Mixing
If null space and the range of a linear mapping is finite dimensional then the departure space is finite...
$begingroup$
There are several questions posted which are looking for solutions and hints. I was hoping that someone could verify my proof.
Let $V,W$ be vector spaces and $A:Vto W$ be a linear mapping. Suppose that $dim ker A = s$ and $dim A(V) = t$. We need to show that $A$ is finite dimensional.
We assume the contrary that $V$ is not finite dimensional, hence we can find an infinite list of linearly independent vectors.
We have been given that $dim ker A = s$, thus, we can find a basis ${ alpha_1 , ldots , alpha_s }$ for $ker A$. Now, here's my proof strategy: I will extend the previous list to ${ alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$ such that ${ A(alpha_{s+1}) , ldots , A(alpha_{s+t}) }$ is a basis for $A(V)$ and this list ${ alpha_1 , ldots , alpha_s , ldots , alpha_{s+t} }$ is linearly independent.
Let $alpha_{s+1} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s }$ be arbitrary.
Now, we can pick $alpha_{s+2} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1} }$ such that $A(alpha_{s+2}) notin text{Sp} { A(alpha_{s+1})}$. For otherwise we would have that for all $alpha_{s+2} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1} }$, $A(alpha_{s+2}) in text{Sp} { A(alpha_{s+1})}$ which implies that $A(V)= text{Sp} { A(alpha_{s+1})}$ and which further would contradict our assumption that $dim A(V) = t$. Again, we can pick $alpha_{s+3} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1}, alpha_{s+2} }$ such that $A(alpha_{s+2}) notin text{Sp} { A(alpha_{s+1}) ,A(alpha_{s+2})}$ (The argument for it is same as the previous one!). If we repeat this argument for $t$ times, we would have found vectors such that ${ alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$ such that the vectors ${ A(alpha_{s+1}) , ldots , A(alpha_{s+t}) }$ are linear independent and this list ${ alpha_1 , ldots , alpha_s , ldots , alpha_{s+t} }$ is linearly independent. Since, we know that linearly independent vectors of right length is the basis, it must be that ${ A(alpha_{s+1}) , ldots , A(alpha_{s+t}) }$ form a basis for $A(V)$.
Now, we bring about a contradiction. Pick a vector $v in V setminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$. $A(v) in A(V)$, thus, $A(v)=c_1 A(alpha_{s+1}) + ldots + c_t A(alpha_{s+t})$. By definition of linear mappings, we have $A(v-c_1
alpha_{s+1}- ldots - c_t alpha_{s+t})=0$. It follows that $v-c_1
alpha_{s+1}- ldots - c_t alpha_{s+t} in ker A$ thus $v-c_1
alpha_{s+1}- ldots - c_t alpha_{s+t}=d_1 alpha_1 + ldots + d_s alpha_s$ which is contradicts our assumption that $v not in text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$. Thus, $V$ must be finite dimensional.
Is this proof correct? I know it's a long proof. Thanks in advance!
linear-algebra proof-verification linear-transformations alternative-proof
asked Jan 31 at 16:16
Ashish KAshish K
923614
$endgroup$
add a comment |
$begingroup$
There are several questions posted which are looking for solutions and hints. I was hoping that someone could verify my proof.
Let $V,W$ be vector spaces and $A:Vto W$ be a linear mapping. Suppose that $dim ker A = s$ and $dim A(V) = t$. We need to show that $A$ is finite dimensional.
We assume the contrary that $V$ is not finite dimensional, hence we can find an infinite list of linearly independent vectors.
We have been given that $dim ker A = s$, thus, we can find a basis ${ alpha_1 , ldots , alpha_s }$ for $ker A$. Now, here's my proof strategy: I will extend the previous list to ${ alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$ such that ${ A(alpha_{s+1}) , ldots , A(alpha_{s+t}) }$ is a basis for $A(V)$ and this list ${ alpha_1 , ldots , alpha_s , ldots , alpha_{s+t} }$ is linearly independent.
Let $alpha_{s+1} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s }$ be arbitrary.
Now, we can pick $alpha_{s+2} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1} }$ such that $A(alpha_{s+2}) notin text{Sp} { A(alpha_{s+1})}$. For otherwise we would have that for all $alpha_{s+2} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1} }$, $A(alpha_{s+2}) in text{Sp} { A(alpha_{s+1})}$ which implies that $A(V)= text{Sp} { A(alpha_{s+1})}$ and which further would contradict our assumption that $dim A(V) = t$. Again, we can pick $alpha_{s+3} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1}, alpha_{s+2} }$ such that $A(alpha_{s+2}) notin text{Sp} { A(alpha_{s+1}) ,A(alpha_{s+2})}$ (The argument for it is same as the previous one!). If we repeat this argument for $t$ times, we would have found vectors such that ${ alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$ such that the vectors ${ A(alpha_{s+1}) , ldots , A(alpha_{s+t}) }$ are linear independent and this list ${ alpha_1 , ldots , alpha_s , ldots , alpha_{s+t} }$ is linearly independent. Since, we know that linearly independent vectors of right length is the basis, it must be that ${ A(alpha_{s+1}) , ldots , A(alpha_{s+t}) }$ form a basis for $A(V)$.
