Is $U$ a subspace of $Bbb R^3$?












0












$begingroup$


$U={(2r,-s^2,t) mid r, s text{ and} t in Bbb R}$



So I know I have to make sure of the following:



(a) The zero vector is in $U$ (replace $r,s,t$ by $0$ and you get $0$)



(b) It is closed under addition (the resultant of adding other vectors, like $(r,s,t)$ to $(2r,-s^2,t)$, is still in $U$)



(c) It is closed under scalar multiplication (any scalar times $U$ is still in $U$)



I found that they all work. But apparently, $U$ is not a subset. Why? Is it because there are three variables, $r,s,t$?






linear-algebra







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    $U={(2r,-s^2,t) mid r, s text{ and} t in Bbb R}$



    So I know I have to make sure of the following:



    (a) The zero vector is in $U$ (replace $r,s,t$ by $0$ and you get $0$)



    (b) It is closed under addition (the resultant of adding other vectors, like $(r,s,t)$ to $(2r,-s^2,t)$, is still in $U$)



    (c) It is closed under scalar multiplication (any scalar times $U$ is still in $U$)



    I found that they all work. But apparently, $U$ is not a subset. Why? Is it because there are three variables, $r,s,t$?






    linear-algebra







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      $U={(2r,-s^2,t) mid r, s text{ and} t in Bbb R}$



      So I know I have to make sure of the following:



      (a) The zero vector is in $U$ (replace $r,s,t$ by $0$ and you get $0$)



      (b) It is closed under addition (the resultant of adding other vectors, like $(r,s,t)$ to $(2r,-s^2,t)$, is still in $U$)



      (c) It is closed under scalar multiplication (any scalar times $U$ is still in $U$)



      I found that they all work. But apparently, $U$ is not a subset. Why? Is it because there are three variables, $r,s,t$?






      linear-algebra







      share|cite|improve this question











      $endgroup$




      $U={(2r,-s^2,t) mid r, s text{ and} t in Bbb R}$



      So I know I have to make sure of the following:



      (a) The zero vector is in $U$ (replace $r,s,t$ by $0$ and you get $0$)



      (b) It is closed under addition (the resultant of adding other vectors, like $(r,s,t)$ to $(2r,-s^2,t)$, is still in $U$)



      (c) It is closed under scalar multiplication (any scalar times $U$ is still in $U$)



      I found that they all work. But apparently, $U$ is not a subset. Why? Is it because there are three variables, $r,s,t$?






      linear-algebra




      linear-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 2 '18 at 5:00









      Tianlalu

      3,08121038




      3,08121038










      asked Dec 2 '18 at 4:41









      Sami JrSami Jr

      72




      72






















          1 Answer
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          active

          oldest

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          1












          $begingroup$

          $(1,-1,1)$ is in $U$. Multiply by $-1$, which is a scalar. The vector $(-1,1,-1)$ is not in $U$. (Notice $-s^2$ can only be a negative number. In my experience such condition is a red flag!!)






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
            $endgroup$
            – Sami Jr
            Dec 2 '18 at 4:47












          • $begingroup$
            As you pointed out, a subspace must be closed under scalar multiplication. If one entry of a vector inside a subset of interest can only be a negative number, multiplying it by a negative number would give us a vector outside that set (because the negative entry now becomes positive,) making the subset not closed under scalar multiplication.
            $endgroup$
            – Ko Byeongmin
            Dec 2 '18 at 4:49












          • $begingroup$
            Similar example: consider our usual xy-plane. The first quadrant is not a subspace of $R^2$, because multiply any nonzero vector inside the first quadrant by a negative number and the multiplied vector is outside the first quadrant.
            $endgroup$
            – Ko Byeongmin
            Dec 2 '18 at 4:53








          • 1




            $begingroup$
            Ohhh thank you so much for the added explanation!!
            $endgroup$
            – Sami Jr
            Dec 2 '18 at 4:57













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          1 Answer
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          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          $(1,-1,1)$ is in $U$. Multiply by $-1$, which is a scalar. The vector $(-1,1,-1)$ is not in $U$. (Notice $-s^2$ can only be a negative number. In my experience such condition is a red flag!!)






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
            $endgroup$
            – Sami Jr
            Dec 2 '18 at 4:47












          • $begingroup$
            As you pointed out, a subspace must be closed under scalar multiplication. If one entry of a vector inside a subset of interest can only be a negative number, multiplying it by a negative number would give us a vector outside that set (because the negative entry now becomes positive,) making the subset not closed under scalar multiplication.
            $endgroup$
            – Ko Byeongmin
            Dec 2 '18 at 4:49












          • $begingroup$
            Similar example: consider our usual xy-plane. The first quadrant is not a subspace of $R^2$, because multiply any nonzero vector inside the first quadrant by a negative number and the multiplied vector is outside the first quadrant.
            $endgroup$
            – Ko Byeongmin
            Dec 2 '18 at 4:53








          • 1




            $begingroup$
            Ohhh thank you so much for the added explanation!!
            $endgroup$
            – Sami Jr
            Dec 2 '18 at 4:57


















          1












          $begingroup$

          $(1,-1,1)$ is in $U$. Multiply by $-1$, which is a scalar. The vector $(-1,1,-1)$ is not in $U$. (Notice $-s^2$ can only be a negative number. In my experience such condition is a red flag!!)






