23:18
$begingroup$
If i would take a matrice 2 by 2 and apply a linear transformation to it, for example reflexion, the end result would still be a matrice 2 by 2 so we stay in the same dimension.
$endgroup$
– xim
Jan 20 at 23:29
$begingroup$
If i would take a matrice 2 by 2 and apply a linear transformation to it, for example reflexion, the end result would still be a matrice 2 by 2 so we stay in the same dimension.
$endgroup$
– xim
Jan 20 at 23:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, this is true. Choose an arbitrary vector $x in V setminus {0}$. It will be mapped to some $y in V$. As $V$ is one-dimensional, $y$ needs to be some multiple of $x$, say $y = ax$. Let $v in V$ and write $v = lambda x$. By linearity, we get $$T(v) = T(lambda x) = lambda T(x) = lambda y = lambda ax = av.$$
edited Jan 20 at 23:26
J. W. Tanner
2,8111217
answered Jan 20 at 23:15
KlausKlaus
2,12711
$endgroup$
add a comment |
$begingroup$
You know everything about a linear operator if you know where it sends basis vectors.
One-dimensional vector space $V$ has a single basis vector $e$, which means that any other vector $v in V$ has a form $v=ke$, where $k in F$. Where can $T : V rightarrow V$ send vector $e$? Only to some vector $ke$, there are no other guys in $V$.
answered Jan 20 at 23:13
ZeeklessZeekless
577111
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
Yes, this is true. Choose an arbitrary vector $x in V setminus {0}$. It will be mapped to some $y in V$. As $V$ is one-dimensional, $y$ needs to be some multiple of $x$, say $y = ax$. Let $v in V$ and write $v = lambda x$. By linearity, we get $$T(v) = T(lambda x) = lambda T(x) = lambda y = lambda ax = av.$$
edited Jan 20 at 23:26
J. W. Tanner
2,8111217
answered Jan 20 at 23:15
KlausKlaus
2,12711
$endgroup$
add a comment |
$begingroup$
Yes, this is true. Choose an arbitrary vector $x in V setminus {0}$. It will be mapped to some $y in V$. As $V$ is one-dimensional, $y$ needs to be some multiple of $x$, say $y = ax$. Let $v in V$ and write $v = lambda x$. By linearity, we get $$T(v) = T(lambda x) = lambda T(x) = lambda y = lambda ax = av.$$
edited Jan 20 at 23:26
J. W. Tanner
2,8111217
answered Jan 20 at 23:15
KlausKlaus
2,12711
$endgroup$
add a comment |
$begingroup$
Yes, this is true. Choose an arbitrary vector $x in V setminus {0}$. It will be mapped to some $y in V$. As $V$ is one-dimensional, $y$ needs to be some multiple of $x$, say $y = ax$. Let $v in V$ and write $v = lambda x$. By linearity, we get $$T(v) = T(lambda x) = lambda T(x) = lambda y = lambda ax = av.$$
edited Jan 20 at 23:26
J. W. Tanner
2,8111217
answered Jan 20 at 23:15
KlausKlaus
2,12711
$endgroup$
Yes, this is true. Choose an arbitrary vector $x in V setminus {0}$. It will be mapped to some $y in V$. As $V$ is one-dimensional, $y$ needs to be some multiple of $x$, say $y = ax$. Let $v in V$ and write $v = lambda x$. By linearity, we get $$T(v) = T(lambda x) = lambda T(x) = lambda y = lambda ax = av.$$
edited Jan 20 at 23:26
J. W. Tanner
2,8111217
answered Jan 20 at 23:15
KlausKlaus
2,12711
edited Jan 20 at 23:26
J. W. Tanner
2,8111217
edited Jan 20 at 23:26
J. W. Tanner
2,8111217
edited Jan 20 at 23:26
J. W. Tanner
2,8111217
2,8111217
answered Jan 20 at 23:15
KlausKlaus
2,12711
answered Jan 20 at 23:15
KlausKlaus
2,12711
answered Jan 20 at 23:15
KlausKlaus
2,12711
2,12711
add a comment |
add a comment |
$begingroup$
You know everything about a linear operator if you know where it sends basis vectors.
One-dimensional vector space $V$ has a single basis vector $e$, which means that any other vector $v in V$ has a form $v=ke$, where $k in F$. Where can $T : V rightarrow V$ send vector $e$? Only to some vector $ke$, there are no other guys in $V$.
answered Jan 20 at 23:13
ZeeklessZeekless
577111
$endgroup$
add a comment |
$begingroup$
You know everything about a linear operator if you know where it sends basis vectors.
One-dimensional vector space $V$ has a single basis vector $e$, which means that any other vector $v in V$ has a form $v=ke$, where $k in F$. Where can $T : V rightarrow V$ send vector $e$? Only to some vector $ke$, there are no other guys in $V$.
answered Jan 20 at 23:13
ZeeklessZeekless
577111
$endgroup$
add a comment |
$begingroup$
You know everything about a linear operator if you know where it sends basis vectors.
One-dimensional vector space $V$ has a single basis vector $e$, which means that any other vector $v in V$ has a form $v=ke$, where $k in F$. Where can $T : V rightarrow V$ send vector $e$? Only to some vector $ke$, there are no other guys in $V$.
answered Jan 20 at 23:13
ZeeklessZeekless
577111
$endgroup$
You know everything about a linear operator if you know where it sends basis vectors.
One-dimensional vector space $V$ has a single basis vector $e$, which means that any other vector $v in V$ has a form $v=ke$, where $k in F$. Where can $T : V rightarrow V$ send vector $e$? Only to some vector $ke$, there are no other guys in $V$.
answered Jan 20 at 23:13
ZeeklessZeekless
577111
answered Jan 20 at 23:13
ZeeklessZeekless
577111
answered Jan 20 at 23:13
ZeeklessZeekless
577111
answered Jan 20 at 23:13
ZeeklessZeekless
577111
577111
add a comment |
add a comment |
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$begingroup$
Do you think that what you wrote is a proof of the statement?
$endgroup$
– José Carlos Santos
Jan 20 at 23:09
$begingroup$
Your answer (yes) is correct but your reasoning is not. There are many ways to prove the statement. Try starting by describing a 1-dimensional vector space. Then look up the definition of a linear operator. You should see what to do after that.
$endgroup$
– Ben W
Jan 20 at 23:12
$begingroup$
I thought that when it's false, you give a counter example and if it's true you state the 'theorem' or the 'theoretical statement'. From what i'm seeing, T : V → V is keeping the same dimension after the linear transformation so it's true that it's one dimensional. And if i would take any scalar from the field and use another vector space in that same field, the linear transformation still keeps the dimension from the preimage.
$endgroup$
– xim
Jan 20 at 23:14
$begingroup$
You need to be more careful. What does "keeping the same dimension" mean? It sounds like you think $V$ and $TV$ must have the same dimension but this is not so. And what do you mean, "another vector space"? Unless you want to try to use the zero subspace (which you shouldn't), you are only working with one vector space $V$.
$endgroup$
– Ben W
Jan 20 at 23:18
$begingroup$
If i would take a matrice 2 by 2 and apply a linear transformation to it, for example reflexion, the end result would still be a matrice 2 by 2 so we stay in the same dimension.
$endgroup$
– xim
Jan 20 at 23:29