Find $a,b$ such that $mathbb R^3 = V oplus W$












1












$begingroup$


I have some doubts with this task:




Let $a,b in mathbb R$. In linear space $mathbb R^3$ we have:
$$V = span([1,1+a,-2]^T,[2,6,-2-a]^T)= span(v_1,v_2) $$ and $$W = span([0,3,-1-b]^T,[2,2+b,-2]^T) = span(w_1,w_2)$$
Find $a,b$ such that $mathbb R^3 = V oplus W$






If $mathbb R^3 = V oplus W$ then $V cap W = left{vec{0}right}$

I think that I should consider these cases:
$1.$ $(v_1,v_2,w_1)$ are linear independent and $w_2 = alpha w_1 $

$2.$ $(v_1,v_2,w_2)$ are linear independent and $w_1 = alpha w_2 $ (but it is equivalent to $1$ so I don't have to check that)



$3.$ $(w_1,w_2,v_1)$ are linear independent and $v_2 = alpha v_1 $

$4.$ $(w_1,w_2,v_2)$ are linear independent and $v_1 = alpha v_2 $ (but it is equivalent to $3$ so I don't have to check that)



but there are a lot of calculus and I am not sure if (i) this is correct way (ii) how to do this quickly



Can somebody help me with this problem? Thanks for your time.






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I have some doubts with this task:




    Let $a,b in mathbb R$. In linear space $mathbb R^3$ we have:
    $$V = span([1,1+a,-2]^T,[2,6,-2-a]^T)= span(v_1,v_2) $$ and $$W = span([0,3,-1-b]^T,[2,2+b,-2]^T) = span(w_1,w_2)$$
    Find $a,b$ such that $mathbb R^3 = V oplus W$






    If $mathbb R^3 = V oplus W$ then $V cap W = left{vec{0}right}$

    I think that I should consider these cases:
    $1.$ $(v_1,v_2,w_1)$ are linear independent and $w_2 = alpha w_1 $

    $2.$ $(v_1,v_2,w_2)$ are linear independent and $w_1 = alpha w_2 $ (but it is equivalent to $1$ so I don't have to check that)



    $3.$ $(w_1,w_2,v_1)$ are linear independent and $v_2 = alpha v_1 $

    $4.$ $(w_1,w_2,v_2)$ are linear independent and $v_1 = alpha v_2 $ (but it is equivalent to $3$ so I don't have to check that)



    but there are a lot of calculus and I am not sure if (i) this is correct way (ii) how to do this quickly



    Can somebody help me with this problem? Thanks for your time.






    linear-algebra vector-spaces







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I have some doubts with this task:




      Let $a,b in mathbb R$. In linear space $mathbb R^3$ we have:
      $$V = span([1,1+a,-2]^T,[2,6,-2-a]^T)= span(v_1,v_2) $$ and $$W = span([0,3,-1-b]^T,[2,2+b,-2]^T) = span(w_1,w_2)$$
      Find $a,b$ such that $mathbb R^3 = V oplus W$






      If $mathbb R^3 = V oplus W$ then $V cap W = left{vec{0}right}$

      I think that I should consider these cases:
      $1.$ $(v_1,v_2,w_1)$ are linear independent and $w_2 = alpha w_1 $

      $2.$ $(v_1,v_2,w_2)$ are linear independent and $w_1 = alpha w_2 $ (but it is equivalent to $1$ so I don't have to check that)



      $3.$ $(w_1,w_2,v_1)$ are linear independent and $v_2 = alpha v_1 $

      $4.$ $(w_1,w_2,v_2)$ are linear independent and $v_1 = alpha v_2 $ (but it is equivalent to $3$ so I don't have to check that)



      but there are a lot of calculus and I am not sure if (i) this is correct way (ii) how to do this quickly



      Can somebody help me with this problem? Thanks for your time.






      linear-algebra vector-spaces







      share|cite|improve this question











      $endgroup$




      I have some doubts with this task:




      Let $a,b in mathbb R$. In linear space $mathbb R^3$ we have:
      $$V = span([1,1+a,-2]^T,[2,6,-2-a]^T)= span(v_1,v_2) $$ and $$W = span([0,3,-1-b]^T,[2,2+b,-2]^T) = span(w_1,w_2)$$
      Find $a,b$ such that $mathbb R^3 = V oplus W$






