Column space and null space

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Let $Ain M_5,7(mathbbR)$ be a matrix such that $Ax=b$ has solution for every $b$.




I have to say what this information tells me about column- and null-space and rows of a matrix. The only thing I can think of is that column-space can have dimension $1$ to $5$ and null-space dimension can be deduced using rank-nullity theorem.



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    favorite













    Let $Ain M_5,7(mathbbR)$ be a matrix such that $Ax=b$ has solution for every $b$.




    I have to say what this information tells me about column- and null-space and rows of a matrix. The only thing I can think of is that column-space can have dimension $1$ to $5$ and null-space dimension can be deduced using rank-nullity theorem.



    Is there something more to see here?










    share|cite|improve this question























      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite












      Let $Ain M_5,7(mathbbR)$ be a matrix such that $Ax=b$ has solution for every $b$.




      I have to say what this information tells me about column- and null-space and rows of a matrix. The only thing I can think of is that column-space can have dimension $1$ to $5$ and null-space dimension can be deduced using rank-nullity theorem.



      Is there something more to see here?










      share|cite|improve this question














      Let $Ain M_5,7(mathbbR)$ be a matrix such that $Ax=b$ has solution for every $b$.




      I have to say what this information tells me about column- and null-space and rows of a matrix. The only thing I can think of is that column-space can have dimension $1$ to $5$ and null-space dimension can be deduced using rank-nullity theorem.



      Is there something more to see here?







      linear-algebra matrices systems-of-equations






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      smiljanic997

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          Column space is all of $mathbbR^5$ since any $bin mathbbR^5$ appears in the range of $A$. Thus rank is 5, and so the rank-nullity theorem says $5+mathrmnullity=7$, thus $mathrmnullity=2$.






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            Column space is all of $mathbbR^5$ since any $bin mathbbR^5$ appears in the range of $A$. Thus rank is 5, and so the rank-nullity theorem says $5+mathrmnullity=7$, thus $mathrmnullity=2$.






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              up vote
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              Column space is all of $mathbbR^5$ since any $bin mathbbR^5$ appears in the range of $A$. Thus rank is 5, and so the rank-nullity theorem says $5+mathrmnullity=7$, thus $mathrmnullity=2$.






              share|cite|improve this answer






















                up vote
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                up vote
                5
                down vote









                Column space is all of $mathbbR^5$ since any $bin mathbbR^5$ appears in the range of $A$. Thus rank is 5, and so the rank-nullity theorem says $5+mathrmnullity=7$, thus $mathrmnullity=2$.






                share|cite|improve this answer












                Column space is all of $mathbbR^5$ since any $bin mathbbR^5$ appears in the range of $A$. Thus rank is 5, and so the rank-nullity theorem says $5+mathrmnullity=7$, thus $mathrmnullity=2$.







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                answered 3 hours ago









                Ben Mares

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