Are there matrices with same column space but different rank

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Is there exist some matrices $A,B$ that they have the same column space but different rank? (I do not get if this matrices is $ntimes n$ or $mtimes n$).



I think that they share the same column space something like this




A=$beginbmatrix
1& 0\
0& 0
endbmatrix$
B=$beginbmatrix
1& 0\
0& 1
endbmatrix$.




Here they have the same one vector from column space, but I do not know is this meaning that they need to have every vector of column space the same or not, what do you mean?






linear-algebra matrices matrix-rank







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  • In your example, $boldsymbol A, boldsymbol B$ has different column spaces.
    – xbh
    Sep 2 at 10:31














up vote
0
down vote

favorite












Is there exist some matrices $A,B$ that they have the same column space but different rank? (I do not get if this matrices is $ntimes n$ or $mtimes n$).



I think that they share the same column space something like this




A=$beginbmatrix
1& 0\
0& 0
endbmatrix$
B=$beginbmatrix
1& 0\
0& 1
endbmatrix$.




Here they have the same one vector from column space, but I do not know is this meaning that they need to have every vector of column space the same or not, what do you mean?






linear-algebra matrices matrix-rank







share|cite|improve this question























  • In your example, $boldsymbol A, boldsymbol B$ has different column spaces.
    – xbh
    Sep 2 at 10:31












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Is there exist some matrices $A,B$ that they have the same column space but different rank? (I do not get if this matrices is $ntimes n$ or $mtimes n$).



I think that they share the same column space something like this




A=$beginbmatrix
1& 0\
0& 0
endbmatrix$
B=$beginbmatrix
1& 0\
0& 1
endbmatrix$.




Here they have the same one vector from column space, but I do not know is this meaning that they need to have every vector of column space the same or not, what do you mean?






linear-algebra matrices matrix-rank







share|cite|improve this question















Is there exist some matrices $A,B$ that they have the same column space but different rank? (I do not get if this matrices is $ntimes n$ or $mtimes n$).



I think that they share the same column space something like this




A=$beginbmatrix
1& 0\
0& 0
endbmatrix$
B=$beginbmatrix
1& 0\
0& 1
endbmatrix$.




Here they have the same one vector from column space, but I do not know is this meaning that they need to have every vector of column space the same or not, what do you mean?






linear-algebra matrices matrix-rank




linear-algebra matrices matrix-rank






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edited Sep 2 at 10:37









José Carlos Santos

121k16101186




121k16101186










asked Sep 2 at 10:28









Marko Škorić

3918




3918











  • In your example, $boldsymbol A, boldsymbol B$ has different column spaces.
    – xbh
    Sep 2 at 10:31
















  • In your example, $boldsymbol A, boldsymbol B$ has different column spaces.
    – xbh
    Sep 2 at 10:31















In your example, $boldsymbol A, boldsymbol B$ has different column spaces.
– xbh
Sep 2 at 10:31




In your example, $boldsymbol A, boldsymbol B$ has different column spaces.
– xbh
Sep 2 at 10:31










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










The rank of a matrix is equal to the dimension of the column space (and also equal to the dimension of the row space). Therefore, the answer is negative.






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  • thank you for your help
    – Marko Å korić
    Sep 2 at 10:32










  • @MarkoÅ korić If my answer was useful, perhaps that you could mark it as the accepted one.
    – José Carlos Santos
    Sep 2 at 11:04










  • I try but I need to wait for 20 minutes, sorry
    – Marko Å korić
    Sep 2 at 11:06










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










The rank of a matrix is equal to the dimension of the column space (and also equal to the dimension of the row space). Therefore, the answer is negative.






share|cite|improve this answer




















  • thank you for your help
    – Marko Å korić
    Sep 2 at 10:32










  • @MarkoÅ korić If my answer was useful, perhaps that you could mark it as the accepted one.
    – José Carlos Santos
    Sep 2 at 11:04










  • I try but I need to wait for 20 minutes, sorry
    – Marko Å korić
    Sep 2 at 11:06














up vote
1
down vote



accepted










The rank of a matrix is equal to the dimension of the column space (and also equal to the dimension of the row space). Therefore, the answer is negative.






share|cite|improve this answer




















  • thank you for your help
    – Marko Å korić
    Sep 2 at 10:32










  • @MarkoÅ korić If my answer was useful, perhaps that you could mark it as the accepted one.
    – José Carlos Santos
    Sep 2 at 11:04










  • I try but I need to wait for 20 minutes, sorry
    – Marko Å korić
    Sep 2 at 11:06












up vote
1
down vote



accepted







up vote
1
down vote



accepted






The rank of a matrix is equal to the dimension of the column space (and also equal to the dimension of the row space). Therefore, the answer is negative.






share|cite|improve this answer












The rank of a matrix is equal to the dimension of the column space (and also equal to the dimension of the row space). Therefore, the answer is negative.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 2 at 10:31









José Carlos Santos

121k16101186




121k16101186











  • thank you for your help
    – Marko Å korić
    Sep 2 at 10:32










  • @MarkoÅ korić If my answer was useful, perhaps that you could mark it as the accepted one.
    – José Carlos Santos
    Sep 2 at 11:04










  • I try but I need to wait for 20 minutes, sorry
    – Marko Å korić
    Sep 2 at 11:06
















  • thank you for your help
    – Marko Å korić
    Sep 2 at 10:32










  • @MarkoÅ korić If my answer was useful, perhaps that you could mark it as the accepted one.
    – José Carlos Santos
    Sep 2 at 11:04










  • I try but I need to wait for 20 minutes, sorry
    – Marko Å korić
    Sep 2 at 11:06















thank you for your help
– Marko Å korić
Sep 2 at 10:32




thank you for your help
– Marko Å korić
Sep 2 at 10:32












@MarkoŠkorić If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Sep 2 at 11:04




@MarkoŠkorić If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Sep 2 at 11:04












I try but I need to wait for 20 minutes, sorry
– Marko Å korić
Sep 2 at 11:06




I try but I need to wait for 20 minutes, sorry
– Marko Å korić
Sep 2 at 11:06

















 

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