      If $mathbb R^3 = V oplus W$ then $V cap W = left{vec{0}right}$

      I think that I should consider these cases:
      $1.$ $(v_1,v_2,w_1)$ are linear independent and $w_2 = alpha w_1 $

      $2.$ $(v_1,v_2,w_2)$ are linear independent and $w_1 = alpha w_2 $ (but it is equivalent to $1$ so I don't have to check that)



      $3.$ $(w_1,w_2,v_1)$ are linear independent and $v_2 = alpha v_1 $

      $4.$ $(w_1,w_2,v_2)$ are linear independent and $v_1 = alpha v_2 $ (but it is equivalent to $3$ so I don't have to check that)



      but there are a lot of calculus and I am not sure if (i) this is correct way (ii) how to do this quickly



      Can somebody help me with this problem? Thanks for your time.






      linear-algebra vector-spaces




      linear-algebra vector-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Jan 8 at 11:16







      VirtualUser

















      asked Jan 8 at 10:41









      VirtualUserVirtualUser

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          $begingroup$

          Observe that both $V$ and $W$ can be at most two dimensional each. If both are two dimensional then their intersection cannot be just ${0}$. If one of them is $0$ dimensional OR both are $1$ dimensional then $Bbb{R}^3$ cannot be their direct sum. So the only two possibilities are:





          1. $V$ is $1-$dim and W is $2-$dim.


          2. $V$ is $2-$dim and W is $1-$dim.


          In the first case $v_1$ and $v_2$ must be dependent vectors and $v_1,w_1,w_2$ must be independent vectors. That means
          $$kbegin{bmatrix}1\1+a\-2end{bmatrix}=begin{bmatrix}2\6\-2-aend{bmatrix} implies k=2 text{ and } a=2$$
          For the independence of ${v_1,w_1,w_2}$, the following matrix should have full rank.
          $$begin{bmatrix}1&3&-2\0&3&-1-b\2&2+b&-2end{bmatrix} longrightarrow begin{bmatrix}1&3&-2\0&3&-1-b\0&b-4&2end{bmatrix} longrightarrow begin{bmatrix}1&3&-2\0&3&-1-b\0&0&frac{(b-2)(b-1)}{3}end{bmatrix}.$$
          Thus for full rank, we want $b neq 1,2$.



          Thus with $a=2$ and $b neq 1,2$, $Bbb{R}^3=V oplus W$.



          Similarly you can do the other case.






          share|cite|improve this answer









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            $begingroup$

            Observe that both $V$ and $W$ can be at most two dimensional each. If both are two dimensional then their intersection cannot be just ${0}$. If one of them is $0$ dimensional OR both are $1$ dimensional then $Bbb{R}^3$ cannot be their direct sum. So the only two possibilities are:





            1. $V$ is $1-$dim and W is $2-$dim.


            2. $V$ is $2-$dim and W is $1-$dim.


            In the first case $v_1$ and $v_2$ must be dependent vectors and $v_1,w_1,w_2$ must be independent vectors. That means
            $$kbegin{bmatrix}1\1+a\-2end{bmatrix}=begin{bmatrix}2\6\-2-aend{bmatrix} implies k=2 text{ and } a=2$$
            For the independence of ${v_1,w_1,w_2}$, the following matrix should have full rank.
            $$begin{bmatrix}1&3&-2\0&3&-1-b\2&2+b&-2end{bmatrix} longrightarrow begin{bmatrix}1&3&-2\0&3&-1-b\0&b-4&2end{bmatrix} longrightarrow begin{bmatrix}1&3&-2\0&3&-1-b\0&0&frac{(b-2)(b-1)}{3}end{bmatrix}.$$
            Thus for full rank, we want $b neq 1,2$.



            Thus with $a=2$ and $b neq 1,2$, $Bbb{R}^3=V oplus W$.



