Mixing
Base of V contains a base of Ker f and a base of Imf
$begingroup$
If we have a linear map $f:Vrightarrow W$, where {$x_1, x_2, ldots, x_k$} is a base of Ker $f$ and we complete this base such that ${x_1, ldots, x_k, x_{k+1}, ldots, x_n}$ is a base in $V$, how we can prove that ${f(x_{k+1}), f(x_{k+2}), ldots, f(x_n)}$ is a base in Im $f$? Where Im $f$ is the image of function $f$.
linear-algebra vector-spaces
edited Jan 10 at 18:47
Viktor Glombik
8211527
asked Jan 10 at 18:34
Raul1998Raul1998
63
$endgroup$
add a comment |
$begingroup$
If we have a linear map $f:Vrightarrow W$, where {$x_1, x_2, ldots, x_k$} is a base of Ker $f$ and we complete this base such that ${x_1, ldots, x_k, x_{k+1}, ldots, x_n}$ is a base in $V$, how we can prove that ${f(x_{k+1}), f(x_{k+2}), ldots, f(x_n)}$ is a base in Im $f$? Where Im $f$ is the image of function $f$.
linear-algebra vector-spaces
edited Jan 10 at 18:47
Viktor Glombik
8211527
asked Jan 10 at 18:34
Raul1998Raul1998
63
$endgroup$
add a comment |
$begingroup$
If we have a linear map $f:Vrightarrow W$, where {$x_1, x_2, ldots, x_k$} is a base of Ker $f$ and we complete this base such that ${x_1, ldots, x_k, x_{k+1}, ldots, x_n}$ is a base in $V$, how we can prove that ${f(x_{k+1}), f(x_{k+2}), ldots, f(x_n)}$ is a base in Im $f$? Where Im $f$ is the image of function $f$.
linear-algebra vector-spaces
edited Jan 10 at 18:47
Viktor Glombik
8211527
asked Jan 10 at 18:34
Raul1998Raul1998
63
$endgroup$
If we have a linear map $f:Vrightarrow W$, where {$x_1, x_2, ldots, x_k$} is a base of Ker $f$ and we complete this base such that ${x_1, ldots, x_k, x_{k+1}, ldots, x_n}$ is a base in $V$, how we can prove that ${f(x_{k+1}), f(x_{k+2}), ldots, f(x_n)}$ is a base in Im $f$? Where Im $f$ is the image of function $f$.
linear-algebra vector-spaces
linear-algebra vector-spaces
edited Jan 10 at 18:47
Viktor Glombik
8211527
asked Jan 10 at 18:34
Raul1998Raul1998
63
edited Jan 10 at 18:47
Viktor Glombik
8211527
asked Jan 10 at 18:34
Raul1998Raul1998
63
edited Jan 10 at 18:47
Viktor Glombik
8211527
edited Jan 10 at 18:47
Viktor Glombik
8211527
edited Jan 10 at 18:47
Viktor Glombik
8211527
8211527
asked Jan 10 at 18:34
Raul1998Raul1998
63
asked Jan 10 at 18:34
Raul1998Raul1998
63
asked Jan 10 at 18:34
Raul1998Raul1998
63
63
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $v_1,v_2,...,v_n$ be the basis of $V$. To begin with let's prove that $Im(V)=Lin{(L(v_1),...,L(v_n))}$
$a_1L(v_1)+a_2L(v_2)+...+a_nL(v_n)=L(a_1v_1+a_2v_2+...+a_nv_n)=u$
By definition what makes the image of L are the vectors u for which there exists a vector s such that $L(s)=u$, so the statement holds.
Now if vectors $v_1,v_2...v_k$ belong to the kernel then $L(v_1),...,L(v_k)$ and their linear combinations are mapped to $0$. That leaves the remaining $Lin{(L(v_{k+1}),...,L(v_n))}$ to span the image. And since $dim(Ker(L))+dim(Im(L))=dim(V)$ and we have $n-k$ vectors then it is also the smallest linearly independent set which makes it the basis.
answered Jan 10 at 18:47
The CatThe Cat
25112
$endgroup$
add a comment |
$begingroup$
Since ${x_1,dots,x_k,x_{k+1},dots,x_n}$ is a spanning set of $V$, the set
$$
{f(x_1),dots,f(x_k),f(x_{k+1}),dots,f(x_n)}
$$
is a spanning set for the image of $f$. However, $f(x_1)=dots=f(x_k)=0$, so also ${f(x_{k+1},dots,f(x_n)}$ is a spanning set for the image of $f$.
By the rank-nullity theorem, $dimoperatorname{im}f=n-k$, so the set has to be a basis.
