How can I find rank of $A=sum_{i=1}^4 x_ix_i^T$ without actually finding the matrix $A$?












1














Suppose $A=sumlimits_{i=1}^4 x_ix_i^T$ where



$x_1=(1,-1,1,0)^T,x_2=(1,1,0,1)^T,x_3=(1,3,1,0)^T$ and $x_4=(1,1,1,0)^T$.




How can I find the rank of $A$ without explicitly finding the matrix $A$ and then transforming $A$ to an echelon matrix?




I know that $operatorname{rank}(A)=3$ proceeding the usual way.



We have $operatorname{rank}(x_ix_i^T)=1$ for each $i$, but that doesn't help me much here. Also, $x_1=2x_4-x_3$. Does that imply there are $3$ linearly independent rows/columns in $A$?



As this question was supposed to be solved in the exam with limited time at hand, I am left to wonder if it can be solved rather quickly.






linear-algebra matrix-rank







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    1














    Suppose $A=sumlimits_{i=1}^4 x_ix_i^T$ where



    $x_1=(1,-1,1,0)^T,x_2=(1,1,0,1)^T,x_3=(1,3,1,0)^T$ and $x_4=(1,1,1,0)^T$.




    How can I find the rank of $A$ without explicitly finding the matrix $A$ and then transforming $A$ to an echelon matrix?




    I know that $operatorname{rank}(A)=3$ proceeding the usual way.



    We have $operatorname{rank}(x_ix_i^T)=1$ for each $i$, but that doesn't help me much here. Also, $x_1=2x_4-x_3$. Does that imply there are $3$ linearly independent rows/columns in $A$?



    As this question was supposed to be solved in the exam with limited time at hand, I am left to wonder if it can be solved rather quickly.






    linear-algebra matrix-rank







    share|cite|improve this question

























      1












      1








      1


      1





      Suppose $A=sumlimits_{i=1}^4 x_ix_i^T$ where



      $x_1=(1,-1,1,0)^T,x_2=(1,1,0,1)^T,x_3=(1,3,1,0)^T$ and $x_4=(1,1,1,0)^T$.




      How can I find the rank of $A$ without explicitly finding the matrix $A$ and then transforming $A$ to an echelon matrix?




      I know that $operatorname{rank}(A)=3$ proceeding the usual way.



      We have $operatorname{rank}(x_ix_i^T)=1$ for each $i$, but that doesn't help me much here. Also, $x_1=2x_4-x_3$. Does that imply there are $3$ linearly independent rows/columns in $A$?



      As this question was supposed to be solved in the exam with limited time at hand, I am left to wonder if it can be solved rather quickly.






      linear-algebra matrix-rank







      share|cite|improve this question













      Suppose $A=sumlimits_{i=1}^4 x_ix_i^T$ where



      $x_1=(1,-1,1,0)^T,x_2=(1,1,0,1)^T,x_3=(1,3,1,0)^T$ and $x_4=(1,1,1,0)^T$.




      How can I find the rank of $A$ without explicitly finding the matrix $A$ and then transforming $A$ to an echelon matrix?




      I know that $operatorname{rank}(A)=3$ proceeding the usual way.



      We have $operatorname{rank}(x_ix_i^T)=1$ for each $i$, but that doesn't help me much here. Also, $x_1=2x_4-x_3$. Does that imply there are $3$ linearly independent rows/columns in $A$?



      As this question was supposed to be solved in the exam with limited time at hand, I am left to wonder if it can be solved rather quickly.






      linear-algebra matrix-rank




      linear-algebra matrix-rank






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked yesterday









      StubbornAtom

      5,34411138




      5,34411138






















          3 Answers
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          The matrix $A$ can be written as $A=XX^T$ where $X=[x_1 x_2 x_3 x_4]$. Now use the fact that $operatorname{rank}XX^T=operatorname{rank}X$, so all you need to do is the find the rank of $X$.






          share|cite|improve this answer





















          • Thank you. I overlooked this decomposition.
            – StubbornAtom
            yesterday



















          0














          The rank of $A$ is by definition the dimension of the range, and the range is spanned by ${x_1, x_2, x_3, x_4}$. Thus it suffices to find the dimension of this span.



          Since both $x_1, x_3, x_4$ has $0$ in the fourth entries while $x_2$ does not. Thus if you can show that $operatorname{span}{x_1, x_3, x_4}$ is two dimensional, then the span of ${x_1, x_2, x_3, x_4}$ is three dimensional.



