Which are linear transformations?












0












$begingroup$


1) Define $T:Vrightarrow V$ for $T(v)=v$, $forall$ $v in V$.



For this one I have:



Let $T(v)=v$ be the identity transformation. Because $T(u+v)=T(u)+T(v)$ and $T(cu)=cu=cT(u)$. $T$ is a linear transformation.



But I don't know if is that quite simple. The other alternative I have is:



If $v={v_1,v_2,...,v_n}$ and is a basis for $V$, then any $vin V$ can be written as the linear combination $v=a_1v_1+a_2v_2+...+a_nv_n$ with $a_i$ scalar. Then



$$T(v)=T(a_1v_1+a_2v_2+...+a_nv_n)$$
$$T(v)=a_1T(v_1)+a_2T(v_2)+...+a_nT(v_n)$$
$$T(v)=a_1v_1+a_2v_2+...+a_nv_n$$
$$T(v)=v$$



Therefore, $T$ is a linear transformation.



Can be any of them?



2) Let $V$, $W$ vector spaces. Define $T:Vrightarrow W$ for $T(v)=0$, $forall$ $v in V$.



For this one I have:



Let $v=c_1v_1+c_2v_2+...+c_nv_n$ be an arbitrary vector in $v$. Then, $$T(v)=T(c_1v_1+c_2v_2+...+c_nv_n)$$
$$T(v)=c_1T(v_1)+c_2T(v_2)+...+c_nT(v_n)$$
$$T(v)=0+0+...+0$$
$$T(v)=0$$



Therefore, $T$ is a linear transformation.



Is this right?



3) Define $T: C[0,1]rightarrow R$ for $T(f)=f(0)+1$.



For this last one I have not clue where to start. Help please!






linear-algebra vector-spaces linear-transformations







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    1) Define $T:Vrightarrow V$ for $T(v)=v$, $forall$ $v in V$.



    For this one I have:



    Let $T(v)=v$ be the identity transformation. Because $T(u+v)=T(u)+T(v)$ and $T(cu)=cu=cT(u)$. $T$ is a linear transformation.



    But I don't know if is that quite simple. The other alternative I have is:



    If $v={v_1,v_2,...,v_n}$ and is a basis for $V$, then any $vin V$ can be written as the linear combination $v=a_1v_1+a_2v_2+...+a_nv_n$ with $a_i$ scalar. Then



    $$T(v)=T(a_1v_1+a_2v_2+...+a_nv_n)$$
    $$T(v)=a_1T(v_1)+a_2T(v_2)+...+a_nT(v_n)$$
    $$T(v)=a_1v_1+a_2v_2+...+a_nv_n$$
    $$T(v)=v$$



    Therefore, $T$ is a linear transformation.



    Can be any of them?



    2) Let $V$, $W$ vector spaces. Define $T:Vrightarrow W$ for $T(v)=0$, $forall$ $v in V$.



    For this one I have:



    Let $v=c_1v_1+c_2v_2+...+c_nv_n$ be an arbitrary vector in $v$. Then, $$T(v)=T(c_1v_1+c_2v_2+...+c_nv_n)$$
    $$T(v)=c_1T(v_1)+c_2T(v_2)+...+c_nT(v_n)$$
    $$T(v)=0+0+...+0$$
    $$T(v)=0$$



    Therefore, $T$ is a linear transformation.



    Is this right?



    3) Define $T: C[0,1]rightarrow R$ for $T(f)=f(0)+1$.



    For this last one I have not clue where to start. Help please!






    linear-algebra vector-spaces linear-transformations







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      1) Define $T:Vrightarrow V$ for $T(v)=v$, $forall$ $v in V$.



      For this one I have:



      Let $T(v)=v$ be the identity transformation. Because $T(u+v)=T(u)+T(v)$ and $T(cu)=cu=cT(u)$. $T$ is a linear transformation.



      But I don't know if is that quite simple. The other alternative I have is:



      If $v={v_1,v_2,...,v_n}$ and is a basis for $V$, then any $vin V$ can be written as the linear combination $v=a_1v_1+a_2v_2+...+a_nv_n$ with $a_i$ scalar. Then



      $$T(v)=T(a_1v_1+a_2v_2+...+a_nv_n)$$
      $$T(v)=a_1T(v_1)+a_2T(v_2)+...+a_nT(v_n)$$
      $$T(v)=a_1v_1+a_2v_2+...+a_nv_n$$
      $$T(v)=v$$



      Therefore, $T$ is a linear transformation.



