If $U_1,U_2subseteq U, U$ linear independent set then $( cap )= $












-1












$begingroup$


If I take an element $v$ in $(<U_1> cap <U_2>)$ why this element can be described as :



$v=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$



?



where $z_1,..,z_nin U_1cap U_2,x_{k+1},...,x_{k+n}in U_1backslash U_2$ and $y_{k+1},...,y_{k+m}in U_2backslash U_1$



I thought every element of $(<U_1> cap <U_2>)$ must be a linear combination of vectors that are bothin $U_1$ and $U_2$



My tought process so far:



Let $xin U_1cap U_2Rightarrow xin U_1$ and $xin U_2$



We have
$$<U_1>cap <U_2>:= {sum_{i=1}^{n}lambda_iu_i,u_iin U_1,ninmathbb{N},lambda_iinmathbb{F}}cap{sum_{j=1}^{k}mu_ju_j,u_jin U_2,kinmathbb{N},mu_jinmathbb{F}}$$



I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $ and then I have proved



$xin <U_1>cap <U_2>Longrightarrow U_1cap U_2subseteq <U_1>cap <U_2>Longrightarrow <U_1cap U_2>subseteq <U_1>cap <U_2>$



Because Intersections of subspaces are again subspaces. And $<A>$ is the smalleset subspace that contains $A$.



To prove the next part the original proof proceeds with the Claim in the head of the Body of the Question.






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    "I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $" But you can not "choose" these, the definition provides you with some, which may be different.
    $endgroup$
    – Did
    Jan 19 at 19:22
















-1












$begingroup$


If I take an element $v$ in $(<U_1> cap <U_2>)$ why this element can be described as :



$v=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$



?



where $z_1,..,z_nin U_1cap U_2,x_{k+1},...,x_{k+n}in U_1backslash U_2$ and $y_{k+1},...,y_{k+m}in U_2backslash U_1$



I thought every element of $(<U_1> cap <U_2>)$ must be a linear combination of vectors that are bothin $U_1$ and $U_2$



My tought process so far:



Let $xin U_1cap U_2Rightarrow xin U_1$ and $xin U_2$



We have
$$<U_1>cap <U_2>:= {sum_{i=1}^{n}lambda_iu_i,u_iin U_1,ninmathbb{N},lambda_iinmathbb{F}}cap{sum_{j=1}^{k}mu_ju_j,u_jin U_2,kinmathbb{N},mu_jinmathbb{F}}$$



I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $ and then I have proved



$xin <U_1>cap <U_2>Longrightarrow U_1cap U_2subseteq <U_1>cap <U_2>Longrightarrow <U_1cap U_2>subseteq <U_1>cap <U_2>$



Because Intersections of subspaces are again subspaces. And $<A>$ is the smalleset subspace that contains $A$.



To prove the next part the original proof proceeds with the Claim in the head of the Body of the Question.






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    "I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $" But you can not "choose" these, the definition provides you with some, which may be different.
    $endgroup$
    – Did
    Jan 19 at 19:22














-1












-1








-1





$begingroup$


If I take an element $v$ in $(<U_1> cap <U_2>)$ why this element can be described as :



$v=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$



?



where $z_1,..,z_nin U_1cap U_2,x_{k+1},...,x_{k+n}in U_1backslash U_2$ and $y_{k+1},...,y_{k+m}in U_2backslash U_1$



I thought every element of $(<U_1> cap <U_2>)$ must be a linear combination of vectors that are bothin $U_1$ and $U_2$



My tought process so far:



Let $xin U_1cap U_2Rightarrow xin U_1$ and $xin U_2$



We have
$$<U_1>cap <U_2>:= {sum_{i=1}^{n}lambda_iu_i,u_iin U_1,ninmathbb{N},lambda_iinmathbb{F}}cap{sum_{j=1}^{k}mu_ju_j,u_jin U_2,kinmathbb{N},mu_jinmathbb{F}}$$



I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $ and then I have proved



$xin <U_1>cap <U_2>Longrightarrow U_1cap U_2subseteq <U_1>cap <U_2>Longrightarrow <U_1cap U_2>subseteq <U_1>cap <U_2>$



Because Intersections of subspaces are again subspaces. And $<A>$ is the smalleset subspace that contains $A$.



