Does the vector space of all $3times3$ matrices include $3times3$ matrices with $0$ as their determinants? [closed]

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I'm still confused as to how matrices, vector spaces, and subspaces relate to each other. If a vector space $V$ contains all $3times3$ matrices wouldn't that also mean that matrices whose determinants are $0$ also be included?

linear-algebra vector-spaces determinant

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edited Jul 22 at 22:38

Daniel Buck

2,3041623

asked Jul 22 at 20:16

edmonda7

163

closed as unclear what you're asking by John Ma, Xander Henderson, Lord Shark the Unknown, Claude Leibovici, Taroccoesbrocco Jul 29 at 7:16

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  And how would that be problematic?
  – Arthur
  Jul 22 at 20:18
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  If it did not, it wouldn't have a zero element, in contradiction of the definition of a vector space.
  – saulspatz
  Jul 22 at 20:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @saulspatz You are referring to the zero matrix (which is always in any subspace) but the OP is asking abut the matrices with determinant equal to zero, those may by or may not in a given subspace.
  – gimusi
  Jul 22 at 20:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @gimusi That's not what he says.
  – saulspatz
  Jul 22 at 20:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @saulspatz Ok you have given the example of the zero matrix to show that there is always at least a matrix with $det=0$?
  – gimusi
  Jul 22 at 20:57
  

  
  
  
  

 | 
show 4 more comments

up vote
0
down vote

favorite

I'm still confused as to how matrices, vector spaces, and subspaces relate to each other. If a vector space $V$ contains all $3times3$ matrices wouldn't that also mean that matrices whose determinants are $0$ also be included?

linear-algebra vector-spaces determinant

share|cite|improve this question

edited Jul 22 at 22:38

Daniel Buck

2,3041623

asked Jul 22 at 20:16

edmonda7

163

closed as unclear what you're asking by John Ma, Xander Henderson, Lord Shark the Unknown, Claude Leibovici, Taroccoesbrocco Jul 29 at 7:16

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  And how would that be problematic?
  – Arthur
  Jul 22 at 20:18
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  If it did not, it wouldn't have a zero element, in contradiction of the definition of a vector space.
  – saulspatz
  Jul 22 at 20:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @saulspatz You are referring to the zero matrix (which is always in any subspace) but the OP is asking abut the matrices with determinant equal to zero, those may by or may not in a given subspace.
  – gimusi
  Jul 22 at 20:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @gimusi That's not what he says.
  – saulspatz
  Jul 22 at 20:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @saulspatz Ok you have given the example of the zero matrix to show that there is always at least a matrix with $det=0$?
  – gimusi
  Jul 22 at 20:57
  

  
  
  
  

 | 
show 4 more comments

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I'm still confused as to how matrices, vector spaces, and subspaces relate to each other. If a vector space $V$ contains all $3times3$ matrices wouldn't that also mean that matrices whose determinants are $0$ also be included?

linear-algebra vector-spaces determinant

share|cite|improve this question

edited Jul 22 at 22:38

Daniel Buck

2,3041623

asked Jul 22 at 20:16

edmonda7

163

I'm still confused as to how matrices, vector spaces, and subspaces relate to each other. If a vector space $V$ contains all $3times3$ matrices wouldn't that also mean that matrices whose determinants are $0$ also be included?

linear-algebra vector-spaces determinant

share|cite|improve this question

edited Jul 22 at 22:38

Daniel Buck

2,3041623

asked Jul 22 at 20:16

edmonda7

163

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 22:38

Daniel Buck

2,3041623

edited Jul 22 at 22:38

Daniel Buck

2,3041623

edited Jul 22 at 22:38

Daniel Buck

2,3041623

2,3041623

asked Jul 22 at 20:16

edmonda7

163

asked Jul 22 at 20:16

edmonda7

163

asked Jul 22 at 20:16

edmonda7

163

163

closed as unclear what you're asking by John Ma, Xander Henderson, Lord Shark the Unknown, Claude Leibovici, Taroccoesbrocco Jul 29 at 7:16

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

closed as unclear what you're asking by John Ma, Xander Henderson, Lord Shark the Unknown, Claude Leibovici, Taroccoesbrocco Jul 29 at 7:16