      Can be any of them?



      2) Let $V$, $W$ vector spaces. Define $T:Vrightarrow W$ for $T(v)=0$, $forall$ $v in V$.



      For this one I have:



      Let $v=c_1v_1+c_2v_2+...+c_nv_n$ be an arbitrary vector in $v$. Then, $$T(v)=T(c_1v_1+c_2v_2+...+c_nv_n)$$
      $$T(v)=c_1T(v_1)+c_2T(v_2)+...+c_nT(v_n)$$
      $$T(v)=0+0+...+0$$
      $$T(v)=0$$



      Therefore, $T$ is a linear transformation.



      Is this right?



      3) Define $T: C[0,1]rightarrow R$ for $T(f)=f(0)+1$.



      For this last one I have not clue where to start. Help please!






      linear-algebra vector-spaces linear-transformations







      share|cite|improve this question









      $endgroup$




      1) Define $T:Vrightarrow V$ for $T(v)=v$, $forall$ $v in V$.



      For this one I have:



      Let $T(v)=v$ be the identity transformation. Because $T(u+v)=T(u)+T(v)$ and $T(cu)=cu=cT(u)$. $T$ is a linear transformation.



      But I don't know if is that quite simple. The other alternative I have is:



      If $v={v_1,v_2,...,v_n}$ and is a basis for $V$, then any $vin V$ can be written as the linear combination $v=a_1v_1+a_2v_2+...+a_nv_n$ with $a_i$ scalar. Then



      $$T(v)=T(a_1v_1+a_2v_2+...+a_nv_n)$$
      $$T(v)=a_1T(v_1)+a_2T(v_2)+...+a_nT(v_n)$$
      $$T(v)=a_1v_1+a_2v_2+...+a_nv_n$$
      $$T(v)=v$$



      Therefore, $T$ is a linear transformation.



      Can be any of them?



      2) Let $V$, $W$ vector spaces. Define $T:Vrightarrow W$ for $T(v)=0$, $forall$ $v in V$.



      For this one I have:



      Let $v=c_1v_1+c_2v_2+...+c_nv_n$ be an arbitrary vector in $v$. Then, $$T(v)=T(c_1v_1+c_2v_2+...+c_nv_n)$$
      $$T(v)=c_1T(v_1)+c_2T(v_2)+...+c_nT(v_n)$$
      $$T(v)=0+0+...+0$$
      $$T(v)=0$$



      Therefore, $T$ is a linear transformation.



      Is this right?



      3) Define $T: C[0,1]rightarrow R$ for $T(f)=f(0)+1$.



      For this last one I have not clue where to start. Help please!






      linear-algebra vector-spaces linear-transformations




      linear-algebra vector-spaces linear-transformations






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 23 '18 at 19:15









      gi2302gi2302

      103




      103






















          2 Answers
          2






          active

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          2












          $begingroup$


          1. Your first approach if fine. The second one isn't, since it assumes that the mape is linear.

          2. Again, you are assuming that $T$ is linear! All you have to do is$$T(u+v)=0=0+0=T(u)+T(v)text{ and }T(cu)=0=ctimes0=cT(u).$$

          3. It is not linear, since $T(0)neq0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            José Carlos gracias. In the first one, in the second approach if I don't assume $T$ is linear is ok?
            $endgroup$
            – gi2302
            Nov 23 '18 at 19:31










          • $begingroup$
            That means not assuming that $T(c_1v_1+cdots+c_nv_n)=c_1T(v_1)+cdots+c_nT(v_n)$.
            $endgroup$
            – José Carlos Santos
            Nov 23 '18 at 19:33



