Now, we bring about a contradiction. Pick a vector $v in V setminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$. $A(v) in A(V)$, thus, $A(v)=c_1 A(alpha_{s+1}) + ldots + c_t A(alpha_{s+t})$. By definition of linear mappings, we have $A(v-c_1
alpha_{s+1}- ldots - c_t alpha_{s+t})=0$. It follows that $v-c_1
alpha_{s+1}- ldots - c_t alpha_{s+t} in ker A$ thus $v-c_1
alpha_{s+1}- ldots - c_t alpha_{s+t}=d_1 alpha_1 + ldots + d_s alpha_s$ which is contradicts our assumption that $v not in text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$. Thus, $V$ must be finite dimensional.
Is this proof correct? I know it's a long proof. Thanks in advance!
linear-algebra proof-verification linear-transformations alternative-proof
asked Jan 31 at 16:16
Ashish KAshish K
923614
$endgroup$
$begingroup$
Yes -- you have the right idea. The sentence "We assume the contrary that V is not finite dimensional, hence we can find an infinite list of linearly independent vectors." isn't really doing anything. You could leave it out, and just turn the contradiction at the end into showing that $V$ is finite-dimensional, which you did anyway.
$endgroup$
– csprun
Jan 31 at 16:27
$begingroup$
@csprun isn't it implicit in that fact the $V$ minus span of vectors is nonempty? For otherwise V could be the span of some vectors and I couldn't find anymore vectors I did in the proof?
$endgroup$
– Ashish K
Jan 31 at 16:37
$begingroup$
You mean at the very end? Instead of picking $v$ outside that span, just pick any $v in V$ and show it's in that span. You hardly have to change anything in your proof.
$endgroup$
– csprun
Jan 31 at 16:40
$begingroup$
@csprun oh yes that's right. There's no need of contradiction anyways. Thank you very much!
$endgroup$
– Ashish K
Jan 31 at 16:48
1
$begingroup$
Sure! Good luck.
$endgroup$
– csprun
Jan 31 at 16:52
add a comment |
$begingroup$
There are several questions posted which are looking for solutions and hints. I was hoping that someone could verify my proof.
Let $V,W$ be vector spaces and $A:Vto W$ be a linear mapping. Suppose that $dim ker A = s$ and $dim A(V) = t$. We need to show that $A$ is finite dimensional.
We assume the contrary that $V$ is not finite dimensional, hence we can find an infinite list of linearly independent vectors.
We have been given that $dim ker A = s$, thus, we can find a basis ${ alpha_1 , ldots , alpha_s }$ for $ker A$. Now, here's my proof strategy: I will extend the previous list to ${ alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$ such that ${ A(alpha_{s+1}) , ldots , A(alpha_{s+t}) }$ is a basis for $A(V)$ and this list ${ alpha_1 , ldots , alpha_s , ldots , alpha_{s+t} }$ is linearly independent.
Let $alpha_{s+1} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s }$ be arbitrary.
Now, we can pick $alpha_{s+2} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1} }$ such that $A(alpha_{s+2}) notin text{Sp} { A(alpha_{s+1})}$. For otherwise we would have that for all $alpha_{s+2} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1} }$, $A(alpha_{s+2}) in text{Sp} { A(alpha_{s+1})}$ which implies that $A(V)= text{Sp} { A(alpha_{s+1})}$ and which further would contradict our assumption that $dim A(V) = t$. Again, we can pick $alpha_{s+3} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1}, alpha_{s+2} }$ such that $A(alpha_{s+2}) notin text{Sp} { A(alpha_{s+1}) ,A(alpha_{s+2})}$ (The argument for it is same as the previous one!). If we repeat this argument for $t$ times, we would have found vectors such that ${ alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$ such that the vectors ${ A(alpha_{s+1}) , ldots , A(alpha_{s+t}) }$ are linear independent and this list ${ alpha_1 , ldots , alpha_s , ldots , alpha_{s+t} }$ is linearly independent. Since, we know that linearly independent vectors of right length is the basis, it must be that ${ A(alpha_{s+1}) , ldots , A(alpha_{s+t}) }$ form a basis for $A(V)$.
