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Can a matrix transformation be onto but not one to one?
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Is there a way for a matrix transformation to be onto and not 1–1, or is that not true?
linear-algebra
asked 1 hour ago
Kevin Robinson
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up vote
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down vote
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Is there a way for a matrix transformation to be onto and not 1–1, or is that not true?
linear-algebra
asked 1 hour ago
Kevin Robinson
161
New contributor
Kevin Robinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In general every function is onto its image. So take any transformation which is not $1-1$ and define the range to be its image. That's it.
– Mark
1 hour ago
1
Any $ntimes n$ square matrix that is onto will necessarily be one-to-one and vice versa. Examples of matrices that are onto but not one-to-one or that are one-to-one but not onto are going to be non-square rectangular matrices.
– JMoravitz
1 hour ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Is there a way for a matrix transformation to be onto and not 1–1, or is that not true?
linear-algebra
asked 1 hour ago
Kevin Robinson
161
New contributor
Kevin Robinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is there a way for a matrix transformation to be onto and not 1–1, or is that not true?
linear-algebra
linear-algebra
asked 1 hour ago
Kevin Robinson
161
New contributor
Kevin Robinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Kevin Robinson
161
New contributor
Kevin Robinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Kevin Robinson
161
New contributor
Kevin Robinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Kevin Robinson
161
asked 1 hour ago
Kevin Robinson
161
161
New contributor
Kevin Robinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kevin Robinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kevin Robinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In general every function is onto its image. So take any transformation which is not $1-1$ and define the range to be its image. That's it.
– Mark
1 hour ago
1
Any $ntimes n$ square matrix that is onto will necessarily be one-to-one and vice versa. Examples of matrices that are onto but not one-to-one or that are one-to-one but not onto are going to be non-square rectangular matrices.
– JMoravitz
1 hour ago
add a comment |Â
In general every function is onto its image. So take any transformation which is not $1-1$ and define the range to be its image. That's it.
– Mark
1 hour ago
1
Any $ntimes n$ square matrix that is onto will necessarily be one-to-one and vice versa. Examples of matrices that are onto but not one-to-one or that are one-to-one but not onto are going to be non-square rectangular matrices.
– JMoravitz
1 hour ago
In general every function is onto its image. So take any transformation which is not $1-1$ and define the range to be its image. That's it.
– Mark
1 hour ago
In general every function is onto its image. So take any transformation which is not $1-1$ and define the range to be its image. That's it.
– Mark
1 hour ago
1
1
Any $ntimes n$ square matrix that is onto will necessarily be one-to-one and vice versa. Examples of matrices that are onto but not one-to-one or that are one-to-one but not onto are going to be non-square rectangular matrices.
– JMoravitz
1 hour ago
Any $ntimes n$ square matrix that is onto will necessarily be one-to-one and vice versa. Examples of matrices that are onto but not one-to-one or that are one-to-one but not onto are going to be non-square rectangular matrices.
– JMoravitz
1 hour ago
add a comment |Â
2 Answers
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up vote
3
down vote
Any nonzero $1 times 2$ matrix will do the job.
Note that for square matrices, onto is equivalent to 1-1 by the rank-nullity theorem.
answered 1 hour ago
Alex Provost
15.1k22250
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up vote
3
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Put $A= beginpmatrix
1 & 1\
endpmatrix.$
Then $A:BbbR^2 longrightarrow BbbR$ is onto, but not one-to-one.
If $bin BbbR$ then $$Abeginpmatrix b\
0 endpmatrix=1b+01=b$$
so $A$ is onto.
But $$Abeginpmatrix 1\
-1 endpmatrix=1(-1)+1(1)=0$$
so $A$ is not one-to-one, since $ker(A)ne vec 0$.
answered 1 hour ago
Chickenmancer
3,141622
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Any nonzero $1 times 2$ matrix will do the job.
Note that for square matrices, onto is equivalent to 1-1 by the rank-nullity theorem.
answered 1 hour ago
Alex Provost
15.1k22250
add a comment |Â
up vote
3
down vote
Any nonzero $1 times 2$ matrix will do the job.