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  And how would that be problematic?
  – Arthur
  Jul 22 at 20:18
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  If it did not, it wouldn't have a zero element, in contradiction of the definition of a vector space.
  – saulspatz
  Jul 22 at 20:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @saulspatz You are referring to the zero matrix (which is always in any subspace) but the OP is asking abut the matrices with determinant equal to zero, those may by or may not in a given subspace.
  – gimusi
  Jul 22 at 20:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @gimusi That's not what he says.
  – saulspatz
  Jul 22 at 20:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @saulspatz Ok you have given the example of the zero matrix to show that there is always at least a matrix with $det=0$?
  – gimusi
  Jul 22 at 20:57
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  And how would that be problematic?
  – Arthur
  Jul 22 at 20:18
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  If it did not, it wouldn't have a zero element, in contradiction of the definition of a vector space.
  – saulspatz
  Jul 22 at 20:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @saulspatz You are referring to the zero matrix (which is always in any subspace) but the OP is asking abut the matrices with determinant equal to zero, those may by or may not in a given subspace.
  – gimusi
  Jul 22 at 20:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @gimusi That's not what he says.
  – saulspatz
  Jul 22 at 20:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @saulspatz Ok you have given the example of the zero matrix to show that there is always at least a matrix with $det=0$?
  – gimusi
  Jul 22 at 20:57
  

  
  
  
  

1

1

And how would that be problematic?
– Arthur
Jul 22 at 20:18

And how would that be problematic?
– Arthur
Jul 22 at 20:18

2

2

If it did not, it wouldn't have a zero element, in contradiction of the definition of a vector space.
– saulspatz
Jul 22 at 20:26

If it did not, it wouldn't have a zero element, in contradiction of the definition of a vector space.
– saulspatz
Jul 22 at 20:26

@saulspatz You are referring to the zero matrix (which is always in any subspace) but the OP is asking abut the matrices with determinant equal to zero, those may by or may not in a given subspace.
– gimusi
Jul 22 at 20:38

@saulspatz You are referring to the zero matrix (which is always in any subspace) but the OP is asking abut the matrices with determinant equal to zero, those may by or may not in a given subspace.
– gimusi
Jul 22 at 20:38

@gimusi That's not what he says.
– saulspatz
Jul 22 at 20:51

@gimusi That's not what he says.
– saulspatz
Jul 22 at 20:51

@saulspatz Ok you have given the example of the zero matrix to show that there is always at least a matrix with $det=0$?
– gimusi
Jul 22 at 20:57

@saulspatz Ok you have given the example of the zero matrix to show that there is always at least a matrix with $det=0$?
– gimusi
Jul 22 at 20:57

 | 
show 4 more comments

2 Answers
2

active

oldest

votes

up vote
3
down vote

Yes of course the vector space of 3-by-3 (or n-by-n) matrices includes all the matrices and thus also those with determinant equal to zero.

What is not true is that the set of all matrices with determinant equal to zero is a subspace, can you see why?

share|cite|improve this answer

answered Jul 22 at 20:17

gimusi

65.2k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  would this mean that only some matrices with determinant equal to zero fall under the vector space while others, if tested by the axioms of closure, would not work?
  – edmonda7
  Jul 22 at 20:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @edmonda7 The set of all the n-by-n matrices is a vector space (it is trival to check it by definition).
  – gimusi
  Jul 22 at 20:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @edmonda7 But the set of all the n-by-n matrices with $det=0$ is not a vector space. Can you find a counterexample?
  – gimusi
  Jul 22 at 20:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @edmonda7 Maybe the set of 3-by-3 matrices was given with some particular propetries? Could you check your notes again?
  – gimusi
  Jul 22 at 20:27
  

  
  
  
  

add a comment | 

up vote
0
down vote

The space of $3×3$ matrices over a field $mathbb F$ is just $mathbb F^9$. This includes all the matrices with determinant zero... The reason is that we get all matrices period, including of any given determinant. That is the same as all $9$-tuples of elements of $mathbb F$. For, there are $9$ entries in a $3×3$ matrix; and each entry can be any element of $mathbb F$...

share|cite|improve this answer

edited Jul 22 at 22:31

answered Jul 22 at 21:11

Chris Custer

5,4082622

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can see this confusing the OP as it doesn't explicitly explain why the space of $3times 3$ matrices over a field $mathbbF$ contains all the matrices with determinant zero. You've just stated that we get $mathbbF^9$ and that it does contain them. But why?
  – Daniel Buck
  Jul 22 at 21:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanielBuck Right... I added a little more.
  – Chris Custer
  Jul 22 at 22:14
  

  
  
  
  

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

Yes of course the vector space of 3-by-3 (or n-by-n) matrices includes all the matrices and thus also those with determinant equal to zero.

