Mixing
Proof of $mathcal{L}(V,W)$ is a vector space
$begingroup$
To prove whether something is a vector space, my understand is to prove the following properties: commutativity, associativity, additive identity, additive inverse, multiplicative identity, distributive properties. However, in the answer, it tries to prove closed under addition, close under scalar multiplication.
are those conditions for a subspace?
linear-algebra
edited Jan 13 at 22:38
caverac
14.6k31130
asked Jan 13 at 22:15
JOHN JOHN
3499
$endgroup$
add a comment |
$begingroup$
To prove whether something is a vector space, my understand is to prove the following properties: commutativity, associativity, additive identity, additive inverse, multiplicative identity, distributive properties. However, in the answer, it tries to prove closed under addition, close under scalar multiplication.
are those conditions for a subspace?
linear-algebra
edited Jan 13 at 22:38
caverac
14.6k31130
asked Jan 13 at 22:15
JOHN JOHN
3499
$endgroup$
$begingroup$
You are right that in the proof, they do not check commutativity, associativity etc. The picture you included shows that the vector space operations of addition and scalar multiplication $$L(V,W) times L(V,W) rightarrow L(V,W), (T,S) mapsto T+S$$ $$mathbb F times L(V,W) rightarrow L(V,W), (lambda,T) mapsto lambda T$$ are well defined functions. This is the hardest part of showing that $L(V,W)$ is a vector space. You should check the other properties yourself.
$endgroup$
– D_S
Jan 13 at 22:20
add a comment |
$begingroup$
To prove whether something is a vector space, my understand is to prove the following properties: commutativity, associativity, additive identity, additive inverse, multiplicative identity, distributive properties. However, in the answer, it tries to prove closed under addition, close under scalar multiplication.
are those conditions for a subspace?
linear-algebra
edited Jan 13 at 22:38
caverac
14.6k31130
asked Jan 13 at 22:15
JOHN JOHN
3499
$endgroup$
To prove whether something is a vector space, my understand is to prove the following properties: commutativity, associativity, additive identity, additive inverse, multiplicative identity, distributive properties. However, in the answer, it tries to prove closed under addition, close under scalar multiplication.
are those conditions for a subspace?
linear-algebra
linear-algebra
edited Jan 13 at 22:38
caverac
14.6k31130
asked Jan 13 at 22:15
JOHN JOHN
3499
edited Jan 13 at 22:38
caverac
14.6k31130
asked Jan 13 at 22:15
JOHN JOHN
3499
edited Jan 13 at 22:38
caverac
14.6k31130
edited Jan 13 at 22:38
caverac
14.6k31130
edited Jan 13 at 22:38
caverac
14.6k31130
14.6k31130
asked Jan 13 at 22:15
JOHN JOHN
3499
asked Jan 13 at 22:15
JOHN JOHN
3499
asked Jan 13 at 22:15
JOHN JOHN
3499
3499
$begingroup$
You are right that in the proof, they do not check commutativity, associativity etc. The picture you included shows that the vector space operations of addition and scalar multiplication $$L(V,W) times L(V,W) rightarrow L(V,W), (T,S) mapsto T+S$$ $$mathbb F times L(V,W) rightarrow L(V,W), (lambda,T) mapsto lambda T$$ are well defined functions. This is the hardest part of showing that $L(V,W)$ is a vector space. You should check the other properties yourself.
$endgroup$
– D_S
Jan 13 at 22:20
add a comment |
$begingroup$
You are right that in the proof, they do not check commutativity, associativity etc. The picture you included shows that the vector space operations of addition and scalar multiplication $$L(V,W) times L(V,W) rightarrow L(V,W), (T,S) mapsto T+S$$ $$mathbb F times L(V,W) rightarrow L(V,W), (lambda,T) mapsto lambda T$$ are well defined functions. This is the hardest part of showing that $L(V,W)$ is a vector space. You should check the other properties yourself.
$endgroup$
– D_S
Jan 13 at 22:20
$begingroup$
You are right that in the proof, they do not check commutativity, associativity etc. The picture you included shows that the vector space operations of addition and scalar multiplication $$L(V,W) times L(V,W) rightarrow L(V,W), (T,S) mapsto T+S$$ $$mathbb F times L(V,W) rightarrow L(V,W), (lambda,T) mapsto lambda T$$ are well defined functions. This is the hardest part of showing that $L(V,W)$ is a vector space. You should check the other properties yourself.
$endgroup$
– D_S
Jan 13 at 22:20
$begingroup$
You are right that in the proof, they do not check commutativity, associativity etc. The picture you included shows that the vector space operations of addition and scalar multiplication $$L(V,W) times L(V,W) rightarrow L(V,W), (T,S) mapsto T+S$$ $$mathbb F times L(V,W) rightarrow L(V,W), (lambda,T) mapsto lambda T$$ are well defined functions. This is the hardest part of showing that $L(V,W)$ is a vector space. You should check the other properties yourself.
$endgroup$
– D_S
Jan 13 at 22:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are 100% correct. It's an easy trap to fall into, to only verify the subspace conditions instead of all the many conditions for a full vector space, but just verifying subspace conditions is definitely not enough! I've seen countless students (as well as a few teachers) fall into this trap.
Now, there is one potential saving grace here: perhaps it was proven that the set $W^V$ of functions from $V$ to $W$ (linear or not) is a vector space. That is, $W^V$ under the given operations, all of the 8 or so properties of vector spaces were proven previously. Then, proving $mathcal{L}(V, W)$ is a subspace $W^V$ is a valid way of proving $mathcal{L}(V, W)$ is a vector space in its own right, as it will inherit almost all the properties from $W^V$.