To prove the next part the original proof proceeds with the Claim in the head of the Body of the Question.






linear-algebra







share|cite|improve this question











$endgroup$




If I take an element $v$ in $(<U_1> cap <U_2>)$ why this element can be described as :



$v=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$



?



where $z_1,..,z_nin U_1cap U_2,x_{k+1},...,x_{k+n}in U_1backslash U_2$ and $y_{k+1},...,y_{k+m}in U_2backslash U_1$



I thought every element of $(<U_1> cap <U_2>)$ must be a linear combination of vectors that are bothin $U_1$ and $U_2$



My tought process so far:



Let $xin U_1cap U_2Rightarrow xin U_1$ and $xin U_2$



We have
$$<U_1>cap <U_2>:= {sum_{i=1}^{n}lambda_iu_i,u_iin U_1,ninmathbb{N},lambda_iinmathbb{F}}cap{sum_{j=1}^{k}mu_ju_j,u_jin U_2,kinmathbb{N},mu_jinmathbb{F}}$$



I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $ and then I have proved



$xin <U_1>cap <U_2>Longrightarrow U_1cap U_2subseteq <U_1>cap <U_2>Longrightarrow <U_1cap U_2>subseteq <U_1>cap <U_2>$



Because Intersections of subspaces are again subspaces. And $<A>$ is the smalleset subspace that contains $A$.



To prove the next part the original proof proceeds with the Claim in the head of the Body of the Question.






linear-algebra




linear-algebra






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 19:58







RM777

















asked Jan 18 at 18:38









RM777RM777

38012




38012












  • $begingroup$
    "I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $" But you can not "choose" these, the definition provides you with some, which may be different.
    $endgroup$
    – Did
    Jan 19 at 19:22


















  • $begingroup$
    "I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $" But you can not "choose" these, the definition provides you with some, which may be different.
    $endgroup$
    – Did
    Jan 19 at 19:22
















$begingroup$
"I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $" But you can not "choose" these, the definition provides you with some, which may be different.
$endgroup$
– Did
Jan 19 at 19:22




$begingroup$
"I choose $m=1,n=1,mu=lambda=1,u_i,u_j=x $" But you can not "choose" these, the definition provides you with some, which may be different.
$endgroup$
– Did
Jan 19 at 19:22










1 Answer
1






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oldest

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$begingroup$

I take a $xin <U_1>cap<U_2>:=W$



If I take a random linear combination $sum_{i=1}^{n}mu_ia_i$ with Elements of $U_1$ (resp $U_2$). Then some of the vectors might be in the intersection, i.e $a_iin U_1cap U_2$ while others might be not, i.e $a_iin U_1backslash U_2$ (resp $U_2backslash U_1$). To be a Vector of the set $W$m $x$ must be written in the form



$$x=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$$



I.e it must be a linear combination of $<U_1>$ as well as $<U_2>$.



The reason why one distinguishes between $a_iin U_1cap U_2$ and $a_iin U_1backslash U_2$ (resp $U_2backslash U_1)$ is because i the next step one can rewrite the equation above to:



$$0=sum_{i=1}^{k}(lambda_i-mu_i)z_i + sum_{i=k+1}^{k+n}lambda_ix_i-sum_{i=k+1}^{k+m}mu_iy_i $$



The linear independens then yields $lambda_1=mu_1,....,lambda_k=mu_k$ and $lambda_{k+1}=…lambda_{k+n}=mu_{k+1}=…=mu_{k+m}=0$. Therefor x is a linear combination of vectors which are both in $U_1$ and $U_2$. I.e: $xin <U_1cap U_2>$






share|cite|improve this answer









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    1 Answer
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    $begingroup$

    I take a $xin <U_1>cap<U_2>:=W$



    If I take a random linear combination $sum_{i=1}^{n}mu_ia_i$ with Elements of $U_1$ (resp $U_2$). Then some of the vectors might be in the intersection, i.e $a_iin U_1cap U_2$ while others might be not, i.e $a_iin U_1backslash U_2$ (resp $U_2backslash U_1$). To be a Vector of the set $W$m $x$ must be written in the form



    $$x=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$$



    I.e it must be a linear combination of $<U_1>$ as well as $<U_2>$.