What is not true is that the set of all matrices with determinant equal to zero is a subspace, can you see why?

share|cite|improve this answer

answered Jul 22 at 20:17

gimusi

65.2k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  would this mean that only some matrices with determinant equal to zero fall under the vector space while others, if tested by the axioms of closure, would not work?
  – edmonda7
  Jul 22 at 20:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @edmonda7 The set of all the n-by-n matrices is a vector space (it is trival to check it by definition).
  – gimusi
  Jul 22 at 20:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @edmonda7 But the set of all the n-by-n matrices with $det=0$ is not a vector space. Can you find a counterexample?
  – gimusi
  Jul 22 at 20:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @edmonda7 Maybe the set of 3-by-3 matrices was given with some particular propetries? Could you check your notes again?
  – gimusi
  Jul 22 at 20:27
  

  
  
  
  

add a comment | 

up vote
3
down vote

Yes of course the vector space of 3-by-3 (or n-by-n) matrices includes all the matrices and thus also those with determinant equal to zero.

What is not true is that the set of all matrices with determinant equal to zero is a subspace, can you see why?

share|cite|improve this answer

answered Jul 22 at 20:17

gimusi

65.2k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  would this mean that only some matrices with determinant equal to zero fall under the vector space while others, if tested by the axioms of closure, would not work?
  – edmonda7
  Jul 22 at 20:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @edmonda7 The set of all the n-by-n matrices is a vector space (it is trival to check it by definition).
  – gimusi
  Jul 22 at 20:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @edmonda7 But the set of all the n-by-n matrices with $det=0$ is not a vector space. Can you find a counterexample?
  – gimusi
  Jul 22 at 20:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @edmonda7 Maybe the set of 3-by-3 matrices was given with some particular propetries? Could you check your notes again?
  – gimusi
  Jul 22 at 20:27
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Yes of course the vector space of 3-by-3 (or n-by-n) matrices includes all the matrices and thus also those with determinant equal to zero.

What is not true is that the set of all matrices with determinant equal to zero is a subspace, can you see why?

share|cite|improve this answer

answered Jul 22 at 20:17

gimusi

65.2k73583

Yes of course the vector space of 3-by-3 (or n-by-n) matrices includes all the matrices and thus also those with determinant equal to zero.

What is not true is that the set of all matrices with determinant equal to zero is a subspace, can you see why?

share|cite|improve this answer

answered Jul 22 at 20:17

gimusi

65.2k73583

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 20:17

gimusi

65.2k73583

answered Jul 22 at 20:17

gimusi

65.2k73583

answered Jul 22 at 20:17

gimusi

65.2k73583

65.2k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  would this mean that only some matrices with determinant equal to zero fall under the vector space while others, if tested by the axioms of closure, would not work?
  – edmonda7
  Jul 22 at 20:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @edmonda7 The set of all the n-by-n matrices is a vector space (it is trival to check it by definition).
  – gimusi
  Jul 22 at 20:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @edmonda7 But the set of all the n-by-n matrices with $det=0$ is not a vector space. Can you find a counterexample?
  – gimusi
  Jul 22 at 20:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @edmonda7 Maybe the set of 3-by-3 matrices was given with some particular propetries? Could you check your notes again?
  – gimusi
  Jul 22 at 20:27
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  would this mean that only some matrices with determinant equal to zero fall under the vector space while others, if tested by the axioms of closure, would not work?
  – edmonda7
  Jul 22 at 20:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @edmonda7 The set of all the n-by-n matrices is a vector space (it is trival to check it by definition).
  – gimusi
  Jul 22 at 20:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @edmonda7 But the set of all the n-by-n matrices with $det=0$ is not a vector space. Can you find a counterexample?
  – gimusi
  Jul 22 at 20:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @edmonda7 Maybe the set of 3-by-3 matrices was given with some particular propetries? Could you check your notes again?
  – gimusi
  Jul 22 at 20:27
  

  
  
  
  

would this mean that only some matrices with determinant equal to zero fall under the vector space while others, if tested by the axioms of closure, would not work?
– edmonda7
Jul 22 at 20:22

would this mean that only some matrices with determinant equal to zero fall under the vector space while others, if tested by the axioms of closure, would not work?
– edmonda7
Jul 22 at 20:22

@edmonda7 The set of all the n-by-n matrices is a vector space (it is trival to check it by definition).
– gimusi
Jul 22 at 20:24

@edmonda7 The set of all the n-by-n matrices is a vector space (it is trival to check it by definition).
– gimusi
Jul 22 at 20:24

@edmonda7 But the set of all the n-by-n matrices with $det=0$ is not a vector space. Can you find a counterexample?
– gimusi
Jul 22 at 20:25

@edmonda7 But the set of all the n-by-n matrices with $det=0$ is not a vector space. Can you find a counterexample?
– gimusi
Jul 22 at 20:25

1

1

@edmonda7 Maybe the set of 3-by-3 matrices was given with some particular propetries? Could you check your notes again?
– gimusi
Jul 22 at 20:27