          You know already that $operatorname{span}{x_1, x_3, x_4}$ is at most two dimensional. That it is exactly two is easy, since obviously (e.g.) ${x_1, x_4}$ is linearly independent.






          share|cite|improve this answer








          New contributor




          Arctic Char is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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            0














            Observe that for any vector $c$,
            $$Ac = sum_{i=1}^{4} x_ix_i^Tc = sum_{i=1}^{4} (x^Tc)x_i$$



            The image of a linear map is a vector space. That means that for every $Ac$ we can find some set of vectors which spans this set. Can you find such a set (even if the set is linearly dependent)? If you can find such a set, then you just need to count how many linearly independent vectors you need to span the same space (i.e. remove vectors which are linearly dependent).






            share|cite|improve this answer





















            • How does $x_i x_i^T c$ equal $(x_i^T c) x_i$?
              – StubbornAtom
              yesterday








            • 1




              $x_i^Tc$ is a scalar, so you can move it around
              – tch
              yesterday












            • Oh I see $c$ is a vector.
              – StubbornAtom
              yesterday











            Your Answer





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            3 Answers
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            3 Answers
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            active

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            active

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            2














            The matrix $A$ can be written as $A=XX^T$ where $X=[x_1 x_2 x_3 x_4]$. Now use the fact that $operatorname{rank}XX^T=operatorname{rank}X$, so all you need to do is the find the rank of $X$.






            share|cite|improve this answer





















            • Thank you. I overlooked this decomposition.
              – StubbornAtom
              yesterday
















            2














            The matrix $A$ can be written as $A=XX^T$ where $X=[x_1 x_2 x_3 x_4]$. Now use the fact that $operatorname{rank}XX^T=operatorname{rank}X$, so all you need to do is the find the rank of $X$.






            share|cite|improve this answer





















            • Thank you. I overlooked this decomposition.
              – StubbornAtom
              yesterday














            2












            2








            2






            The matrix $A$ can be written as $A=XX^T$ where $X=[x_1 x_2 x_3 x_4]$. Now use the fact that $operatorname{rank}XX^T=operatorname{rank}X$, so all you need to do is the find the rank of $X$.






            share|cite|improve this answer












            The matrix $A$ can be written as $A=XX^T$ where $X=[x_1 x_2 x_3 x_4]$. Now use the fact that $operatorname{rank}XX^T=operatorname{rank}X$, so all you need to do is the find the rank of $X$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            A.Γ.

            22.6k32656




            22.6k32656












            • Thank you. I overlooked this decomposition.
              – StubbornAtom
              yesterday


















            • Thank you. I overlooked this decomposition.
              – StubbornAtom
              yesterday
















            Thank you. I overlooked this decomposition.
            – StubbornAtom
            yesterday




            Thank you. I overlooked this decomposition.
            – StubbornAtom
            yesterday











            0














            The rank of $A$ is by definition the dimension of the range, and the range is spanned by ${x_1, x_2, x_3, x_4}$. Thus it suffices to find the dimension of this span.



            Since both $x_1, x_3, x_4$ has $0$ in the fourth entries while $x_2$ does not. Thus if you can show that $operatorname{span}{x_1, x_3, x_4}$ is two dimensional, then the span of ${x_1, x_2, x_3, x_4}$ is three dimensional.



            You know already that $operatorname{span}{x_1, x_3, x_4}$ is at most two dimensional. That it is exactly two is easy, since obviously (e.g.) ${x_1, x_4}$ is linearly independent.






            share|cite|improve this answer








            New contributor




            Arctic Char is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.























              0














              The rank of $A$ is by definition the dimension of the range, and the range is spanned by ${x_1, x_2, x_3, x_4}$. Thus it suffices to find the dimension of this span.



              Since both $x_1, x_3, x_4$ has $0$ in the fourth entries while $x_2$ does not. Thus if you can show that $operatorname{span}{x_1, x_3, x_4}$ is two dimensional, then the span of ${x_1, x_2, x_3, x_4}$ is three dimensional.



              You know already that $operatorname{span}{x_1, x_3, x_4}$ is at most two dimensional. That it is exactly two is easy, since obviously (e.g.) ${x_1, x_4}$ is linearly independent.






              share|cite|improve this answer








              New contributor




              Arctic Char is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.





















                0












                0








                0






                The rank of $A$ is by definition the dimension of the range, and the range is spanned by ${x_1, x_2, x_3, x_4}$. Thus it suffices to find the dimension of this span.



                Since both $x_1, x_3, x_4$ has $0$ in the fourth entries while $x_2$ does not. Thus if you can show that $operatorname{span}{x_1, x_3, x_4}$ is two dimensional, then the span of ${x_1, x_2, x_3, x_4}$ is three dimensional.



                You know already that $operatorname{span}{x_1, x_3, x_4}$ is at most two dimensional. That it is exactly two is easy, since obviously (e.g.) ${x_1, x_4}$ is linearly independent.






                share|cite|improve this answer








                New contributor




                Arctic Char is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                The rank of $A$ is by definition the dimension of the range, and the range is spanned by ${x_1, x_2, x_3, x_4}$. Thus it suffices to find the dimension of this span.