Now, we bring about a contradiction. Pick a vector $v in V setminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$. $A(v) in A(V)$, thus, $A(v)=c_1 A(alpha_{s+1}) + ldots + c_t A(alpha_{s+t})$. By definition of linear mappings, we have $A(v-c_1
alpha_{s+1}- ldots - c_t alpha_{s+t})=0$. It follows that $v-c_1
alpha_{s+1}- ldots - c_t alpha_{s+t} in ker A$ thus $v-c_1
alpha_{s+1}- ldots - c_t alpha_{s+t}=d_1 alpha_1 + ldots + d_s alpha_s$ which is contradicts our assumption that $v not in text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$. Thus, $V$ must be finite dimensional.
Is this proof correct? I know it's a long proof. Thanks in advance!
linear-algebra proof-verification linear-transformations alternative-proof
asked Jan 31 at 16:16
Ashish KAshish K
923614
$endgroup$
There are several questions posted which are looking for solutions and hints. I was hoping that someone could verify my proof.
Let $V,W$ be vector spaces and $A:Vto W$ be a linear mapping. Suppose that $dim ker A = s$ and $dim A(V) = t$. We need to show that $A$ is finite dimensional.
We assume the contrary that $V$ is not finite dimensional, hence we can find an infinite list of linearly independent vectors.
We have been given that $dim ker A = s$, thus, we can find a basis ${ alpha_1 , ldots , alpha_s }$ for $ker A$. Now, here's my proof strategy: I will extend the previous list to ${ alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$ such that ${ A(alpha_{s+1}) , ldots , A(alpha_{s+t}) }$ is a basis for $A(V)$ and this list ${ alpha_1 , ldots , alpha_s , ldots , alpha_{s+t} }$ is linearly independent.
Let $alpha_{s+1} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s }$ be arbitrary.
Now, we can pick $alpha_{s+2} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1} }$ such that $A(alpha_{s+2}) notin text{Sp} { A(alpha_{s+1})}$. For otherwise we would have that for all $alpha_{s+2} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1} }$, $A(alpha_{s+2}) in text{Sp} { A(alpha_{s+1})}$ which implies that $A(V)= text{Sp} { A(alpha_{s+1})}$ and which further would contradict our assumption that $dim A(V) = t$. Again, we can pick $alpha_{s+3} in Vsetminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1}, alpha_{s+2} }$ such that $A(alpha_{s+2}) notin text{Sp} { A(alpha_{s+1}) ,A(alpha_{s+2})}$ (The argument for it is same as the previous one!). If we repeat this argument for $t$ times, we would have found vectors such that ${ alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$ such that the vectors ${ A(alpha_{s+1}) , ldots , A(alpha_{s+t}) }$ are linear independent and this list ${ alpha_1 , ldots , alpha_s , ldots , alpha_{s+t} }$ is linearly independent. Since, we know that linearly independent vectors of right length is the basis, it must be that ${ A(alpha_{s+1}) , ldots , A(alpha_{s+t}) }$ form a basis for $A(V)$.
Now, we bring about a contradiction. Pick a vector $v in V setminus text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$. $A(v) in A(V)$, thus, $A(v)=c_1 A(alpha_{s+1}) + ldots + c_t A(alpha_{s+t})$. By definition of linear mappings, we have $A(v-c_1
alpha_{s+1}- ldots - c_t alpha_{s+t})=0$. It follows that $v-c_1
alpha_{s+1}- ldots - c_t alpha_{s+t} in ker A$ thus $v-c_1
alpha_{s+1}- ldots - c_t alpha_{s+t}=d_1 alpha_1 + ldots + d_s alpha_s$ which is contradicts our assumption that $v not in text{Sp} { alpha_1 , ldots , alpha_s , alpha_{s+1}, ldots , alpha_{s+t} }$. Thus, $V$ must be finite dimensional.
Is this proof correct? I know it's a long proof. Thanks in advance!
linear-algebra proof-verification linear-transformations alternative-proof
linear-algebra proof-verification linear-transformations alternative-proof
asked Jan 31 at 16:16
Ashish KAshish K
923614
asked Jan 31 at 16:16
Ashish KAshish K
923614
asked Jan 31 at 16:16
Ashish KAshish K
923614
asked Jan 31 at 16:16
Ashish KAshish K
923614
asked Jan 31 at 16:16
Ashish KAshish K
923614
923614
$begingroup$
Yes -- you have the right idea. The sentence "We assume the contrary that V is not finite dimensional, hence we can find an infinite list of linearly independent vectors." isn't really doing anything. You could leave it out, and just turn the contradiction at the end into showing that $V$ is finite-dimensional, which you did anyway.
$endgroup$
– csprun
Jan 31 at 16:27
$begingroup$
@csprun isn't it implicit in that fact the $V$ minus span of vectors is nonempty? For otherwise V could be the span of some vectors and I couldn't find anymore vectors I did in the proof?
$endgroup$
– Ashish K
Jan 31 at 16:37
$begingroup$
You mean at the very end? Instead of picking $v$ outside that span, just pick any $v in V$ and show it's in that span. You hardly have to change anything in your proof.
