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Let $f:Vto BbbR $, such that $fneq 0$. Prove that $f$ is surjective.
Clash Royale CLAN TAG#URR8PPP
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Let $V$ be a vector space over $BbbR$ such that $dim V=n,;ngeq 2$ and $$f:Vto BbbR $$ be linear such that $fneq 0.$
I want to
prove that $f$ is surjective,
prove or disprove that $f$ is injective,
I also want to find $dim ker f.$
I think that it suffices to prove that $operatornameImf= BbbR.$ We have the following inclusion:
$$operatornameImf= f(x)inBbbR:xin Vsubset BbbR,$$ but I'm finding it hard to prove that $$BbbRsubsetoperatornameImf= f(x):xin V.$$
For kernel, we have that
$ker f= xin V:f(x)=0,$ but I am stuck at this point.
No.3 is not a problem, since $dim V=dim ker f+ dim operatornameImf.$ Can anyone help me out?
linear-algebra
asked Jul 30 at 15:47
Mike
58712
add a comment |Â
up vote
0
down vote
favorite
Let $V$ be a vector space over $BbbR$ such that $dim V=n,;ngeq 2$ and $$f:Vto BbbR $$ be linear such that $fneq 0.$
I want to
prove that $f$ is surjective,
prove or disprove that $f$ is injective,
I also want to find $dim ker f.$
I think that it suffices to prove that $operatornameImf= BbbR.$ We have the following inclusion:
$$operatornameImf= f(x)inBbbR:xin Vsubset BbbR,$$ but I'm finding it hard to prove that $$BbbRsubsetoperatornameImf= f(x):xin V.$$
For kernel, we have that
$ker f= xin V:f(x)=0,$ but I am stuck at this point.
No.3 is not a problem, since $dim V=dim ker f+ dim operatornameImf.$ Can anyone help me out?
linear-algebra
asked Jul 30 at 15:47
Mike
58712
$V = BbbR^n $ and observe what happens in those case and then generalize
– Panchal Shamsundar
Jul 30 at 16:03
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $V$ be a vector space over $BbbR$ such that $dim V=n,;ngeq 2$ and $$f:Vto BbbR $$ be linear such that $fneq 0.$
I want to
prove that $f$ is surjective,
prove or disprove that $f$ is injective,
I also want to find $dim ker f.$
I think that it suffices to prove that $operatornameImf= BbbR.$ We have the following inclusion:
$$operatornameImf= f(x)inBbbR:xin Vsubset BbbR,$$ but I'm finding it hard to prove that $$BbbRsubsetoperatornameImf= f(x):xin V.$$
For kernel, we have that
$ker f= xin V:f(x)=0,$ but I am stuck at this point.
No.3 is not a problem, since $dim V=dim ker f+ dim operatornameImf.$ Can anyone help me out?
linear-algebra
asked Jul 30 at 15:47
Mike
58712
Let $V$ be a vector space over $BbbR$ such that $dim V=n,;ngeq 2$ and $$f:Vto BbbR $$ be linear such that $fneq 0.$
I want to
prove that $f$ is surjective,
prove or disprove that $f$ is injective,
I also want to find $dim ker f.$
I think that it suffices to prove that $operatornameImf= BbbR.$ We have the following inclusion:
$$operatornameImf= f(x)inBbbR:xin Vsubset BbbR,$$ but I'm finding it hard to prove that $$BbbRsubsetoperatornameImf= f(x):xin V.$$
For kernel, we have that
$ker f= xin V:f(x)=0,$ but I am stuck at this point.
No.3 is not a problem, since $dim V=dim ker f+ dim operatornameImf.$ Can anyone help me out?
linear-algebra
asked Jul 30 at 15:47
Mike
58712
asked Jul 30 at 15:47
Mike
58712
asked Jul 30 at 15:47
Mike
58712
asked Jul 30 at 15:47
Mike
58712
58712
$V = BbbR^n $ and observe what happens in those case and then generalize
– Panchal Shamsundar
Jul 30 at 16:03
add a comment |Â
$V = BbbR^n $ and observe what happens in those case and then generalize
– Panchal Shamsundar
Jul 30 at 16:03
$V = BbbR^n $ and observe what happens in those case and then generalize
– Panchal Shamsundar
Jul 30 at 16:03
$V = BbbR^n $ and observe what happens in those case and then generalize
– Panchal Shamsundar
Jul 30 at 16:03
add a comment |Â
2 Answers
2
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up vote
3
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For $1$, use the fact that $f(cx)=cf(x)$ for all $cinBbbR$.
For $2$, use the fact that $f$ is injective if and only if $dimker f=0$.
answered Jul 30 at 15:51
Servaes
20k33484
add a comment |Â
up vote
2
down vote
$f$ is surjective: Since $fneq 0$, there is $x in mathbbR$ such that there is $vin V: f(v)=x$. We have to find a preimage for an arbitrary $ain mathbbR$:
Just calculate $f(fracaxcdot v)$ by using linearity.
About injectiveness: A surjective linear map that is also injective is an isomorphism. Compare the dimensions of $mathbbR$ and $V$.
answered Jul 30 at 15:55
Babelfish
408112
additional you say more about dual map.
– Ninja hatori
Jul 30 at 15:57
1
@Ninjahatori Excuse me, I don't really understand your comment. There is no use of dual maps in my answer. I guess that the TO is not familiar to the dual space, given his question. Do you really think that incorporating dual maps would improve the answer?