But otherwise, the proof is incorrect.
answered Jan 13 at 22:51
Theo BenditTheo Bendit
18.3k12152
$endgroup$
1
$begingroup$
Note also that the alleged "proof" above didn't even check all the subspace conditions. The existence of zero vector is also important.
$endgroup$
– BigbearZzz
Jan 13 at 23:02
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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$begingroup$
You are 100% correct. It's an easy trap to fall into, to only verify the subspace conditions instead of all the many conditions for a full vector space, but just verifying subspace conditions is definitely not enough! I've seen countless students (as well as a few teachers) fall into this trap.
Now, there is one potential saving grace here: perhaps it was proven that the set $W^V$ of functions from $V$ to $W$ (linear or not) is a vector space. That is, $W^V$ under the given operations, all of the 8 or so properties of vector spaces were proven previously. Then, proving $mathcal{L}(V, W)$ is a subspace $W^V$ is a valid way of proving $mathcal{L}(V, W)$ is a vector space in its own right, as it will inherit almost all the properties from $W^V$.
But otherwise, the proof is incorrect.
answered Jan 13 at 22:51
Theo BenditTheo Bendit
18.3k12152
$endgroup$
1
$begingroup$
Note also that the alleged "proof" above didn't even check all the subspace conditions. The existence of zero vector is also important.
$endgroup$
– BigbearZzz
Jan 13 at 23:02
add a comment |
$begingroup$
You are 100% correct. It's an easy trap to fall into, to only verify the subspace conditions instead of all the many conditions for a full vector space, but just verifying subspace conditions is definitely not enough! I've seen countless students (as well as a few teachers) fall into this trap.
Now, there is one potential saving grace here: perhaps it was proven that the set $W^V$ of functions from $V$ to $W$ (linear or not) is a vector space. That is, $W^V$ under the given operations, all of the 8 or so properties of vector spaces were proven previously. Then, proving $mathcal{L}(V, W)$ is a subspace $W^V$ is a valid way of proving $mathcal{L}(V, W)$ is a vector space in its own right, as it will inherit almost all the properties from $W^V$.
But otherwise, the proof is incorrect.
answered Jan 13 at 22:51
Theo BenditTheo Bendit
18.3k12152
$endgroup$
1
$begingroup$
Note also that the alleged "proof" above didn't even check all the subspace conditions. The existence of zero vector is also important.
$endgroup$
– BigbearZzz
Jan 13 at 23:02
add a comment |
$begingroup$
You are 100% correct. It's an easy trap to fall into, to only verify the subspace conditions instead of all the many conditions for a full vector space, but just verifying subspace conditions is definitely not enough! I've seen countless students (as well as a few teachers) fall into this trap.
Now, there is one potential saving grace here: perhaps it was proven that the set $W^V$ of functions from $V$ to $W$ (linear or not) is a vector space. That is, $W^V$ under the given operations, all of the 8 or so properties of vector spaces were proven previously. Then, proving $mathcal{L}(V, W)$ is a subspace $W^V$ is a valid way of proving $mathcal{L}(V, W)$ is a vector space in its own right, as it will inherit almost all the properties from $W^V$.
But otherwise, the proof is incorrect.
answered Jan 13 at 22:51
Theo BenditTheo Bendit
18.3k12152
$endgroup$
You are 100% correct. It's an easy trap to fall into, to only verify the subspace conditions instead of all the many conditions for a full vector space, but just verifying subspace conditions is definitely not enough! I've seen countless students (as well as a few teachers) fall into this trap.
Now, there is one potential saving grace here: perhaps it was proven that the set $W^V$ of functions from $V$ to $W$ (linear or not) is a vector space. That is, $W^V$ under the given operations, all of the 8 or so properties of vector spaces were proven previously. Then, proving $mathcal{L}(V, W)$ is a subspace $W^V$ is a valid way of proving $mathcal{L}(V, W)$ is a vector space in its own right, as it will inherit almost all the properties from $W^V$.
But otherwise, the proof is incorrect.
answered Jan 13 at 22:51
Theo BenditTheo Bendit
18.3k12152
answered Jan 13 at 22:51
Theo BenditTheo Bendit
18.3k12152
answered Jan 13 at 22:51
Theo BenditTheo Bendit
18.3k12152
answered Jan 13 at 22:51
Theo BenditTheo Bendit
18.3k12152
18.3k12152
1
$begingroup$
Note also that the alleged "proof" above didn't even check all the subspace conditions. The existence of zero vector is also important.
$endgroup$
– BigbearZzz
Jan 13 at 23:02
add a comment |
1
$begingroup$
Note also that the alleged "proof" above didn't even check all the subspace conditions. The existence of zero vector is also important.
$endgroup$
– BigbearZzz
Jan 13 at 23:02
1
1
$begingroup$
Note also that the alleged "proof" above didn't even check all the subspace conditions. The existence of zero vector is also important.
$endgroup$
– BigbearZzz
Jan 13 at 23:02
$begingroup$
Note also that the alleged "proof" above didn't even check all the subspace conditions. The existence of zero vector is also important.
$endgroup$
– BigbearZzz
Jan 13 at 23:02
add a comment |
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$begingroup$
You are right that in the proof, they do not check commutativity, associativity etc. The picture you included shows that the vector space operations of addition and scalar multiplication $$L(V,W) times L(V,W) rightarrow L(V,W), (T,S) mapsto T+S$$ $$mathbb F times L(V,W) rightarrow L(V,W), (lambda,T) mapsto lambda T$$ are well defined functions. This is the hardest part of showing that $L(V,W)$ is a vector space. You should check the other properties yourself.
$endgroup$
– D_S
Jan 13 at 22:20