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Exist a endomorphism $f$ on $V$ such that $Im (f) = W$
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Let $V$ is a finite vector space, $W$ is a subspace of $V$. Show that:
- Exist a endomorphism $f$ on $V$ such that $Im (f) = W$.
- Exist a endomorphism $g$ on $V$ such that $Ker (g) = W$.
My attempt:
Consider a homomorphism $h$ on $V$. Let $f = h|_W: W rightarrow V$, we have $Im(f) = f(W)$. The remainder of this problem is $f(W) = W$ implies $f = id(W)$.
On the one hand, consider $g = h|_W: W rightarrow V$ such that $g(w)=0, forall w in W$. Q.E.D
Is my solution right?
linear-algebra linear-transformations
asked Aug 2 at 15:28
Minh
727
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up vote
0
down vote
favorite
Let $V$ is a finite vector space, $W$ is a subspace of $V$. Show that:
- Exist a endomorphism $f$ on $V$ such that $Im (f) = W$.
- Exist a endomorphism $g$ on $V$ such that $Ker (g) = W$.
My attempt:
Consider a homomorphism $h$ on $V$. Let $f = h|_W: W rightarrow V$, we have $Im(f) = f(W)$. The remainder of this problem is $f(W) = W$ implies $f = id(W)$.
On the one hand, consider $g = h|_W: W rightarrow V$ such that $g(w)=0, forall w in W$. Q.E.D
Is my solution right?
linear-algebra linear-transformations
asked Aug 2 at 15:28
Minh
727
1
The sentence $f(W)=W$ does not imply that $f$ is the identity on $W$. For instance, the linear transformation $xrightarrow 2x$ from $mathbbR$ to $mathbbR$ sends $mathbbR$ to itself but is not the identity. Moreover, you need $f$ to be an endormorphism on $V$. However, the $f$ you try to construct is a linear mapping from $W$ to $V$. It does not meet the requirement of the question. As for the second question, how do you precisely justify the existence of an endomorphism that is $0$ on $W$ and only on $W$?
– Suzet
Aug 2 at 15:31
I believe that's exists an endomorphism like the problem remainded, but I can not find it.
– Minh
Aug 2 at 15:35
1
Hint: use the fact that $V$ is finite dimensional. Consider a basis for it that is adapted to $W$. Then, try to build endomorphisms using this.
– Suzet
Aug 2 at 15:39
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $V$ is a finite vector space, $W$ is a subspace of $V$. Show that:
- Exist a endomorphism $f$ on $V$ such that $Im (f) = W$.
- Exist a endomorphism $g$ on $V$ such that $Ker (g) = W$.
My attempt:
Consider a homomorphism $h$ on $V$. Let $f = h|_W: W rightarrow V$, we have $Im(f) = f(W)$. The remainder of this problem is $f(W) = W$ implies $f = id(W)$.
On the one hand, consider $g = h|_W: W rightarrow V$ such that $g(w)=0, forall w in W$. Q.E.D
Is my solution right?
linear-algebra linear-transformations
asked Aug 2 at 15:28
Minh
727
Let $V$ is a finite vector space, $W$ is a subspace of $V$. Show that:
- Exist a endomorphism $f$ on $V$ such that $Im (f) = W$.
- Exist a endomorphism $g$ on $V$ such that $Ker (g) = W$.
My attempt:
Consider a homomorphism $h$ on $V$. Let $f = h|_W: W rightarrow V$, we have $Im(f) = f(W)$. The remainder of this problem is $f(W) = W$ implies $f = id(W)$.
On the one hand, consider $g = h|_W: W rightarrow V$ such that $g(w)=0, forall w in W$. Q.E.D
Is my solution right?
linear-algebra linear-transformations
asked Aug 2 at 15:28
Minh
727
asked Aug 2 at 15:28
Minh
727
asked Aug 2 at 15:28
Minh
727
asked Aug 2 at 15:28
Minh
727
727
1
The sentence $f(W)=W$ does not imply that $f$ is the identity on $W$. For instance, the linear transformation $xrightarrow 2x$ from $mathbbR$ to $mathbbR$ sends $mathbbR$ to itself but is not the identity. Moreover, you need $f$ to be an endormorphism on $V$. However, the $f$ you try to construct is a linear mapping from $W$ to $V$. It does not meet the requirement of the question. As for the second question, how do you precisely justify the existence of an endomorphism that is $0$ on $W$ and only on $W$?
– Suzet
Aug 2 at 15:31
I believe that's exists an endomorphism like the problem remainded, but I can not find it.
– Minh
Aug 2 at 15:35
1
Hint: use the fact that $V$ is finite dimensional. Consider a basis for it that is adapted to $W$. Then, try to build endomorphisms using this.
