prove that 2 vectors is linearly independent in the vector space of vectors of length 2 with entries of...












0














I have the following vectors
$V_1=(e^t,te^t), V_2=(1,t)$
now I want to prove that this two vectors are linearly independent in the vector space of vectors of length 2 with entries of real-valued functions.



when I'm writing the formula:



$C_1*V_1^T+C_2*V_2^T=0$ and writing it in matrix notation I can't get to prove that the coefficients needs to be 0.



I do note that if I'm fixing t for a given value then they are linearly dependent which is somewhat weird because i know i need to prove that for general t they are not.



Would appreciate some thoughts.






linear-algebra vector-spaces







share|cite|improve this question



























    0














    I have the following vectors
    $V_1=(e^t,te^t), V_2=(1,t)$
    now I want to prove that this two vectors are linearly independent in the vector space of vectors of length 2 with entries of real-valued functions.



    when I'm writing the formula:



    $C_1*V_1^T+C_2*V_2^T=0$ and writing it in matrix notation I can't get to prove that the coefficients needs to be 0.



    I do note that if I'm fixing t for a given value then they are linearly dependent which is somewhat weird because i know i need to prove that for general t they are not.



    Would appreciate some thoughts.






    linear-algebra vector-spaces







    share|cite|improve this question

























      0












      0








      0







      I have the following vectors
      $V_1=(e^t,te^t), V_2=(1,t)$
      now I want to prove that this two vectors are linearly independent in the vector space of vectors of length 2 with entries of real-valued functions.



      when I'm writing the formula:



      $C_1*V_1^T+C_2*V_2^T=0$ and writing it in matrix notation I can't get to prove that the coefficients needs to be 0.



      I do note that if I'm fixing t for a given value then they are linearly dependent which is somewhat weird because i know i need to prove that for general t they are not.



      Would appreciate some thoughts.






      linear-algebra vector-spaces







      share|cite|improve this question













      I have the following vectors
      $V_1=(e^t,te^t), V_2=(1,t)$
      now I want to prove that this two vectors are linearly independent in the vector space of vectors of length 2 with entries of real-valued functions.



      when I'm writing the formula:



      $C_1*V_1^T+C_2*V_2^T=0$ and writing it in matrix notation I can't get to prove that the coefficients needs to be 0.



      I do note that if I'm fixing t for a given value then they are linearly dependent which is somewhat weird because i know i need to prove that for general t they are not.



      Would appreciate some thoughts.






      linear-algebra vector-spaces




      linear-algebra vector-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 9 at 15:47









      chemist

      144




      144






















          2 Answers
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          0














          You get the linear equations:



          $$begin{cases}I&c_1e^t+c_2=0\{}\II&c_1te^t+c_2t=0end{cases}$$



          The above two equations are true for any calue of $;tinBbb R;$ (I'm assuming the functions are defined on the whole real line), so take some particular values:



          $$t=0stackrel Iimplies c_1+c_2=0implies c_2=-c_1stackrel {II}implies c_1tleft(e^t-1right)=0$$



          Take now $;t=1;$ and you get $;c_1(e-1)=0implies c_1=0;$ and thus also $;c_2=0;$ and we're done.






          share|cite|improve this answer





















          • so this is a valid proof, isn't any robust method for this kind of cases? i did it but for some reason it seems to simple for a proof
            – chemist
            Dec 9 at 16:16










          • I can't see anything "robust" in this proof. It is just going by the very simple definition ...
            – DonAntonio
            Dec 9 at 16:23










          • Yeah i get that, but what if there were like k vectors, this would be the only way cause if i can represent in a matrix and isolate the part that depends on t, then its just matrix reduction
            – chemist
            Dec 9 at 16:36










          • @chemist I don't quite follow you. The above is a proof. Maybe there are other ones though I suspect not a simple and elementary as this one. If there were $;k;$ one could probably do the same, although then it would probably be a little more difficult.
            – DonAntonio
            Dec 9 at 17:33



















          0














          Linear dependence of ${bf v}_1$ and ${bf v}_2$ in the vector space $X$ of functions ${bf x}:>tmapstobigl(x_1(t),x_2(t)bigr)$ means that there are fixed numbers $lambda$, $mu in{mathbb R}$, not both zero, such that $$lambda {bf v}_1(t)+mu {bf v}_2(t)={bf 0}qquadforall> t .tag{1}$$
          Inspecting the first component of $(1)$ we obtain the condition $lambda e^t+mu=0$. This condition cannot be satisfied for all $t$ with fixed numbers $lambda$, $mu$, other than $lambda=mu=0$. It follows that the two given vectors are linearly independent in $X$.






