Column Space, Rank and Matrix Concatenation












1












$begingroup$


I have the following question:



Given Matrices $A$ and $B$, the following relation exists between their column spaces:



$$text{col}(B) subseteq text{col}(A)$$



Then, which of the following is true for Matrix $C=[A,,,,,B]$?



A) $text{rank}(C)=text{rank}(A)$



B) $text{rank}(C)=text{rank}(B)$



C) It is not possible to specify $text{rank}(C)$ in terms of $text{rank}(A)$ and $text{rank}(B)$



My guess, and it seems a reasonable one, would be alternative (A), but I don't know how to solve/express it mathematically.



Is my guess correct? Could you walk me through the steps to prove it?



Thanks in advance






linear-algebra matrices matrix-rank







share|cite|improve this question









$endgroup$












  • $begingroup$
    It is true that $mathrm{rank}(C) = mathrm{rank}(A)$, since $mathrm{col}(C)subsetmathrm{col}(A)$ and vice versa.
    $endgroup$
    – Math1000
    Jan 4 at 22:19






  • 1




    $begingroup$
    What is the column space of C?
    $endgroup$
    – Doug M
    Jan 4 at 22:20






  • 1




    $begingroup$
    $C binom{x}{y} = Ax + By$. Hence ${cal R}C = {cal R}A + {cal R} B$. You are given ${cal R}B subset {cal R} A$.
    $endgroup$
    – copper.hat
    Jan 4 at 22:31












  • $begingroup$
    Is it correct that if $text{col}(B) subseteq text{col}(A)$, then $text{rank}(B) leq text{rank}(A)$ and the rank of a concatenated matrix is equal to the maximum rank among the concatenating matrices? Is this a way to solve it?
    $endgroup$
    – bertozzijr
    Jan 4 at 23:11
















1












$begingroup$


I have the following question:



Given Matrices $A$ and $B$, the following relation exists between their column spaces:



$$text{col}(B) subseteq text{col}(A)$$



Then, which of the following is true for Matrix $C=[A,,,,,B]$?



A) $text{rank}(C)=text{rank}(A)$



B) $text{rank}(C)=text{rank}(B)$



C) It is not possible to specify $text{rank}(C)$ in terms of $text{rank}(A)$ and $text{rank}(B)$



My guess, and it seems a reasonable one, would be alternative (A), but I don't know how to solve/express it mathematically.



Is my guess correct? Could you walk me through the steps to prove it?



Thanks in advance






linear-algebra matrices matrix-rank







share|cite|improve this question









$endgroup$












  • $begingroup$
    It is true that $mathrm{rank}(C) = mathrm{rank}(A)$, since $mathrm{col}(C)subsetmathrm{col}(A)$ and vice versa.
    $endgroup$
    – Math1000
    Jan 4 at 22:19






  • 1




    $begingroup$
    What is the column space of C?
    $endgroup$
    – Doug M
    Jan 4 at 22:20






  • 1




    $begingroup$
    $C binom{x}{y} = Ax + By$. Hence ${cal R}C = {cal R}A + {cal R} B$. You are given ${cal R}B subset {cal R} A$.
    $endgroup$
    – copper.hat
    Jan 4 at 22:31












  • $begingroup$
    Is it correct that if $text{col}(B) subseteq text{col}(A)$, then $text{rank}(B) leq text{rank}(A)$ and the rank of a concatenated matrix is equal to the maximum rank among the concatenating matrices? Is this a way to solve it?
    $endgroup$
    – bertozzijr
    Jan 4 at 23:11














1












1








1





$begingroup$


I have the following question:



Given Matrices $A$ and $B$, the following relation exists between their column spaces:



$$text{col}(B) subseteq text{col}(A)$$



Then, which of the following is true for Matrix $C=[A,,,,,B]$?



A) $text{rank}(C)=text{rank}(A)$



B) $text{rank}(C)=text{rank}(B)$



C) It is not possible to specify $text{rank}(C)$ in terms of $text{rank}(A)$ and $text{rank}(B)$



My guess, and it seems a reasonable one, would be alternative (A), but I don't know how to solve/express it mathematically.



Is my guess correct? Could you walk me through the steps to prove it?



Thanks in advance






linear-algebra matrices matrix-rank







share|cite|improve this question









$endgroup$




I have the following question:



Given Matrices $A$ and $B$, the following relation exists between their column spaces:



$$text{col}(B) subseteq text{col}(A)$$



Then, which of the following is true for Matrix $C=[A,,,,,B]$?



A) $text{rank}(C)=text{rank}(A)$



B) $text{rank}(C)=text{rank}(B)$



C) It is not possible to specify $text{rank}(C)$ in terms of $text{rank}(A)$ and $text{rank}(B)$



My guess, and it seems a reasonable one, would be alternative (A), but I don't know how to solve/express it mathematically.



Is my guess correct? Could you walk me through the steps to prove it?