          0












          $begingroup$

          You have rightly concluded the first two ones. For the third one is not a linear mapping. For any linear mapping $T$ we must have $T(0)=0$. Let $f(x)=0$. Therefore $f$ lies inside the domain of $T$ since it is always continuous but $$T(0)=T(f(x))=f(0)+1=1ne 0$$furthermore if $$T(f)=f(0)+1$$then $$T(kf)=kf(0)+1ne kf(0)+k=kT(f)qquadqquad kne 1$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks!!!!!!!!!
            $endgroup$
            – gi2302
            Nov 23 '18 at 19:32










          • $begingroup$
            You're welcome!!
            $endgroup$
            – Mostafa Ayaz
            Nov 23 '18 at 19:36











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$


          1. Your first approach if fine. The second one isn't, since it assumes that the mape is linear.

          2. Again, you are assuming that $T$ is linear! All you have to do is$$T(u+v)=0=0+0=T(u)+T(v)text{ and }T(cu)=0=ctimes0=cT(u).$$

          3. It is not linear, since $T(0)neq0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            José Carlos gracias. In the first one, in the second approach if I don't assume $T$ is linear is ok?
            $endgroup$
            – gi2302
            Nov 23 '18 at 19:31










          • $begingroup$
            That means not assuming that $T(c_1v_1+cdots+c_nv_n)=c_1T(v_1)+cdots+c_nT(v_n)$.
            $endgroup$
            – José Carlos Santos
            Nov 23 '18 at 19:33
















          2












          $begingroup$


          1. Your first approach if fine. The second one isn't, since it assumes that the mape is linear.

          2. Again, you are assuming that $T$ is linear! All you have to do is$$T(u+v)=0=0+0=T(u)+T(v)text{ and }T(cu)=0=ctimes0=cT(u).$$

          3. It is not linear, since $T(0)neq0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            José Carlos gracias. In the first one, in the second approach if I don't assume $T$ is linear is ok?
            $endgroup$
            – gi2302
            Nov 23 '18 at 19:31










          • $begingroup$
            That means not assuming that $T(c_1v_1+cdots+c_nv_n)=c_1T(v_1)+cdots+c_nT(v_n)$.
            $endgroup$
            – José Carlos Santos
            Nov 23 '18 at 19:33














          2












          2








          2





          $begingroup$


          1. Your first approach if fine. The second one isn't, since it assumes that the mape is linear.

          2. Again, you are assuming that $T$ is linear! All you have to do is$$T(u+v)=0=0+0=T(u)+T(v)text{ and }T(cu)=0=ctimes0=cT(u).$$

          3. It is not linear, since $T(0)neq0$.






          share|cite|improve this answer









          $endgroup$




          1. Your first approach if fine. The second one isn't, since it assumes that the mape is linear.

          2. Again, you are assuming that $T$ is linear! All you have to do is$$T(u+v)=0=0+0=T(u)+T(v)text{ and }T(cu)=0=ctimes0=cT(u).$$

          3. It is not linear, since $T(0)neq0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 '18 at 19:18









          José Carlos SantosJosé Carlos Santos

          153k22123225




          153k22123225












          • $begingroup$
            José Carlos gracias. In the first one, in the second approach if I don't assume $T$ is linear is ok?
            $endgroup$
            – gi2302
            Nov 23 '18 at 19:31










          • $begingroup$
            That means not assuming that $T(c_1v_1+cdots+c_nv_n)=c_1T(v_1)+cdots+c_nT(v_n)$.
            $endgroup$
            – José Carlos Santos
            Nov 23 '18 at 19:33


















          • $begingroup$
            José Carlos gracias. In the first one, in the second approach if I don't assume $T$ is linear is ok?
            $endgroup$
            – gi2302
            Nov 23 '18 at 19:31










          • $begingroup$
            That means not assuming that $T(c_1v_1+cdots+c_nv_n)=c_1T(v_1)+cdots+c_nT(v_n)$.
            $endgroup$
            – José Carlos Santos
            Nov 23 '18 at 19:33
















          $begingroup$
          José Carlos gracias. In the first one, in the second approach if I don't assume $T$ is linear is ok?
          $endgroup$
          – gi2302
          Nov 23 '18 at 19:31




          $begingroup$
          José Carlos gracias. In the first one, in the second approach if I don't assume $T$ is linear is ok?
          $endgroup$
          – gi2302
          Nov 23 '18 at 19:31