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
            $endgroup$
            – Sami Jr
            Dec 2 '18 at 4:47












          • $begingroup$
            As you pointed out, a subspace must be closed under scalar multiplication. If one entry of a vector inside a subset of interest can only be a negative number, multiplying it by a negative number would give us a vector outside that set (because the negative entry now becomes positive,) making the subset not closed under scalar multiplication.
            $endgroup$
            – Ko Byeongmin
            Dec 2 '18 at 4:49












          • $begingroup$
            Similar example: consider our usual xy-plane. The first quadrant is not a subspace of $R^2$, because multiply any nonzero vector inside the first quadrant by a negative number and the multiplied vector is outside the first quadrant.
            $endgroup$
            – Ko Byeongmin
            Dec 2 '18 at 4:53








          • 1




            $begingroup$
            Ohhh thank you so much for the added explanation!!
            $endgroup$
            – Sami Jr
            Dec 2 '18 at 4:57
















          1












          1








          1





          $begingroup$

          $(1,-1,1)$ is in $U$. Multiply by $-1$, which is a scalar. The vector $(-1,1,-1)$ is not in $U$. (Notice $-s^2$ can only be a negative number. In my experience such condition is a red flag!!)






          share|cite|improve this answer











          $endgroup$



          $(1,-1,1)$ is in $U$. Multiply by $-1$, which is a scalar. The vector $(-1,1,-1)$ is not in $U$. (Notice $-s^2$ can only be a negative number. In my experience such condition is a red flag!!)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 2 '18 at 5:01









          Tianlalu

          3,08121038




          3,08121038










          answered Dec 2 '18 at 4:45









          Ko ByeongminKo Byeongmin

          1326




          1326








          • 1




            $begingroup$
            But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
            $endgroup$
            – Sami Jr
            Dec 2 '18 at 4:47












          • $begingroup$
            As you pointed out, a subspace must be closed under scalar multiplication. If one entry of a vector inside a subset of interest can only be a negative number, multiplying it by a negative number would give us a vector outside that set (because the negative entry now becomes positive,) making the subset not closed under scalar multiplication.
            $endgroup$
            – Ko Byeongmin
            Dec 2 '18 at 4:49












          • $begingroup$
            Similar example: consider our usual xy-plane. The first quadrant is not a subspace of $R^2$, because multiply any nonzero vector inside the first quadrant by a negative number and the multiplied vector is outside the first quadrant.
            $endgroup$
            – Ko Byeongmin
            Dec 2 '18 at 4:53








          • 1




            $begingroup$
            Ohhh thank you so much for the added explanation!!
            $endgroup$
            – Sami Jr
            Dec 2 '18 at 4:57
















          • 1




            $begingroup$
            But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
            $endgroup$
            – Sami Jr
            Dec 2 '18 at 4:47












          • $begingroup$
            As you pointed out, a subspace must be closed under scalar multiplication. If one entry of a vector inside a subset of interest can only be a negative number, multiplying it by a negative number would give us a vector outside that set (because the negative entry now becomes positive,) making the subset not closed under scalar multiplication.
            $endgroup$
            – Ko Byeongmin
            Dec 2 '18 at 4:49












          • $begingroup$
            Similar example: consider our usual xy-plane. The first quadrant is not a subspace of $R^2$, because multiply any nonzero vector inside the first quadrant by a negative number and the multiplied vector is outside the first quadrant.
            $endgroup$
            – Ko Byeongmin
            Dec 2 '18 at 4:53








          • 1




            $begingroup$
            Ohhh thank you so much for the added explanation!!
            $endgroup$
            – Sami Jr
            Dec 2 '18 at 4:57










          1




          1




          $begingroup$
          But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
          $endgroup$
          – Sami Jr
          Dec 2 '18 at 4:47






          $begingroup$
          But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
          $endgroup$
          – Sami Jr
          Dec 2 '18 at 4:47














          $begingroup$
          As you pointed out, a subspace must be closed under scalar multiplication. If one entry of a vector inside a subset of interest can only be a negative number, multiplying it by a negative number would give us a vector outside that set (because the negative entry now becomes positive,) making the subset not closed under scalar multiplication.
          $endgroup$
          – Ko Byeongmin
          Dec 2 '18 at 4:49






          $begingroup$
          As you pointed out, a subspace must be closed under scalar multiplication. If one entry of a vector inside a subset of interest can only be a negative number, multiplying it by a negative number would give us a vector outside that set (because the negative entry now becomes positive,) making the subset not closed under scalar multiplication.
          $endgroup$
          – Ko Byeongmin
          Dec 2 '18 at 4:49














          $begingroup$
          Similar example: consider our usual xy-plane. The first quadrant is not a subspace of $R^2$, because multiply any nonzero vector inside the first quadrant by a negative number and the multiplied vector is outside the first quadrant.
          $endgroup$
          – Ko Byeongmin
          Dec 2 '18 at 4:53






          $begingroup$
          Similar example: consider our usual xy-plane. The first quadrant is not a subspace of $R^2$, because multiply any nonzero vector inside the first quadrant by a negative number and the multiplied vector is outside the first quadrant.
          $endgroup$
          – Ko Byeongmin
          Dec 2 '18 at 4:53






          1




          1




          $begingroup$
          Ohhh thank you so much for the added explanation!!
          $endgroup$
          – Sami Jr
          Dec 2 '18 at 4:57






          $begingroup$
          Ohhh thank you so much for the added explanation!!
          $endgroup$
          – Sami Jr
          Dec 2 '18 at 4:57




















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