Note that for square matrices, onto is equivalent to 1-1 by the rank-nullity theorem.
answered 1 hour ago
Alex Provost
15.1k22250
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Any nonzero $1 times 2$ matrix will do the job.
Note that for square matrices, onto is equivalent to 1-1 by the rank-nullity theorem.
answered 1 hour ago
Alex Provost
15.1k22250
Any nonzero $1 times 2$ matrix will do the job.
Note that for square matrices, onto is equivalent to 1-1 by the rank-nullity theorem.
answered 1 hour ago
Alex Provost
15.1k22250
answered 1 hour ago
Alex Provost
15.1k22250
answered 1 hour ago
Alex Provost
15.1k22250
answered 1 hour ago
Alex Provost
15.1k22250
15.1k22250
add a comment |Â
add a comment |Â
up vote
3
down vote
Put $A= beginpmatrix
1 & 1\
endpmatrix.$
Then $A:BbbR^2 longrightarrow BbbR$ is onto, but not one-to-one.
If $bin BbbR$ then $$Abeginpmatrix b\
0 endpmatrix=1b+01=b$$
so $A$ is onto.
But $$Abeginpmatrix 1\
-1 endpmatrix=1(-1)+1(1)=0$$
so $A$ is not one-to-one, since $ker(A)ne vec 0$.
answered 1 hour ago
Chickenmancer
3,141622
add a comment |Â
up vote
3
down vote
Put $A= beginpmatrix
1 & 1\
endpmatrix.$
Then $A:BbbR^2 longrightarrow BbbR$ is onto, but not one-to-one.
If $bin BbbR$ then $$Abeginpmatrix b\
0 endpmatrix=1b+01=b$$
so $A$ is onto.
But $$Abeginpmatrix 1\
-1 endpmatrix=1(-1)+1(1)=0$$
so $A$ is not one-to-one, since $ker(A)ne vec 0$.
answered 1 hour ago
Chickenmancer
3,141622
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Put $A= beginpmatrix
1 & 1\
endpmatrix.$
Then $A:BbbR^2 longrightarrow BbbR$ is onto, but not one-to-one.
If $bin BbbR$ then $$Abeginpmatrix b\
0 endpmatrix=1b+01=b$$
so $A$ is onto.
But $$Abeginpmatrix 1\
-1 endpmatrix=1(-1)+1(1)=0$$
so $A$ is not one-to-one, since $ker(A)ne vec 0$.
answered 1 hour ago
Chickenmancer
3,141622
Put $A= beginpmatrix
1 & 1\
endpmatrix.$
Then $A:BbbR^2 longrightarrow BbbR$ is onto, but not one-to-one.
If $bin BbbR$ then $$Abeginpmatrix b\
0 endpmatrix=1b+01=b$$
so $A$ is onto.
But $$Abeginpmatrix 1\
-1 endpmatrix=1(-1)+1(1)=0$$
so $A$ is not one-to-one, since $ker(A)ne vec 0$.
answered 1 hour ago
Chickenmancer
3,141622
answered 1 hour ago
Chickenmancer
3,141622
answered 1 hour ago
Chickenmancer
3,141622
answered 1 hour ago
Chickenmancer
3,141622
3,141622
add a comment |Â
add a comment |Â
Kevin Robinson is a new contributor. Be nice, and check out our Code of Conduct.
Kevin Robinson is a new contributor. Be nice, and check out our Code of Conduct.
Kevin Robinson is a new contributor. Be nice, and check out our Code of Conduct.
Kevin Robinson is a new contributor. Be nice, and check out our Code of Conduct.
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In general every function is onto its image. So take any transformation which is not $1-1$ and define the range to be its image. That's it.
– Mark
1 hour ago
1
Any $ntimes n$ square matrix that is onto will necessarily be one-to-one and vice versa. Examples of matrices that are onto but not one-to-one or that are one-to-one but not onto are going to be non-square rectangular matrices.
– JMoravitz
1 hour ago