Proving a matrix is surjective
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Is a $n times n$ matrix always surjective? If it is, how can I prove it using the rank-nullity theorem?
linear-algebra matrices
edited Nov 24 at 17:10
gimusi
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asked Nov 24 at 17:06
Elliot Silver
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Is a $n times n$ matrix always surjective? If it is, how can I prove it using the rank-nullity theorem?
linear-algebra matrices
edited Nov 24 at 17:10
gimusi
88.5k74394
asked Nov 24 at 17:06
Elliot Silver
92
New contributor
Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
Small language point for the future. Matrices are not surjective The linear functions they determine may be.
– Ethan Bolker
Nov 24 at 23:15
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is a $n times n$ matrix always surjective? If it is, how can I prove it using the rank-nullity theorem?
linear-algebra matrices
edited Nov 24 at 17:10
gimusi
88.5k74394
asked Nov 24 at 17:06
Elliot Silver
92
New contributor
Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is a $n times n$ matrix always surjective? If it is, how can I prove it using the rank-nullity theorem?
linear-algebra matrices
linear-algebra matrices
edited Nov 24 at 17:10
gimusi
88.5k74394
asked Nov 24 at 17:06
Elliot Silver
92
New contributor
Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 24 at 17:10
gimusi
88.5k74394
asked Nov 24 at 17:06
Elliot Silver
92
New contributor
Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 24 at 17:10
gimusi
88.5k74394
edited Nov 24 at 17:10
gimusi
88.5k74394
edited Nov 24 at 17:10
gimusi
88.5k74394
88.5k74394
asked Nov 24 at 17:06
Elliot Silver
92
New contributor
Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 24 at 17:06
Elliot Silver
92
asked Nov 24 at 17:06
Elliot Silver
92
92
New contributor
Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
Small language point for the future. Matrices are not surjective The linear functions they determine may be.
– Ethan Bolker
Nov 24 at 23:15
add a comment |
5
Small language point for the future. Matrices are not surjective The linear functions they determine may be.
– Ethan Bolker
Nov 24 at 23:15
5
5
Small language point for the future. Matrices are not surjective The linear functions they determine may be.
– Ethan Bolker
Nov 24 at 23:15
Small language point for the future. Matrices are not surjective The linear functions they determine may be.
– Ethan Bolker
Nov 24 at 23:15
add a comment |
3 Answers
3
active
oldest
votes
up vote
5
down vote
No it's not. To see this consider the zero matrix as a map from $mathbb{R}^nrightarrow mathbb{R}^n$ which is clearly not surjective.
answered Nov 24 at 17:09
CyclotomicField
2,1141312
add a comment |
up vote
4
down vote
Recall that surjective means that for any $bin mathbb{R^n}$ the system
$$Ax=b$$
has solution, that is true if and only if $dim(Im(A))=n$ that if and only if $A$ is full rank.
answered Nov 24 at 17:08
gimusi
88.5k74394
add a comment |
up vote
1
down vote
Maybe your getting confused because there is a fact that says that a $ntimes n$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=dim Ker (A) + dim Im (A)$$ and its clear that $dim Ker(A)=0 implies dim Im(A)=n$ which in this case means that $A$ is surjective.
answered Nov 24 at 23:13
Robson
725221
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
No it's not. To see this consider the zero matrix as a map from $mathbb{R}^nrightarrow mathbb{R}^n$ which is clearly not surjective.
answered Nov 24 at 17:09
CyclotomicField
2,1141312
add a comment |
up vote
5
down vote
No it's not. To see this consider the zero matrix as a map from $mathbb{R}^nrightarrow mathbb{R}^n$ which is clearly not surjective.
answered Nov 24 at 17:09
CyclotomicField
2,1141312
add a comment |
up vote
5
down vote
up vote
5
down vote
No it's not. To see this consider the zero matrix as a map from $mathbb{R}^nrightarrow mathbb{R}^n$ which is clearly not surjective.
answered Nov 24 at 17:09
CyclotomicField
2,1141312
No it's not. To see this consider the zero matrix as a map from $mathbb{R}^nrightarrow mathbb{R}^n$ which is clearly not surjective.
answered Nov 24 at 17:09
CyclotomicField
2,1141312
answered Nov 24 at 17:09
CyclotomicField
2,1141312
answered Nov 24 at 17:09
CyclotomicField
2,1141312
answered Nov 24 at 17:09
CyclotomicField
2,1141312
2,1141312
add a comment |
add a comment |
up vote
4
down vote
Recall that surjective means that for any $bin mathbb{R^n}$ the system
$$Ax=b$$
has solution, that is true if and only if $dim(Im(A))=n$ that if and only if $A$ is full rank.
answered Nov 24 at 17:08
gimusi
88.5k74394
add a comment |
up vote
4
down vote
Recall that surjective means that for any $bin mathbb{R^n}$ the system
$$Ax=b$$
has solution, that is true if and only if $dim(Im(A))=n$ that if and only if $A$ is full rank.
answered Nov 24 at 17:08
gimusi
88.5k74394
add a comment |
up vote
4
down vote
up vote
4
down vote
Recall that surjective means that for any $bin mathbb{R^n}$ the system
$$Ax=b$$
has solution, that is true if and only if $dim(Im(A))=n$ that if and only if $A$ is full rank.
answered Nov 24 at 17:08
gimusi
88.5k74394
Recall that surjective means that for any $bin mathbb{R^n}$ the system
$$Ax=b$$
has solution, that is true if and only if $dim(Im(A))=n$ that if and only if $A$ is full rank.
answered Nov 24 at 17:08
gimusi
88.5k74394
edited Nov 24 at 17:16
answered Nov 24 at 17:08
gimusi
88.5k74394
answered Nov 24 at 17:08
gimusi
88.5k74394
answered Nov 24 at 17:08
gimusi
88.5k74394
88.5k74394
add a comment |
add a comment |
up vote
1
down vote
Maybe your getting confused because there is a fact that says that a $ntimes n$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=dim Ker (A) + dim Im (A)$$ and its clear that $dim Ker(A)=0 implies dim Im(A)=n$ which in this case means that $A$ is surjective.
answered Nov 24 at 23:13
Robson
725221
add a comment |
up vote
1
down vote
Maybe your getting confused because there is a fact that says that a $ntimes n$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=dim Ker (A) + dim Im (A)$$ and its clear that $dim Ker(A)=0 implies dim Im(A)=n$ which in this case means that $A$ is surjective.
answered Nov 24 at 23:13
Robson
725221
add a comment |
up vote
1
down vote
up vote
1
down vote
Maybe your getting confused because there is a fact that says that a $ntimes n$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=dim Ker (A) + dim Im (A)$$ and its clear that $dim Ker(A)=0 implies dim Im(A)=n$ which in this case means that $A$ is surjective.
answered Nov 24 at 23:13
Robson
725221
Maybe your getting confused because there is a fact that says that a $ntimes n$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=dim Ker (A) + dim Im (A)$$ and its clear that $dim Ker(A)=0 implies dim Im(A)=n$ which in this case means that $A$ is surjective.
answered Nov 24 at 23:13
Robson
725221
edited Nov 24 at 23:19
answered Nov 24 at 23:13
Robson
725221
answered Nov 24 at 23:13
Robson
725221
answered Nov 24 at 23:13
Robson
725221
725221
add a comment |
add a comment |
Elliot Silver is a new contributor. Be nice, and check out our Code of Conduct.
Elliot Silver is a new contributor. Be nice, and check out our Code of Conduct.
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5
Small language point for the future. Matrices are not surjective The linear functions they determine may be.
– Ethan Bolker
Nov 24 at 23:15