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 '18 at 17:10
gimusi
1
asked Nov 24 '18 at 17:06
Elliot SilverElliot Silver
304
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add a comment |
$begingroup$
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 '18 at 17:10
gimusi
1
asked Nov 24 '18 at 17:06
Elliot SilverElliot Silver
304
$endgroup$
5
$begingroup$
Small language point for the future. Matrices are not surjective The linear functions they determine may be.
$endgroup$
– Ethan Bolker
Nov 24 '18 at 23:15
add a comment |
$begingroup$
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 '18 at 17:10
gimusi
1
asked Nov 24 '18 at 17:06
Elliot SilverElliot Silver
304
$endgroup$
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 '18 at 17:10
gimusi
1
asked Nov 24 '18 at 17:06
Elliot SilverElliot Silver
304
edited Nov 24 '18 at 17:10
gimusi
1
asked Nov 24 '18 at 17:06
Elliot SilverElliot Silver
304
edited Nov 24 '18 at 17:10
gimusi
1
edited Nov 24 '18 at 17:10
gimusi
1
edited Nov 24 '18 at 17:10
gimusi
1
1
asked Nov 24 '18 at 17:06
Elliot SilverElliot Silver
304
asked Nov 24 '18 at 17:06
Elliot SilverElliot Silver
304
asked Nov 24 '18 at 17:06
Elliot SilverElliot Silver
304
304
5
$begingroup$
Small language point for the future. Matrices are not surjective The linear functions they determine may be.
$endgroup$
– Ethan Bolker
Nov 24 '18 at 23:15
add a comment |
5
$begingroup$
Small language point for the future. Matrices are not surjective The linear functions they determine may be.
$endgroup$
– Ethan Bolker
Nov 24 '18 at 23:15
5
5
$begingroup$
Small language point for the future. Matrices are not surjective The linear functions they determine may be.
$endgroup$
– Ethan Bolker
Nov 24 '18 at 23:15
$begingroup$
Small language point for the future. Matrices are not surjective The linear functions they determine may be.
$endgroup$
– Ethan Bolker
Nov 24 '18 at 23:15
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 '18 at 17:09
CyclotomicFieldCyclotomicField
2,2021313
$endgroup$
add a comment |
$begingroup$
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 '18 at 17:08
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
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 '18 at 23:13
RobsonRobson
769221
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
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active
oldest
votes
$begingroup$
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 '18 at 17:09
CyclotomicFieldCyclotomicField
2,2021313
$endgroup$
add a comment |
$begingroup$
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 '18 at 17:09
CyclotomicFieldCyclotomicField
2,2021313
$endgroup$
add a comment |
$begingroup$
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 '18 at 17:09
CyclotomicFieldCyclotomicField
2,2021313
$endgroup$
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 '18 at 17:09
CyclotomicFieldCyclotomicField
2,2021313
answered Nov 24 '18 at 17:09
CyclotomicFieldCyclotomicField
2,2021313
answered Nov 24 '18 at 17:09
CyclotomicFieldCyclotomicField
2,2021313
answered Nov 24 '18 at 17:09
CyclotomicFieldCyclotomicField
2,2021313
2,2021313
add a comment |
add a comment |
$begingroup$
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 '18 at 17:08
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
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 '18 at 17:08
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
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 '18 at 17:08
gimusigimusi
1
$endgroup$
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 '18 at 17:08
gimusigimusi
1
edited Nov 24 '18 at 17:16
answered Nov 24 '18 at 17:08
gimusigimusi
1
answered Nov 24 '18 at 17:08
gimusigimusi
1
answered Nov 24 '18 at 17:08
gimusigimusi
1
1
add a comment |
add a comment |
$begingroup$
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 '18 at 23:13
RobsonRobson
769221
$endgroup$
add a comment |
$begingroup$
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 '18 at 23:13
RobsonRobson
769221
$endgroup$
add a comment |
$begingroup$
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 '18 at 23:13
RobsonRobson
769221
$endgroup$
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 '18 at 23:13
RobsonRobson
769221
edited Nov 24 '18 at 23:19
answered Nov 24 '18 at 23:13
RobsonRobson
769221
answered Nov 24 '18 at 23:13
RobsonRobson
769221
answered Nov 24 '18 at 23:13
RobsonRobson
769221
769221
add a comment |
add a comment |
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Small language point for the future. Matrices are not surjective The linear functions they determine may be.
$endgroup$
– Ethan Bolker
Nov 24 '18 at 23:15