– Suzet
Aug 2 at 15:39
add a comment |Â
1
The sentence $f(W)=W$ does not imply that $f$ is the identity on $W$. For instance, the linear transformation $xrightarrow 2x$ from $mathbbR$ to $mathbbR$ sends $mathbbR$ to itself but is not the identity. Moreover, you need $f$ to be an endormorphism on $V$. However, the $f$ you try to construct is a linear mapping from $W$ to $V$. It does not meet the requirement of the question. As for the second question, how do you precisely justify the existence of an endomorphism that is $0$ on $W$ and only on $W$?
– Suzet
Aug 2 at 15:31
I believe that's exists an endomorphism like the problem remainded, but I can not find it.
– Minh
Aug 2 at 15:35
1
Hint: use the fact that $V$ is finite dimensional. Consider a basis for it that is adapted to $W$. Then, try to build endomorphisms using this.
– Suzet
Aug 2 at 15:39
1
1
The sentence $f(W)=W$ does not imply that $f$ is the identity on $W$. For instance, the linear transformation $xrightarrow 2x$ from $mathbbR$ to $mathbbR$ sends $mathbbR$ to itself but is not the identity. Moreover, you need $f$ to be an endormorphism on $V$. However, the $f$ you try to construct is a linear mapping from $W$ to $V$. It does not meet the requirement of the question. As for the second question, how do you precisely justify the existence of an endomorphism that is $0$ on $W$ and only on $W$?
– Suzet
Aug 2 at 15:31
The sentence $f(W)=W$ does not imply that $f$ is the identity on $W$. For instance, the linear transformation $xrightarrow 2x$ from $mathbbR$ to $mathbbR$ sends $mathbbR$ to itself but is not the identity. Moreover, you need $f$ to be an endormorphism on $V$. However, the $f$ you try to construct is a linear mapping from $W$ to $V$. It does not meet the requirement of the question. As for the second question, how do you precisely justify the existence of an endomorphism that is $0$ on $W$ and only on $W$?
– Suzet
Aug 2 at 15:31
I believe that's exists an endomorphism like the problem remainded, but I can not find it.
– Minh
Aug 2 at 15:35
I believe that's exists an endomorphism like the problem remainded, but I can not find it.
– Minh
Aug 2 at 15:35
1
1
Hint: use the fact that $V$ is finite dimensional. Consider a basis for it that is adapted to $W$. Then, try to build endomorphisms using this.
– Suzet
Aug 2 at 15:39
Hint: use the fact that $V$ is finite dimensional. Consider a basis for it that is adapted to $W$. Then, try to build endomorphisms using this.
– Suzet
Aug 2 at 15:39
add a comment |Â
1 Answer
1
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up vote
2
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Let $w_1,ldots,w_m$ be a basis for $W.$ Extend this basis to a basis for $V,$ $w_1,ldots,w_m,v_1,ldots,v_p,$ where $m+p = dim V.$
Define $f : w_1,ldots,w_m,v_1,ldots,v_plongrightarrow V$ by $f(w_i)=w_i$ for $1leq i leq m$ and $0$ otherwise. Then $f$ extends uniquely to a linear map $L_f,$ and $textImg(L_f) = W.$
Define $g : w_1,ldots,w_m,v_1,ldots,v_plongrightarrow V$ by $g(w_i)=0$ for $1leq i leq m$ and $g(v_j)=v_j$ for $1leq p$ . Then $g$ extends uniquely to a linear map $L_g,$ and $ker(L_g) = W.$
answered Aug 2 at 15:43
Chickenmancer
2,937621
Do you mean $g(v_j)=v_j$ for $1leq jleq p$?
– Batominovski
Aug 2 at 17:03
Yes, thank you. I have corrected it.
– Chickenmancer
Aug 2 at 17:25
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $w_1,ldots,w_m$ be a basis for $W.$ Extend this basis to a basis for $V,$ $w_1,ldots,w_m,v_1,ldots,v_p,$ where $m+p = dim V.$
Define $f : w_1,ldots,w_m,v_1,ldots,v_plongrightarrow V$ by $f(w_i)=w_i$ for $1leq i leq m$ and $0$ otherwise. Then $f$ extends uniquely to a linear map $L_f,$ and $textImg(L_f) = W.$
Define $g : w_1,ldots,w_m,v_1,ldots,v_plongrightarrow V$ by $g(w_i)=0$ for $1leq i leq m$ and $g(v_j)=v_j$ for $1leq p$ . Then $g$ extends uniquely to a linear map $L_g,$ and $ker(L_g) = W.$
answered Aug 2 at 15:43
Chickenmancer
2,937621
Do you mean $g(v_j)=v_j$ for $1leq jleq p$?
– Batominovski
Aug 2 at 17:03
Yes, thank you. I have corrected it.