If you can't use the rank-nullity theorem, directly prove that the set is linearly independent: if $alpha_{m+1}f(x_{m+1})+dots+alpha_nf(x_n)=0$, then
$$
alpha_{m+1}x_{m+1}+dots+alpha_nx_ninker f
$$
and it's easy to finish.
answered Jan 10 at 22:12
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
Hint:
$V/Ker(f) cong Im(f)$.
Then completing your basis in the way you mentioned, what is a basis for $V/Ker(f)$? Then, consider how the above isomorphism is given.
answered Jan 10 at 18:39
MariahMariah
1,4871618
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $v_1,v_2,...,v_n$ be the basis of $V$. To begin with let's prove that $Im(V)=Lin{(L(v_1),...,L(v_n))}$
$a_1L(v_1)+a_2L(v_2)+...+a_nL(v_n)=L(a_1v_1+a_2v_2+...+a_nv_n)=u$
By definition what makes the image of L are the vectors u for which there exists a vector s such that $L(s)=u$, so the statement holds.
Now if vectors $v_1,v_2...v_k$ belong to the kernel then $L(v_1),...,L(v_k)$ and their linear combinations are mapped to $0$. That leaves the remaining $Lin{(L(v_{k+1}),...,L(v_n))}$ to span the image. And since $dim(Ker(L))+dim(Im(L))=dim(V)$ and we have $n-k$ vectors then it is also the smallest linearly independent set which makes it the basis.
answered Jan 10 at 18:47
The CatThe Cat
25112
$endgroup$
add a comment |
$begingroup$
Let $v_1,v_2,...,v_n$ be the basis of $V$. To begin with let's prove that $Im(V)=Lin{(L(v_1),...,L(v_n))}$
$a_1L(v_1)+a_2L(v_2)+...+a_nL(v_n)=L(a_1v_1+a_2v_2+...+a_nv_n)=u$
By definition what makes the image of L are the vectors u for which there exists a vector s such that $L(s)=u$, so the statement holds.
Now if vectors $v_1,v_2...v_k$ belong to the kernel then $L(v_1),...,L(v_k)$ and their linear combinations are mapped to $0$. That leaves the remaining $Lin{(L(v_{k+1}),...,L(v_n))}$ to span the image. And since $dim(Ker(L))+dim(Im(L))=dim(V)$ and we have $n-k$ vectors then it is also the smallest linearly independent set which makes it the basis.
answered Jan 10 at 18:47
The CatThe Cat
25112
$endgroup$
add a comment |
$begingroup$
Let $v_1,v_2,...,v_n$ be the basis of $V$. To begin with let's prove that $Im(V)=Lin{(L(v_1),...,L(v_n))}$
$a_1L(v_1)+a_2L(v_2)+...+a_nL(v_n)=L(a_1v_1+a_2v_2+...+a_nv_n)=u$
By definition what makes the image of L are the vectors u for which there exists a vector s such that $L(s)=u$, so the statement holds.
Now if vectors $v_1,v_2...v_k$ belong to the kernel then $L(v_1),...,L(v_k)$ and their linear combinations are mapped to $0$. That leaves the remaining $Lin{(L(v_{k+1}),...,L(v_n))}$ to span the image. And since $dim(Ker(L))+dim(Im(L))=dim(V)$ and we have $n-k$ vectors then it is also the smallest linearly independent set which makes it the basis.
answered Jan 10 at 18:47
The CatThe Cat
25112
$endgroup$
Let $v_1,v_2,...,v_n$ be the basis of $V$. To begin with let's prove that $Im(V)=Lin{(L(v_1),...,L(v_n))}$
$a_1L(v_1)+a_2L(v_2)+...+a_nL(v_n)=L(a_1v_1+a_2v_2+...+a_nv_n)=u$
By definition what makes the image of L are the vectors u for which there exists a vector s such that $L(s)=u$, so the statement holds.
Now if vectors $v_1,v_2...v_k$ belong to the kernel then $L(v_1),...,L(v_k)$ and their linear combinations are mapped to $0$. That leaves the remaining $Lin{(L(v_{k+1}),...,L(v_n))}$ to span the image. And since $dim(Ker(L))+dim(Im(L))=dim(V)$ and we have $n-k$ vectors then it is also the smallest linearly independent set which makes it the basis.
answered Jan 10 at 18:47
The CatThe Cat
25112
edited Jan 10 at 18:58
answered Jan 10 at 18:47
The CatThe Cat
25112
answered Jan 10 at 18:47
The CatThe Cat
25112
answered Jan 10 at 18:47
The CatThe Cat
25112
25112
add a comment |
add a comment |
$begingroup$
Since ${x_1,dots,x_k,x_{k+1},dots,x_n}$ is a spanning set of $V$, the set
$$
{f(x_1),dots,f(x_k),f(x_{k+1}),dots,f(x_n)}
$$
is a spanning set for the image of $f$. However, $f(x_1)=dots=f(x_k)=0$, so also ${f(x_{k+1},dots,f(x_n)}$ is a spanning set for the image of $f$.