          share|cite|improve this answer





















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            2 Answers
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            active

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            2 Answers
            2






            active

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            active

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            active

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            0














            You get the linear equations:



            $$begin{cases}I&c_1e^t+c_2=0\{}\II&c_1te^t+c_2t=0end{cases}$$



            The above two equations are true for any calue of $;tinBbb R;$ (I'm assuming the functions are defined on the whole real line), so take some particular values:



            $$t=0stackrel Iimplies c_1+c_2=0implies c_2=-c_1stackrel {II}implies c_1tleft(e^t-1right)=0$$



            Take now $;t=1;$ and you get $;c_1(e-1)=0implies c_1=0;$ and thus also $;c_2=0;$ and we're done.






            share|cite|improve this answer





















            • so this is a valid proof, isn't any robust method for this kind of cases? i did it but for some reason it seems to simple for a proof
              – chemist
              Dec 9 at 16:16










            • I can't see anything "robust" in this proof. It is just going by the very simple definition ...
              – DonAntonio
              Dec 9 at 16:23










            • Yeah i get that, but what if there were like k vectors, this would be the only way cause if i can represent in a matrix and isolate the part that depends on t, then its just matrix reduction
              – chemist
              Dec 9 at 16:36










            • @chemist I don't quite follow you. The above is a proof. Maybe there are other ones though I suspect not a simple and elementary as this one. If there were $;k;$ one could probably do the same, although then it would probably be a little more difficult.
              – DonAntonio
              Dec 9 at 17:33
















            0














            You get the linear equations:



            $$begin{cases}I&c_1e^t+c_2=0\{}\II&c_1te^t+c_2t=0end{cases}$$



            The above two equations are true for any calue of $;tinBbb R;$ (I'm assuming the functions are defined on the whole real line), so take some particular values:



            $$t=0stackrel Iimplies c_1+c_2=0implies c_2=-c_1stackrel {II}implies c_1tleft(e^t-1right)=0$$



            Take now $;t=1;$ and you get $;c_1(e-1)=0implies c_1=0;$ and thus also $;c_2=0;$ and we're done.






            share|cite|improve this answer





















            • so this is a valid proof, isn't any robust method for this kind of cases? i did it but for some reason it seems to simple for a proof
              – chemist
              Dec 9 at 16:16










            • I can't see anything "robust" in this proof. It is just going by the very simple definition ...
              – DonAntonio
              Dec 9 at 16:23










            • Yeah i get that, but what if there were like k vectors, this would be the only way cause if i can represent in a matrix and isolate the part that depends on t, then its just matrix reduction
              – chemist
              Dec 9 at 16:36










            • @chemist I don't quite follow you. The above is a proof. Maybe there are other ones though I suspect not a simple and elementary as this one. If there were $;k;$ one could probably do the same, although then it would probably be a little more difficult.
              – DonAntonio
              Dec 9 at 17:33














            0












            0








            0






            You get the linear equations:



            $$begin{cases}I&c_1e^t+c_2=0\{}\II&c_1te^t+c_2t=0end{cases}$$



            The above two equations are true for any calue of $;tinBbb R;$ (I'm assuming the functions are defined on the whole real line), so take some particular values:



            $$t=0stackrel Iimplies c_1+c_2=0implies c_2=-c_1stackrel {II}implies c_1tleft(e^t-1right)=0$$



            Take now $;t=1;$ and you get $;c_1(e-1)=0implies c_1=0;$ and thus also $;c_2=0;$ and we're done.






            share|cite|improve this answer












            You get the linear equations:



            $$begin{cases}I&c_1e^t+c_2=0\{}\II&c_1te^t+c_2t=0end{cases}$$



            The above two equations are true for any calue of $;tinBbb R;$ (I'm assuming the functions are defined on the whole real line), so take some particular values:



            $$t=0stackrel Iimplies c_1+c_2=0implies c_2=-c_1stackrel {II}implies c_1tleft(e^t-1right)=0$$



            Take now $;t=1;$ and you get $;c_1(e-1)=0implies c_1=0;$ and thus also $;c_2=0;$ and we're done.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 9 at 16:00









            DonAntonio

            177k1491225




            177k1491225












            • so this is a valid proof, isn't any robust method for this kind of cases? i did it but for some reason it seems to simple for a proof
              – chemist
              Dec 9 at 16:16