– Chickenmancer
Aug 2 at 17:25
add a comment |Â
up vote
2
down vote
accepted
Let $w_1,ldots,w_m$ be a basis for $W.$ Extend this basis to a basis for $V,$ $w_1,ldots,w_m,v_1,ldots,v_p,$ where $m+p = dim V.$
Define $f : w_1,ldots,w_m,v_1,ldots,v_plongrightarrow V$ by $f(w_i)=w_i$ for $1leq i leq m$ and $0$ otherwise. Then $f$ extends uniquely to a linear map $L_f,$ and $textImg(L_f) = W.$
Define $g : w_1,ldots,w_m,v_1,ldots,v_plongrightarrow V$ by $g(w_i)=0$ for $1leq i leq m$ and $g(v_j)=v_j$ for $1leq p$ . Then $g$ extends uniquely to a linear map $L_g,$ and $ker(L_g) = W.$
answered Aug 2 at 15:43
Chickenmancer
2,937621
Do you mean $g(v_j)=v_j$ for $1leq jleq p$?
– Batominovski
Aug 2 at 17:03
Yes, thank you. I have corrected it.
– Chickenmancer
Aug 2 at 17:25
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $w_1,ldots,w_m$ be a basis for $W.$ Extend this basis to a basis for $V,$ $w_1,ldots,w_m,v_1,ldots,v_p,$ where $m+p = dim V.$
Define $f : w_1,ldots,w_m,v_1,ldots,v_plongrightarrow V$ by $f(w_i)=w_i$ for $1leq i leq m$ and $0$ otherwise. Then $f$ extends uniquely to a linear map $L_f,$ and $textImg(L_f) = W.$
Define $g : w_1,ldots,w_m,v_1,ldots,v_plongrightarrow V$ by $g(w_i)=0$ for $1leq i leq m$ and $g(v_j)=v_j$ for $1leq p$ . Then $g$ extends uniquely to a linear map $L_g,$ and $ker(L_g) = W.$
answered Aug 2 at 15:43
Chickenmancer
2,937621
Let $w_1,ldots,w_m$ be a basis for $W.$ Extend this basis to a basis for $V,$ $w_1,ldots,w_m,v_1,ldots,v_p,$ where $m+p = dim V.$
Define $f : w_1,ldots,w_m,v_1,ldots,v_plongrightarrow V$ by $f(w_i)=w_i$ for $1leq i leq m$ and $0$ otherwise. Then $f$ extends uniquely to a linear map $L_f,$ and $textImg(L_f) = W.$
Define $g : w_1,ldots,w_m,v_1,ldots,v_plongrightarrow V$ by $g(w_i)=0$ for $1leq i leq m$ and $g(v_j)=v_j$ for $1leq p$ . Then $g$ extends uniquely to a linear map $L_g,$ and $ker(L_g) = W.$
answered Aug 2 at 15:43
Chickenmancer
2,937621
edited Aug 2 at 17:25
answered Aug 2 at 15:43
Chickenmancer
2,937621
answered Aug 2 at 15:43
Chickenmancer
2,937621
answered Aug 2 at 15:43
Chickenmancer
2,937621
2,937621
Do you mean $g(v_j)=v_j$ for $1leq jleq p$?
– Batominovski
Aug 2 at 17:03
Yes, thank you. I have corrected it.
– Chickenmancer
Aug 2 at 17:25
add a comment |Â
Do you mean $g(v_j)=v_j$ for $1leq jleq p$?
– Batominovski
Aug 2 at 17:03
Yes, thank you. I have corrected it.
– Chickenmancer
Aug 2 at 17:25
Do you mean $g(v_j)=v_j$ for $1leq jleq p$?
– Batominovski
Aug 2 at 17:03
Do you mean $g(v_j)=v_j$ for $1leq jleq p$?
– Batominovski
Aug 2 at 17:03
Yes, thank you. I have corrected it.
– Chickenmancer
Aug 2 at 17:25
Yes, thank you. I have corrected it.
– Chickenmancer
Aug 2 at 17:25
add a comment |Â
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1
The sentence $f(W)=W$ does not imply that $f$ is the identity on $W$. For instance, the linear transformation $xrightarrow 2x$ from $mathbbR$ to $mathbbR$ sends $mathbbR$ to itself but is not the identity. Moreover, you need $f$ to be an endormorphism on $V$. However, the $f$ you try to construct is a linear mapping from $W$ to $V$. It does not meet the requirement of the question. As for the second question, how do you precisely justify the existence of an endomorphism that is $0$ on $W$ and only on $W$?
– Suzet
Aug 2 at 15:31
I believe that's exists an endomorphism like the problem remainded, but I can not find it.
– Minh
Aug 2 at 15:35
1
Hint: use the fact that $V$ is finite dimensional. Consider a basis for it that is adapted to $W$. Then, try to build endomorphisms using this.
– Suzet
Aug 2 at 15:39