Thanks in advance






linear-algebra matrices matrix-rank




linear-algebra matrices matrix-rank






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 22:16









bertozzijrbertozzijr

5911618




5911618












  • $begingroup$
    It is true that $mathrm{rank}(C) = mathrm{rank}(A)$, since $mathrm{col}(C)subsetmathrm{col}(A)$ and vice versa.
    $endgroup$
    – Math1000
    Jan 4 at 22:19






  • 1




    $begingroup$
    What is the column space of C?
    $endgroup$
    – Doug M
    Jan 4 at 22:20






  • 1




    $begingroup$
    $C binom{x}{y} = Ax + By$. Hence ${cal R}C = {cal R}A + {cal R} B$. You are given ${cal R}B subset {cal R} A$.
    $endgroup$
    – copper.hat
    Jan 4 at 22:31












  • $begingroup$
    Is it correct that if $text{col}(B) subseteq text{col}(A)$, then $text{rank}(B) leq text{rank}(A)$ and the rank of a concatenated matrix is equal to the maximum rank among the concatenating matrices? Is this a way to solve it?
    $endgroup$
    – bertozzijr
    Jan 4 at 23:11


















  • $begingroup$
    It is true that $mathrm{rank}(C) = mathrm{rank}(A)$, since $mathrm{col}(C)subsetmathrm{col}(A)$ and vice versa.
    $endgroup$
    – Math1000
    Jan 4 at 22:19






  • 1




    $begingroup$
    What is the column space of C?
    $endgroup$
    – Doug M
    Jan 4 at 22:20






  • 1




    $begingroup$
    $C binom{x}{y} = Ax + By$. Hence ${cal R}C = {cal R}A + {cal R} B$. You are given ${cal R}B subset {cal R} A$.
    $endgroup$
    – copper.hat
    Jan 4 at 22:31












  • $begingroup$
    Is it correct that if $text{col}(B) subseteq text{col}(A)$, then $text{rank}(B) leq text{rank}(A)$ and the rank of a concatenated matrix is equal to the maximum rank among the concatenating matrices? Is this a way to solve it?
    $endgroup$
    – bertozzijr
    Jan 4 at 23:11
















$begingroup$
It is true that $mathrm{rank}(C) = mathrm{rank}(A)$, since $mathrm{col}(C)subsetmathrm{col}(A)$ and vice versa.
$endgroup$
– Math1000
Jan 4 at 22:19




$begingroup$
It is true that $mathrm{rank}(C) = mathrm{rank}(A)$, since $mathrm{col}(C)subsetmathrm{col}(A)$ and vice versa.
$endgroup$
– Math1000
Jan 4 at 22:19




1




1




$begingroup$
What is the column space of C?
$endgroup$
– Doug M
Jan 4 at 22:20




$begingroup$
What is the column space of C?
$endgroup$
– Doug M
Jan 4 at 22:20




1




1




$begingroup$
$C binom{x}{y} = Ax + By$. Hence ${cal R}C = {cal R}A + {cal R} B$. You are given ${cal R}B subset {cal R} A$.
$endgroup$
– copper.hat
Jan 4 at 22:31






$begingroup$
$C binom{x}{y} = Ax + By$. Hence ${cal R}C = {cal R}A + {cal R} B$. You are given ${cal R}B subset {cal R} A$.
$endgroup$
– copper.hat
Jan 4 at 22:31














$begingroup$
Is it correct that if $text{col}(B) subseteq text{col}(A)$, then $text{rank}(B) leq text{rank}(A)$ and the rank of a concatenated matrix is equal to the maximum rank among the concatenating matrices? Is this a way to solve it?
$endgroup$
– bertozzijr
Jan 4 at 23:11




$begingroup$
Is it correct that if $text{col}(B) subseteq text{col}(A)$, then $text{rank}(B) leq text{rank}(A)$ and the rank of a concatenated matrix is equal to the maximum rank among the concatenating matrices? Is this a way to solve it?
$endgroup$
– bertozzijr
Jan 4 at 23:11










1 Answer
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$begingroup$

Yes, that's correct.



The columns of $C$ are just the columns of $A$ followed by the columns of $B$, which are all included in $mathrm{col}(A)$, hence
$$mathrm{col}(C) = mathrm{col}(A)$$
so their dimensions - the ranks - coincide.






share|cite|improve this answer









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    $begingroup$

    Yes, that's correct.



    The columns of $C$ are just the columns of $A$ followed by the columns of $B$, which are all included in $mathrm{col}(A)$, hence
    $$mathrm{col}(C) = mathrm{col}(A)$$
    so their dimensions - the ranks - coincide.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Yes, that's correct.



      The columns of $C$ are just the columns of $A$ followed by the columns of $B$, which are all included in $mathrm{col}(A)$, hence
      $$mathrm{col}(C) = mathrm{col}(A)$$
      so their dimensions - the ranks - coincide.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Yes, that's correct.



        The columns of $C$ are just the columns of $A$ followed by the columns of $B$, which are all included in $mathrm{col}(A)$, hence
        $$mathrm{col}(C) = mathrm{col}(A)$$
        so their dimensions - the ranks - coincide.






        share|cite|improve this answer









        $endgroup$



        Yes, that's correct.



        The columns of $C$ are just the columns of $A$ followed by the columns of $B$, which are all included in $mathrm{col}(A)$, hence
        $$mathrm{col}(C) = mathrm{col}(A)$$
        so their dimensions - the ranks - coincide.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 22:52









        BerciBerci

        61.5k23674




        61.5k23674






























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