@edmonda7 Maybe the set of 3-by-3 matrices was given with some particular propetries? Could you check your notes again?
– gimusi
Jul 22 at 20:27

add a comment | 

up vote
0
down vote

The space of $3×3$ matrices over a field $mathbb F$ is just $mathbb F^9$. This includes all the matrices with determinant zero... The reason is that we get all matrices period, including of any given determinant. That is the same as all $9$-tuples of elements of $mathbb F$. For, there are $9$ entries in a $3×3$ matrix; and each entry can be any element of $mathbb F$...

share|cite|improve this answer

edited Jul 22 at 22:31

answered Jul 22 at 21:11

Chris Custer

5,4082622

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can see this confusing the OP as it doesn't explicitly explain why the space of $3times 3$ matrices over a field $mathbbF$ contains all the matrices with determinant zero. You've just stated that we get $mathbbF^9$ and that it does contain them. But why?
  – Daniel Buck
  Jul 22 at 21:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanielBuck Right... I added a little more.
  – Chris Custer
  Jul 22 at 22:14
  

  
  
  
  

add a comment | 

up vote
0
down vote

The space of $3×3$ matrices over a field $mathbb F$ is just $mathbb F^9$. This includes all the matrices with determinant zero... The reason is that we get all matrices period, including of any given determinant. That is the same as all $9$-tuples of elements of $mathbb F$. For, there are $9$ entries in a $3×3$ matrix; and each entry can be any element of $mathbb F$...

share|cite|improve this answer

edited Jul 22 at 22:31

answered Jul 22 at 21:11

Chris Custer

5,4082622

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can see this confusing the OP as it doesn't explicitly explain why the space of $3times 3$ matrices over a field $mathbbF$ contains all the matrices with determinant zero. You've just stated that we get $mathbbF^9$ and that it does contain them. But why?
  – Daniel Buck
  Jul 22 at 21:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanielBuck Right... I added a little more.
  – Chris Custer
  Jul 22 at 22:14
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

The space of $3×3$ matrices over a field $mathbb F$ is just $mathbb F^9$. This includes all the matrices with determinant zero... The reason is that we get all matrices period, including of any given determinant. That is the same as all $9$-tuples of elements of $mathbb F$. For, there are $9$ entries in a $3×3$ matrix; and each entry can be any element of $mathbb F$...

share|cite|improve this answer

edited Jul 22 at 22:31

answered Jul 22 at 21:11

Chris Custer

5,4082622

The space of $3×3$ matrices over a field $mathbb F$ is just $mathbb F^9$. This includes all the matrices with determinant zero... The reason is that we get all matrices period, including of any given determinant. That is the same as all $9$-tuples of elements of $mathbb F$. For, there are $9$ entries in a $3×3$ matrix; and each entry can be any element of $mathbb F$...

share|cite|improve this answer

edited Jul 22 at 22:31

answered Jul 22 at 21:11

Chris Custer

5,4082622

share|cite|improve this answer

share|cite|improve this answer

edited Jul 22 at 22:31

edited Jul 22 at 22:31

edited Jul 22 at 22:31

answered Jul 22 at 21:11

Chris Custer

5,4082622

answered Jul 22 at 21:11

Chris Custer

5,4082622

answered Jul 22 at 21:11

Chris Custer

5,4082622

5,4082622

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can see this confusing the OP as it doesn't explicitly explain why the space of $3times 3$ matrices over a field $mathbbF$ contains all the matrices with determinant zero. You've just stated that we get $mathbbF^9$ and that it does contain them. But why?
  – Daniel Buck
  Jul 22 at 21:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanielBuck Right... I added a little more.
  – Chris Custer
  Jul 22 at 22:14
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can see this confusing the OP as it doesn't explicitly explain why the space of $3times 3$ matrices over a field $mathbbF$ contains all the matrices with determinant zero. You've just stated that we get $mathbbF^9$ and that it does contain them. But why?
  – Daniel Buck
  Jul 22 at 21:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanielBuck Right... I added a little more.
  – Chris Custer
  Jul 22 at 22:14
  

  
  
  
  

I can see this confusing the OP as it doesn't explicitly explain why the space of $3times 3$ matrices over a field $mathbbF$ contains all the matrices with determinant zero. You've just stated that we get $mathbbF^9$ and that it does contain them. But why?
– Daniel Buck
Jul 22 at 21:47

I can see this confusing the OP as it doesn't explicitly explain why the space of $3times 3$ matrices over a field $mathbbF$ contains all the matrices with determinant zero. You've just stated that we get $mathbbF^9$ and that it does contain them. But why?
– Daniel Buck
Jul 22 at 21:47

@DanielBuck Right... I added a little more.
– Chris Custer
Jul 22 at 22:14

@DanielBuck Right... I added a little more.
– Chris Custer
Jul 22 at 22:14

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