$endgroup$
– csprun
Jan 31 at 16:40
$begingroup$
@csprun oh yes that's right. There's no need of contradiction anyways. Thank you very much!
$endgroup$
– Ashish K
Jan 31 at 16:48
1
$begingroup$
Sure! Good luck.
$endgroup$
– csprun
Jan 31 at 16:52
add a comment |
$begingroup$
Yes -- you have the right idea. The sentence "We assume the contrary that V is not finite dimensional, hence we can find an infinite list of linearly independent vectors." isn't really doing anything. You could leave it out, and just turn the contradiction at the end into showing that $V$ is finite-dimensional, which you did anyway.
$endgroup$
– csprun
Jan 31 at 16:27
$begingroup$
@csprun isn't it implicit in that fact the $V$ minus span of vectors is nonempty? For otherwise V could be the span of some vectors and I couldn't find anymore vectors I did in the proof?
$endgroup$
– Ashish K
Jan 31 at 16:37
$begingroup$
You mean at the very end? Instead of picking $v$ outside that span, just pick any $v in V$ and show it's in that span. You hardly have to change anything in your proof.
$endgroup$
– csprun
Jan 31 at 16:40
$begingroup$
@csprun oh yes that's right. There's no need of contradiction anyways. Thank you very much!
$endgroup$
– Ashish K
Jan 31 at 16:48
1
$begingroup$
Sure! Good luck.
$endgroup$
– csprun
Jan 31 at 16:52
$begingroup$
Yes -- you have the right idea. The sentence "We assume the contrary that V is not finite dimensional, hence we can find an infinite list of linearly independent vectors." isn't really doing anything. You could leave it out, and just turn the contradiction at the end into showing that $V$ is finite-dimensional, which you did anyway.
$endgroup$
– csprun
Jan 31 at 16:27
$begingroup$
Yes -- you have the right idea. The sentence "We assume the contrary that V is not finite dimensional, hence we can find an infinite list of linearly independent vectors." isn't really doing anything. You could leave it out, and just turn the contradiction at the end into showing that $V$ is finite-dimensional, which you did anyway.
$endgroup$
– csprun
Jan 31 at 16:27
$begingroup$
@csprun isn't it implicit in that fact the $V$ minus span of vectors is nonempty? For otherwise V could be the span of some vectors and I couldn't find anymore vectors I did in the proof?
$endgroup$
– Ashish K
Jan 31 at 16:37
$begingroup$
@csprun isn't it implicit in that fact the $V$ minus span of vectors is nonempty? For otherwise V could be the span of some vectors and I couldn't find anymore vectors I did in the proof?
$endgroup$
– Ashish K
Jan 31 at 16:37
$begingroup$
You mean at the very end? Instead of picking $v$ outside that span, just pick any $v in V$ and show it's in that span. You hardly have to change anything in your proof.
$endgroup$
– csprun
Jan 31 at 16:40
$begingroup$
You mean at the very end? Instead of picking $v$ outside that span, just pick any $v in V$ and show it's in that span. You hardly have to change anything in your proof.
$endgroup$
– csprun
Jan 31 at 16:40
$begingroup$
@csprun oh yes that's right. There's no need of contradiction anyways. Thank you very much!
$endgroup$
– Ashish K
Jan 31 at 16:48
$begingroup$
@csprun oh yes that's right. There's no need of contradiction anyways. Thank you very much!
$endgroup$
– Ashish K
Jan 31 at 16:48
1
1
$begingroup$
Sure! Good luck.
$endgroup$
– csprun
Jan 31 at 16:52
$begingroup$
Sure! Good luck.
$endgroup$
– csprun
Jan 31 at 16:52
add a comment |
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$begingroup$
Yes -- you have the right idea. The sentence "We assume the contrary that V is not finite dimensional, hence we can find an infinite list of linearly independent vectors." isn't really doing anything. You could leave it out, and just turn the contradiction at the end into showing that $V$ is finite-dimensional, which you did anyway.
$endgroup$
– csprun
Jan 31 at 16:27
$begingroup$
@csprun isn't it implicit in that fact the $V$ minus span of vectors is nonempty? For otherwise V could be the span of some vectors and I couldn't find anymore vectors I did in the proof?
$endgroup$
– Ashish K
Jan 31 at 16:37
$begingroup$
You mean at the very end? Instead of picking $v$ outside that span, just pick any $v in V$ and show it's in that span. You hardly have to change anything in your proof.
$endgroup$
– csprun
Jan 31 at 16:40
$begingroup$
@csprun oh yes that's right. There's no need of contradiction anyways. Thank you very much!
$endgroup$
– Ashish K
Jan 31 at 16:48
1
$begingroup$
Sure! Good luck.
$endgroup$
– csprun
Jan 31 at 16:52