            Similarly you can do the other case.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Observe that both $V$ and $W$ can be at most two dimensional each. If both are two dimensional then their intersection cannot be just ${0}$. If one of them is $0$ dimensional OR both are $1$ dimensional then $Bbb{R}^3$ cannot be their direct sum. So the only two possibilities are:





              1. $V$ is $1-$dim and W is $2-$dim.


              2. $V$ is $2-$dim and W is $1-$dim.


              In the first case $v_1$ and $v_2$ must be dependent vectors and $v_1,w_1,w_2$ must be independent vectors. That means
              $$kbegin{bmatrix}1\1+a\-2end{bmatrix}=begin{bmatrix}2\6\-2-aend{bmatrix} implies k=2 text{ and } a=2$$
              For the independence of ${v_1,w_1,w_2}$, the following matrix should have full rank.
              $$begin{bmatrix}1&3&-2\0&3&-1-b\2&2+b&-2end{bmatrix} longrightarrow begin{bmatrix}1&3&-2\0&3&-1-b\0&b-4&2end{bmatrix} longrightarrow begin{bmatrix}1&3&-2\0&3&-1-b\0&0&frac{(b-2)(b-1)}{3}end{bmatrix}.$$
              Thus for full rank, we want $b neq 1,2$.



              Thus with $a=2$ and $b neq 1,2$, $Bbb{R}^3=V oplus W$.



              Similarly you can do the other case.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Observe that both $V$ and $W$ can be at most two dimensional each. If both are two dimensional then their intersection cannot be just ${0}$. If one of them is $0$ dimensional OR both are $1$ dimensional then $Bbb{R}^3$ cannot be their direct sum. So the only two possibilities are:





                1. $V$ is $1-$dim and W is $2-$dim.


                2. $V$ is $2-$dim and W is $1-$dim.


                In the first case $v_1$ and $v_2$ must be dependent vectors and $v_1,w_1,w_2$ must be independent vectors. That means
                $$kbegin{bmatrix}1\1+a\-2end{bmatrix}=begin{bmatrix}2\6\-2-aend{bmatrix} implies k=2 text{ and } a=2$$
                For the independence of ${v_1,w_1,w_2}$, the following matrix should have full rank.
                $$begin{bmatrix}1&3&-2\0&3&-1-b\2&2+b&-2end{bmatrix} longrightarrow begin{bmatrix}1&3&-2\0&3&-1-b\0&b-4&2end{bmatrix} longrightarrow begin{bmatrix}1&3&-2\0&3&-1-b\0&0&frac{(b-2)(b-1)}{3}end{bmatrix}.$$
                Thus for full rank, we want $b neq 1,2$.



                Thus with $a=2$ and $b neq 1,2$, $Bbb{R}^3=V oplus W$.



                Similarly you can do the other case.






                share|cite|improve this answer









                $endgroup$



                Observe that both $V$ and $W$ can be at most two dimensional each. If both are two dimensional then their intersection cannot be just ${0}$. If one of them is $0$ dimensional OR both are $1$ dimensional then $Bbb{R}^3$ cannot be their direct sum. So the only two possibilities are:





                1. $V$ is $1-$dim and W is $2-$dim.


                2. $V$ is $2-$dim and W is $1-$dim.


                In the first case $v_1$ and $v_2$ must be dependent vectors and $v_1,w_1,w_2$ must be independent vectors. That means
                $$kbegin{bmatrix}1\1+a\-2end{bmatrix}=begin{bmatrix}2\6\-2-aend{bmatrix} implies k=2 text{ and } a=2$$
                For the independence of ${v_1,w_1,w_2}$, the following matrix should have full rank.
                $$begin{bmatrix}1&3&-2\0&3&-1-b\2&2+b&-2end{bmatrix} longrightarrow begin{bmatrix}1&3&-2\0&3&-1-b\0&b-4&2end{bmatrix} longrightarrow begin{bmatrix}1&3&-2\0&3&-1-b\0&0&frac{(b-2)(b-1)}{3}end{bmatrix}.$$
                Thus for full rank, we want $b neq 1,2$.



                Thus with $a=2$ and $b neq 1,2$, $Bbb{R}^3=V oplus W$.



                Similarly you can do the other case.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 11:44









                Anurag AAnurag A

                26k12249




                26k12249






























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