          $begingroup$
          That means not assuming that $T(c_1v_1+cdots+c_nv_n)=c_1T(v_1)+cdots+c_nT(v_n)$.
          $endgroup$
          – José Carlos Santos
          Nov 23 '18 at 19:33




          $begingroup$
          That means not assuming that $T(c_1v_1+cdots+c_nv_n)=c_1T(v_1)+cdots+c_nT(v_n)$.
          $endgroup$
          – José Carlos Santos
          Nov 23 '18 at 19:33











          0












          $begingroup$

          You have rightly concluded the first two ones. For the third one is not a linear mapping. For any linear mapping $T$ we must have $T(0)=0$. Let $f(x)=0$. Therefore $f$ lies inside the domain of $T$ since it is always continuous but $$T(0)=T(f(x))=f(0)+1=1ne 0$$furthermore if $$T(f)=f(0)+1$$then $$T(kf)=kf(0)+1ne kf(0)+k=kT(f)qquadqquad kne 1$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks!!!!!!!!!
            $endgroup$
            – gi2302
            Nov 23 '18 at 19:32










          • $begingroup$
            You're welcome!!
            $endgroup$
            – Mostafa Ayaz
            Nov 23 '18 at 19:36
















          0












          $begingroup$

          You have rightly concluded the first two ones. For the third one is not a linear mapping. For any linear mapping $T$ we must have $T(0)=0$. Let $f(x)=0$. Therefore $f$ lies inside the domain of $T$ since it is always continuous but $$T(0)=T(f(x))=f(0)+1=1ne 0$$furthermore if $$T(f)=f(0)+1$$then $$T(kf)=kf(0)+1ne kf(0)+k=kT(f)qquadqquad kne 1$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks!!!!!!!!!
            $endgroup$
            – gi2302
            Nov 23 '18 at 19:32










          • $begingroup$
            You're welcome!!
            $endgroup$
            – Mostafa Ayaz
            Nov 23 '18 at 19:36














          0












          0








          0





          $begingroup$

          You have rightly concluded the first two ones. For the third one is not a linear mapping. For any linear mapping $T$ we must have $T(0)=0$. Let $f(x)=0$. Therefore $f$ lies inside the domain of $T$ since it is always continuous but $$T(0)=T(f(x))=f(0)+1=1ne 0$$furthermore if $$T(f)=f(0)+1$$then $$T(kf)=kf(0)+1ne kf(0)+k=kT(f)qquadqquad kne 1$$






          share|cite|improve this answer









          $endgroup$



          You have rightly concluded the first two ones. For the third one is not a linear mapping. For any linear mapping $T$ we must have $T(0)=0$. Let $f(x)=0$. Therefore $f$ lies inside the domain of $T$ since it is always continuous but $$T(0)=T(f(x))=f(0)+1=1ne 0$$furthermore if $$T(f)=f(0)+1$$then $$T(kf)=kf(0)+1ne kf(0)+k=kT(f)qquadqquad kne 1$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 '18 at 19:19









          Mostafa AyazMostafa Ayaz

          14.7k3937




          14.7k3937












          • $begingroup$
            Thanks!!!!!!!!!
            $endgroup$
            – gi2302
            Nov 23 '18 at 19:32










          • $begingroup$
            You're welcome!!
            $endgroup$
            – Mostafa Ayaz
            Nov 23 '18 at 19:36


















          • $begingroup$
            Thanks!!!!!!!!!
            $endgroup$
            – gi2302
            Nov 23 '18 at 19:32










          • $begingroup$
            You're welcome!!
            $endgroup$
            – Mostafa Ayaz
            Nov 23 '18 at 19:36
















          $begingroup$
          Thanks!!!!!!!!!
          $endgroup$
          – gi2302
          Nov 23 '18 at 19:32




          $begingroup$
          Thanks!!!!!!!!!
          $endgroup$
          – gi2302
          Nov 23 '18 at 19:32












          $begingroup$
          You're welcome!!
          $endgroup$
          – Mostafa Ayaz
          Nov 23 '18 at 19:36




          $begingroup$
          You're welcome!!
          $endgroup$
          – Mostafa Ayaz
          Nov 23 '18 at 19:36


















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