By the rank-nullity theorem, $dimoperatorname{im}f=n-k$, so the set has to be a basis.
If you can't use the rank-nullity theorem, directly prove that the set is linearly independent: if $alpha_{m+1}f(x_{m+1})+dots+alpha_nf(x_n)=0$, then
$$
alpha_{m+1}x_{m+1}+dots+alpha_nx_ninker f
$$
and it's easy to finish.
answered Jan 10 at 22:12
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
Since ${x_1,dots,x_k,x_{k+1},dots,x_n}$ is a spanning set of $V$, the set
$$
{f(x_1),dots,f(x_k),f(x_{k+1}),dots,f(x_n)}
$$
is a spanning set for the image of $f$. However, $f(x_1)=dots=f(x_k)=0$, so also ${f(x_{k+1},dots,f(x_n)}$ is a spanning set for the image of $f$.
By the rank-nullity theorem, $dimoperatorname{im}f=n-k$, so the set has to be a basis.
If you can't use the rank-nullity theorem, directly prove that the set is linearly independent: if $alpha_{m+1}f(x_{m+1})+dots+alpha_nf(x_n)=0$, then
$$
alpha_{m+1}x_{m+1}+dots+alpha_nx_ninker f
$$
and it's easy to finish.
answered Jan 10 at 22:12
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
Since ${x_1,dots,x_k,x_{k+1},dots,x_n}$ is a spanning set of $V$, the set
$$
{f(x_1),dots,f(x_k),f(x_{k+1}),dots,f(x_n)}
$$
is a spanning set for the image of $f$. However, $f(x_1)=dots=f(x_k)=0$, so also ${f(x_{k+1},dots,f(x_n)}$ is a spanning set for the image of $f$.
By the rank-nullity theorem, $dimoperatorname{im}f=n-k$, so the set has to be a basis.
If you can't use the rank-nullity theorem, directly prove that the set is linearly independent: if $alpha_{m+1}f(x_{m+1})+dots+alpha_nf(x_n)=0$, then
$$
alpha_{m+1}x_{m+1}+dots+alpha_nx_ninker f
$$
and it's easy to finish.
answered Jan 10 at 22:12
egregegreg
181k1485203
$endgroup$
Since ${x_1,dots,x_k,x_{k+1},dots,x_n}$ is a spanning set of $V$, the set
$$
{f(x_1),dots,f(x_k),f(x_{k+1}),dots,f(x_n)}
$$
is a spanning set for the image of $f$. However, $f(x_1)=dots=f(x_k)=0$, so also ${f(x_{k+1},dots,f(x_n)}$ is a spanning set for the image of $f$.
By the rank-nullity theorem, $dimoperatorname{im}f=n-k$, so the set has to be a basis.
If you can't use the rank-nullity theorem, directly prove that the set is linearly independent: if $alpha_{m+1}f(x_{m+1})+dots+alpha_nf(x_n)=0$, then
$$
alpha_{m+1}x_{m+1}+dots+alpha_nx_ninker f
$$
and it's easy to finish.
answered Jan 10 at 22:12
egregegreg
181k1485203
answered Jan 10 at 22:12
egregegreg
181k1485203
answered Jan 10 at 22:12
egregegreg
181k1485203
answered Jan 10 at 22:12
egregegreg
181k1485203
181k1485203
add a comment |
add a comment |
$begingroup$
Hint:
$V/Ker(f) cong Im(f)$.
Then completing your basis in the way you mentioned, what is a basis for $V/Ker(f)$? Then, consider how the above isomorphism is given.
answered Jan 10 at 18:39
MariahMariah
1,4871618
$endgroup$
add a comment |
$begingroup$
Hint:
$V/Ker(f) cong Im(f)$.
Then completing your basis in the way you mentioned, what is a basis for $V/Ker(f)$? Then, consider how the above isomorphism is given.
answered Jan 10 at 18:39
MariahMariah
1,4871618
$endgroup$
add a comment |
$begingroup$
Hint:
$V/Ker(f) cong Im(f)$.
Then completing your basis in the way you mentioned, what is a basis for $V/Ker(f)$? Then, consider how the above isomorphism is given.
answered Jan 10 at 18:39
MariahMariah
1,4871618
$endgroup$
Hint:
$V/Ker(f) cong Im(f)$.
Then completing your basis in the way you mentioned, what is a basis for $V/Ker(f)$? Then, consider how the above isomorphism is given.
answered Jan 10 at 18:39
MariahMariah
1,4871618
answered Jan 10 at 18:39
MariahMariah
1,4871618
answered Jan 10 at 18:39
MariahMariah
1,4871618
answered Jan 10 at 18:39
MariahMariah
1,4871618
1,4871618
add a comment |
add a comment |
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