    The reason why one distinguishes between $a_iin U_1cap U_2$ and $a_iin U_1backslash U_2$ (resp $U_2backslash U_1)$ is because i the next step one can rewrite the equation above to:



    $$0=sum_{i=1}^{k}(lambda_i-mu_i)z_i + sum_{i=k+1}^{k+n}lambda_ix_i-sum_{i=k+1}^{k+m}mu_iy_i $$



    The linear independens then yields $lambda_1=mu_1,....,lambda_k=mu_k$ and $lambda_{k+1}=…lambda_{k+n}=mu_{k+1}=…=mu_{k+m}=0$. Therefor x is a linear combination of vectors which are both in $U_1$ and $U_2$. I.e: $xin <U_1cap U_2>$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I take a $xin <U_1>cap<U_2>:=W$



      If I take a random linear combination $sum_{i=1}^{n}mu_ia_i$ with Elements of $U_1$ (resp $U_2$). Then some of the vectors might be in the intersection, i.e $a_iin U_1cap U_2$ while others might be not, i.e $a_iin U_1backslash U_2$ (resp $U_2backslash U_1$). To be a Vector of the set $W$m $x$ must be written in the form



      $$x=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$$



      I.e it must be a linear combination of $<U_1>$ as well as $<U_2>$.



      The reason why one distinguishes between $a_iin U_1cap U_2$ and $a_iin U_1backslash U_2$ (resp $U_2backslash U_1)$ is because i the next step one can rewrite the equation above to:



      $$0=sum_{i=1}^{k}(lambda_i-mu_i)z_i + sum_{i=k+1}^{k+n}lambda_ix_i-sum_{i=k+1}^{k+m}mu_iy_i $$



      The linear independens then yields $lambda_1=mu_1,....,lambda_k=mu_k$ and $lambda_{k+1}=…lambda_{k+n}=mu_{k+1}=…=mu_{k+m}=0$. Therefor x is a linear combination of vectors which are both in $U_1$ and $U_2$. I.e: $xin <U_1cap U_2>$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I take a $xin <U_1>cap<U_2>:=W$



        If I take a random linear combination $sum_{i=1}^{n}mu_ia_i$ with Elements of $U_1$ (resp $U_2$). Then some of the vectors might be in the intersection, i.e $a_iin U_1cap U_2$ while others might be not, i.e $a_iin U_1backslash U_2$ (resp $U_2backslash U_1$). To be a Vector of the set $W$m $x$ must be written in the form



        $$x=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$$



        I.e it must be a linear combination of $<U_1>$ as well as $<U_2>$.



        The reason why one distinguishes between $a_iin U_1cap U_2$ and $a_iin U_1backslash U_2$ (resp $U_2backslash U_1)$ is because i the next step one can rewrite the equation above to:



        $$0=sum_{i=1}^{k}(lambda_i-mu_i)z_i + sum_{i=k+1}^{k+n}lambda_ix_i-sum_{i=k+1}^{k+m}mu_iy_i $$



        The linear independens then yields $lambda_1=mu_1,....,lambda_k=mu_k$ and $lambda_{k+1}=…lambda_{k+n}=mu_{k+1}=…=mu_{k+m}=0$. Therefor x is a linear combination of vectors which are both in $U_1$ and $U_2$. I.e: $xin <U_1cap U_2>$






        share|cite|improve this answer









        $endgroup$



        I take a $xin <U_1>cap<U_2>:=W$



        If I take a random linear combination $sum_{i=1}^{n}mu_ia_i$ with Elements of $U_1$ (resp $U_2$). Then some of the vectors might be in the intersection, i.e $a_iin U_1cap U_2$ while others might be not, i.e $a_iin U_1backslash U_2$ (resp $U_2backslash U_1$). To be a Vector of the set $W$m $x$ must be written in the form



        $$x=sum_{i=1}^{k}lambda_iz_i+sum_{i=k+1}^{k+n}lambda_{i}x_i=sum_{i=1}^{k}mu_iz_i+sum_{i=k+1}^{k+m}mu_i y_i$$



        I.e it must be a linear combination of $<U_1>$ as well as $<U_2>$.



        The reason why one distinguishes between $a_iin U_1cap U_2$ and $a_iin U_1backslash U_2$ (resp $U_2backslash U_1)$ is because i the next step one can rewrite the equation above to:



        $$0=sum_{i=1}^{k}(lambda_i-mu_i)z_i + sum_{i=k+1}^{k+n}lambda_ix_i-sum_{i=k+1}^{k+m}mu_iy_i $$



        The linear independens then yields $lambda_1=mu_1,....,lambda_k=mu_k$ and $lambda_{k+1}=…lambda_{k+n}=mu_{k+1}=…=mu_{k+m}=0$. Therefor x is a linear combination of vectors which are both in $U_1$ and $U_2$. I.e: $xin <U_1cap U_2>$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 18 at 20:24









        RM777RM777

        38012




        38012






























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