For which numbers $n in mathbb N$ there is a linear map $f_{n}: mathbb R^{n} rightarrow mathbb R^{n}$ such...
$begingroup$
My idea:
When $ker f_{n}=operatorname{image}(f_{n})$ then $dim(ker(f_{n}))=dim(operatorname{image}(f_{n}))$. Moreover I know that for $g: V rightarrow W$ I have $dim(V)=dim (ker g) + dim(operatorname{image} g)$, so in this case: $n=2dim(ker f)$, so this task is true for even $n$.
Unfortunately I am afraid that it is incorrect solution and please rate it.
linear-algebra
edited Jan 13 at 3:16
Namaste
1
asked Jan 12 at 23:23
MP3129MP3129
854211
$endgroup$
add a comment |
$begingroup$
My idea:
When $ker f_{n}=operatorname{image}(f_{n})$ then $dim(ker(f_{n}))=dim(operatorname{image}(f_{n}))$. Moreover I know that for $g: V rightarrow W$ I have $dim(V)=dim (ker g) + dim(operatorname{image} g)$, so in this case: $n=2dim(ker f)$, so this task is true for even $n$.
Unfortunately I am afraid that it is incorrect solution and please rate it.
linear-algebra
edited Jan 13 at 3:16
Namaste
1
asked Jan 12 at 23:23
MP3129MP3129
854211
$endgroup$
add a comment |
$begingroup$
My idea:
When $ker f_{n}=operatorname{image}(f_{n})$ then $dim(ker(f_{n}))=dim(operatorname{image}(f_{n}))$. Moreover I know that for $g: V rightarrow W$ I have $dim(V)=dim (ker g) + dim(operatorname{image} g)$, so in this case: $n=2dim(ker f)$, so this task is true for even $n$.
Unfortunately I am afraid that it is incorrect solution and please rate it.
linear-algebra
edited Jan 13 at 3:16
Namaste
1
asked Jan 12 at 23:23
MP3129MP3129
854211
$endgroup$
My idea:
When $ker f_{n}=operatorname{image}(f_{n})$ then $dim(ker(f_{n}))=dim(operatorname{image}(f_{n}))$. Moreover I know that for $g: V rightarrow W$ I have $dim(V)=dim (ker g) + dim(operatorname{image} g)$, so in this case: $n=2dim(ker f)$, so this task is true for even $n$.
Unfortunately I am afraid that it is incorrect solution and please rate it.
linear-algebra
linear-algebra
edited Jan 13 at 3:16
Namaste
1
asked Jan 12 at 23:23
MP3129MP3129
854211
edited Jan 13 at 3:16
Namaste
1
asked Jan 12 at 23:23
MP3129MP3129
854211
edited Jan 13 at 3:16
Namaste
1
edited Jan 13 at 3:16
Namaste
1
edited Jan 13 at 3:16
Namaste
1
1
asked Jan 12 at 23:23
MP3129MP3129
854211
asked Jan 12 at 23:23
MP3129MP3129
854211
asked Jan 12 at 23:23
MP3129MP3129
854211
854211
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
The assertion is correct. It has been shown already that $n$ must be even. For the converse part consider $V=mathbb R^{2m}$. Define $f: V to V$ by $f(x_1,x_2,cdots,x_{2m})=(x_{m+1},x_{m+2},cdots, x_{2m},0,0, cdots,0)$. Then $f$ is linear and its range coincides with the kernel.
answered Jan 12 at 23:48
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Why I have to consider the converse part?
$endgroup$
– MP3129
Jan 13 at 0:05
1
$begingroup$
'For which numbers $n$' means you have to precisely identify all $n$ with this property. It is not enough to say that if the result is true then $n$ must be even. You also have prove that such a linear functional exists for every even $n$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 0:10
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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$begingroup$
The assertion is correct. It has been shown already that $n$ must be even. For the converse part consider $V=mathbb R^{2m}$. Define $f: V to V$ by $f(x_1,x_2,cdots,x_{2m})=(x_{m+1},x_{m+2},cdots, x_{2m},0,0, cdots,0)$. Then $f$ is linear and its range coincides with the kernel.
answered Jan 12 at 23:48
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Why I have to consider the converse part?
$endgroup$
– MP3129
Jan 13 at 0:05
1
$begingroup$
'For which numbers $n$' means you have to precisely identify all $n$ with this property. It is not enough to say that if the result is true then $n$ must be even. You also have prove that such a linear functional exists for every even $n$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 0:10
add a comment |
$begingroup$
The assertion is correct. It has been shown already that $n$ must be even. For the converse part consider $V=mathbb R^{2m}$. Define $f: V to V$ by $f(x_1,x_2,cdots,x_{2m})=(x_{m+1},x_{m+2},cdots, x_{2m},0,0, cdots,0)$. Then $f$ is linear and its range coincides with the kernel.
answered Jan 12 at 23:48
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Why I have to consider the converse part?
$endgroup$
– MP3129
Jan 13 at 0:05
1
$begingroup$
'For which numbers $n$' means you have to precisely identify all $n$ with this property. It is not enough to say that if the result is true then $n$ must be even. You also have prove that such a linear functional exists for every even $n$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 0:10
add a comment |
$begingroup$
The assertion is correct. It has been shown already that $n$ must be even. For the converse part consider $V=mathbb R^{2m}$. Define $f: V to V$ by $f(x_1,x_2,cdots,x_{2m})=(x_{m+1},x_{m+2},cdots, x_{2m},0,0, cdots,0)$. Then $f$ is linear and its range coincides with the kernel.
answered Jan 12 at 23:48
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
The assertion is correct. It has been shown already that $n$ must be even. For the converse part consider $V=mathbb R^{2m}$. Define $f: V to V$ by $f(x_1,x_2,cdots,x_{2m})=(x_{m+1},x_{m+2},cdots, x_{2m},0,0, cdots,0)$. Then $f$ is linear and its range coincides with the kernel.
answered Jan 12 at 23:48
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Jan 12 at 23:48
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Jan 12 at 23:48
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Jan 12 at 23:48
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
74.9k53270
$begingroup$
Why I have to consider the converse part?
$endgroup$
– MP3129
Jan 13 at 0:05
1
$begingroup$
'For which numbers $n$' means you have to precisely identify all $n$ with this property. It is not enough to say that if the result is true then $n$ must be even. You also have prove that such a linear functional exists for every even $n$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 0:10
add a comment |
$begingroup$
Why I have to consider the converse part?
$endgroup$
– MP3129
Jan 13 at 0:05
1
$begingroup$
'For which numbers $n$' means you have to precisely identify all $n$ with this property. It is not enough to say that if the result is true then $n$ must be even. You also have prove that such a linear functional exists for every even $n$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 0:10
$begingroup$
Why I have to consider the converse part?
$endgroup$
– MP3129
Jan 13 at 0:05
$begingroup$
Why I have to consider the converse part?
$endgroup$
– MP3129
Jan 13 at 0:05
1
1
$begingroup$
'For which numbers $n$' means you have to precisely identify all $n$ with this property. It is not enough to say that if the result is true then $n$ must be even. You also have prove that such a linear functional exists for every even $n$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 0:10
$begingroup$
'For which numbers $n$' means you have to precisely identify all $n$ with this property. It is not enough to say that if the result is true then $n$ must be even. You also have prove that such a linear functional exists for every even $n$.
$endgroup$
– Kavi Rama Murthy
Jan 13 at 0:10
add a comment |
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