                Since both $x_1, x_3, x_4$ has $0$ in the fourth entries while $x_2$ does not. Thus if you can show that $operatorname{span}{x_1, x_3, x_4}$ is two dimensional, then the span of ${x_1, x_2, x_3, x_4}$ is three dimensional.



                You know already that $operatorname{span}{x_1, x_3, x_4}$ is at most two dimensional. That it is exactly two is easy, since obviously (e.g.) ${x_1, x_4}$ is linearly independent.







                share|cite|improve this answer








                New contributor




                Arctic Char is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                share|cite|improve this answer



                share|cite|improve this answer






                New contributor




                Arctic Char is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                answered yesterday









                Arctic Char

                10314




                10314




                New contributor




                Arctic Char is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.





                New contributor





                Arctic Char is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                Arctic Char is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.























                    0














                    Observe that for any vector $c$,
                    $$Ac = sum_{i=1}^{4} x_ix_i^Tc = sum_{i=1}^{4} (x^Tc)x_i$$



                    The image of a linear map is a vector space. That means that for every $Ac$ we can find some set of vectors which spans this set. Can you find such a set (even if the set is linearly dependent)? If you can find such a set, then you just need to count how many linearly independent vectors you need to span the same space (i.e. remove vectors which are linearly dependent).






                    share|cite|improve this answer





















                    • How does $x_i x_i^T c$ equal $(x_i^T c) x_i$?
                      – StubbornAtom
                      yesterday








                    • 1




                      $x_i^Tc$ is a scalar, so you can move it around
                      – tch
                      yesterday












                    • Oh I see $c$ is a vector.
                      – StubbornAtom
                      yesterday
















                    0














                    Observe that for any vector $c$,
                    $$Ac = sum_{i=1}^{4} x_ix_i^Tc = sum_{i=1}^{4} (x^Tc)x_i$$



                    The image of a linear map is a vector space. That means that for every $Ac$ we can find some set of vectors which spans this set. Can you find such a set (even if the set is linearly dependent)? If you can find such a set, then you just need to count how many linearly independent vectors you need to span the same space (i.e. remove vectors which are linearly dependent).






                    share|cite|improve this answer





















                    • How does $x_i x_i^T c$ equal $(x_i^T c) x_i$?
                      – StubbornAtom
                      yesterday








                    • 1




                      $x_i^Tc$ is a scalar, so you can move it around
                      – tch
                      yesterday












                    • Oh I see $c$ is a vector.
                      – StubbornAtom
                      yesterday














                    0












                    0








                    0






                    Observe that for any vector $c$,
                    $$Ac = sum_{i=1}^{4} x_ix_i^Tc = sum_{i=1}^{4} (x^Tc)x_i$$



                    The image of a linear map is a vector space. That means that for every $Ac$ we can find some set of vectors which spans this set. Can you find such a set (even if the set is linearly dependent)? If you can find such a set, then you just need to count how many linearly independent vectors you need to span the same space (i.e. remove vectors which are linearly dependent).






                    share|cite|improve this answer












                    Observe that for any vector $c$,
                    $$Ac = sum_{i=1}^{4} x_ix_i^Tc = sum_{i=1}^{4} (x^Tc)x_i$$



                    The image of a linear map is a vector space. That means that for every $Ac$ we can find some set of vectors which spans this set. Can you find such a set (even if the set is linearly dependent)? If you can find such a set, then you just need to count how many linearly independent vectors you need to span the same space (i.e. remove vectors which are linearly dependent).







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered yesterday









                    tch

                    584210




                    584210












                    • How does $x_i x_i^T c$ equal $(x_i^T c) x_i$?
                      – StubbornAtom
                      yesterday








                    • 1




                      $x_i^Tc$ is a scalar, so you can move it around
                      – tch
                      yesterday












                    • Oh I see $c$ is a vector.
                      – StubbornAtom
                      yesterday


















                    • How does $x_i x_i^T c$ equal $(x_i^T c) x_i$?
                      – StubbornAtom
                      yesterday








                    • 1




                      $x_i^Tc$ is a scalar, so you can move it around
                      – tch
                      yesterday












                    • Oh I see $c$ is a vector.
                      – StubbornAtom
                      yesterday
















                    How does $x_i x_i^T c$ equal $(x_i^T c) x_i$?
                    – StubbornAtom
                    yesterday






                    How does $x_i x_i^T c$ equal $(x_i^T c) x_i$?
                    – StubbornAtom
                    yesterday






                    1




                    1




                    $x_i^Tc$ is a scalar, so you can move it around
                    – tch
                    yesterday






                    $x_i^Tc$ is a scalar, so you can move it around
                    – tch
                    yesterday














                    Oh I see $c$ is a vector.
                    – StubbornAtom
                    yesterday




                    Oh I see $c$ is a vector.
                    – StubbornAtom
                    yesterday


















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