– Babelfish
Jul 30 at 16:03
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
For $1$, use the fact that $f(cx)=cf(x)$ for all $cinBbbR$.
For $2$, use the fact that $f$ is injective if and only if $dimker f=0$.
answered Jul 30 at 15:51
Servaes
20k33484
add a comment |Â
up vote
3
down vote
For $1$, use the fact that $f(cx)=cf(x)$ for all $cinBbbR$.
For $2$, use the fact that $f$ is injective if and only if $dimker f=0$.
answered Jul 30 at 15:51
Servaes
20k33484
add a comment |Â
up vote
3
down vote
up vote
3
down vote
For $1$, use the fact that $f(cx)=cf(x)$ for all $cinBbbR$.
For $2$, use the fact that $f$ is injective if and only if $dimker f=0$.
answered Jul 30 at 15:51
Servaes
20k33484
For $1$, use the fact that $f(cx)=cf(x)$ for all $cinBbbR$.
For $2$, use the fact that $f$ is injective if and only if $dimker f=0$.
answered Jul 30 at 15:51
Servaes
20k33484
answered Jul 30 at 15:51
Servaes
20k33484
answered Jul 30 at 15:51
Servaes
20k33484
answered Jul 30 at 15:51
Servaes
20k33484
20k33484
add a comment |Â
add a comment |Â
up vote
2
down vote
$f$ is surjective: Since $fneq 0$, there is $x in mathbbR$ such that there is $vin V: f(v)=x$. We have to find a preimage for an arbitrary $ain mathbbR$:
Just calculate $f(fracaxcdot v)$ by using linearity.
About injectiveness: A surjective linear map that is also injective is an isomorphism. Compare the dimensions of $mathbbR$ and $V$.
answered Jul 30 at 15:55
Babelfish
408112
additional you say more about dual map.
– Ninja hatori
Jul 30 at 15:57
1
@Ninjahatori Excuse me, I don't really understand your comment. There is no use of dual maps in my answer. I guess that the TO is not familiar to the dual space, given his question. Do you really think that incorporating dual maps would improve the answer?
– Babelfish
Jul 30 at 16:03
add a comment |Â
up vote
2
down vote
$f$ is surjective: Since $fneq 0$, there is $x in mathbbR$ such that there is $vin V: f(v)=x$. We have to find a preimage for an arbitrary $ain mathbbR$:
Just calculate $f(fracaxcdot v)$ by using linearity.
About injectiveness: A surjective linear map that is also injective is an isomorphism. Compare the dimensions of $mathbbR$ and $V$.
answered Jul 30 at 15:55
Babelfish
408112
additional you say more about dual map.
– Ninja hatori
Jul 30 at 15:57
1
@Ninjahatori Excuse me, I don't really understand your comment. There is no use of dual maps in my answer. I guess that the TO is not familiar to the dual space, given his question. Do you really think that incorporating dual maps would improve the answer?
– Babelfish
Jul 30 at 16:03
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$f$ is surjective: Since $fneq 0$, there is $x in mathbbR$ such that there is $vin V: f(v)=x$. We have to find a preimage for an arbitrary $ain mathbbR$:
Just calculate $f(fracaxcdot v)$ by using linearity.
About injectiveness: A surjective linear map that is also injective is an isomorphism. Compare the dimensions of $mathbbR$ and $V$.
answered Jul 30 at 15:55
Babelfish
408112
$f$ is surjective: Since $fneq 0$, there is $x in mathbbR$ such that there is $vin V: f(v)=x$. We have to find a preimage for an arbitrary $ain mathbbR$:
Just calculate $f(fracaxcdot v)$ by using linearity.
About injectiveness: A surjective linear map that is also injective is an isomorphism. Compare the dimensions of $mathbbR$ and $V$.
answered Jul 30 at 15:55
Babelfish
408112
answered Jul 30 at 15:55
Babelfish
408112
answered Jul 30 at 15:55
Babelfish
408112
answered Jul 30 at 15:55
Babelfish
408112
408112
additional you say more about dual map.
– Ninja hatori
Jul 30 at 15:57
1
@Ninjahatori Excuse me, I don't really understand your comment. There is no use of dual maps in my answer. I guess that the TO is not familiar to the dual space, given his question. Do you really think that incorporating dual maps would improve the answer?
– Babelfish
Jul 30 at 16:03
add a comment |Â
additional you say more about dual map.
– Ninja hatori
Jul 30 at 15:57
1
@Ninjahatori Excuse me, I don't really understand your comment. There is no use of dual maps in my answer. I guess that the TO is not familiar to the dual space, given his question. Do you really think that incorporating dual maps would improve the answer?
– Babelfish
Jul 30 at 16:03
additional you say more about dual map.
– Ninja hatori
Jul 30 at 15:57
additional you say more about dual map.
– Ninja hatori
Jul 30 at 15:57
1
1
@Ninjahatori Excuse me, I don't really understand your comment. There is no use of dual maps in my answer. I guess that the TO is not familiar to the dual space, given his question. Do you really think that incorporating dual maps would improve the answer?
– Babelfish
Jul 30 at 16:03
@Ninjahatori Excuse me, I don't really understand your comment. There is no use of dual maps in my answer. I guess that the TO is not familiar to the dual space, given his question. Do you really think that incorporating dual maps would improve the answer?
– Babelfish
Jul 30 at 16:03
add a comment |Â
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$V = BbbR^n $ and observe what happens in those case and then generalize
– Panchal Shamsundar
Jul 30 at 16:03