            • I can't see anything "robust" in this proof. It is just going by the very simple definition ...
              – DonAntonio
              Dec 9 at 16:23










            • Yeah i get that, but what if there were like k vectors, this would be the only way cause if i can represent in a matrix and isolate the part that depends on t, then its just matrix reduction
              – chemist
              Dec 9 at 16:36










            • @chemist I don't quite follow you. The above is a proof. Maybe there are other ones though I suspect not a simple and elementary as this one. If there were $;k;$ one could probably do the same, although then it would probably be a little more difficult.
              – DonAntonio
              Dec 9 at 17:33


















            • so this is a valid proof, isn't any robust method for this kind of cases? i did it but for some reason it seems to simple for a proof
              – chemist
              Dec 9 at 16:16










            • I can't see anything "robust" in this proof. It is just going by the very simple definition ...
              – DonAntonio
              Dec 9 at 16:23










            • Yeah i get that, but what if there were like k vectors, this would be the only way cause if i can represent in a matrix and isolate the part that depends on t, then its just matrix reduction
              – chemist
              Dec 9 at 16:36










            • @chemist I don't quite follow you. The above is a proof. Maybe there are other ones though I suspect not a simple and elementary as this one. If there were $;k;$ one could probably do the same, although then it would probably be a little more difficult.
              – DonAntonio
              Dec 9 at 17:33
















            so this is a valid proof, isn't any robust method for this kind of cases? i did it but for some reason it seems to simple for a proof
            – chemist
            Dec 9 at 16:16




            so this is a valid proof, isn't any robust method for this kind of cases? i did it but for some reason it seems to simple for a proof
            – chemist
            Dec 9 at 16:16












            I can't see anything "robust" in this proof. It is just going by the very simple definition ...
            – DonAntonio
            Dec 9 at 16:23




            I can't see anything "robust" in this proof. It is just going by the very simple definition ...
            – DonAntonio
            Dec 9 at 16:23












            Yeah i get that, but what if there were like k vectors, this would be the only way cause if i can represent in a matrix and isolate the part that depends on t, then its just matrix reduction
            – chemist
            Dec 9 at 16:36




            Yeah i get that, but what if there were like k vectors, this would be the only way cause if i can represent in a matrix and isolate the part that depends on t, then its just matrix reduction
            – chemist
            Dec 9 at 16:36












            @chemist I don't quite follow you. The above is a proof. Maybe there are other ones though I suspect not a simple and elementary as this one. If there were $;k;$ one could probably do the same, although then it would probably be a little more difficult.
            – DonAntonio
            Dec 9 at 17:33




            @chemist I don't quite follow you. The above is a proof. Maybe there are other ones though I suspect not a simple and elementary as this one. If there were $;k;$ one could probably do the same, although then it would probably be a little more difficult.
            – DonAntonio
            Dec 9 at 17:33











            0














            Linear dependence of ${bf v}_1$ and ${bf v}_2$ in the vector space $X$ of functions ${bf x}:>tmapstobigl(x_1(t),x_2(t)bigr)$ means that there are fixed numbers $lambda$, $mu in{mathbb R}$, not both zero, such that $$lambda {bf v}_1(t)+mu {bf v}_2(t)={bf 0}qquadforall> t .tag{1}$$
            Inspecting the first component of $(1)$ we obtain the condition $lambda e^t+mu=0$. This condition cannot be satisfied for all $t$ with fixed numbers $lambda$, $mu$, other than $lambda=mu=0$. It follows that the two given vectors are linearly independent in $X$.






            share|cite|improve this answer


























              0














              Linear dependence of ${bf v}_1$ and ${bf v}_2$ in the vector space $X$ of functions ${bf x}:>tmapstobigl(x_1(t),x_2(t)bigr)$ means that there are fixed numbers $lambda$, $mu in{mathbb R}$, not both zero, such that $$lambda {bf v}_1(t)+mu {bf v}_2(t)={bf 0}qquadforall> t .tag{1}$$
              Inspecting the first component of $(1)$ we obtain the condition $lambda e^t+mu=0$. This condition cannot be satisfied for all $t$ with fixed numbers $lambda$, $mu$, other than $lambda=mu=0$. It follows that the two given vectors are linearly independent in $X$.






              share|cite|improve this answer
























                0












                0








                0






                Linear dependence of ${bf v}_1$ and ${bf v}_2$ in the vector space $X$ of functions ${bf x}:>tmapstobigl(x_1(t),x_2(t)bigr)$ means that there are fixed numbers $lambda$, $mu in{mathbb R}$, not both zero, such that $$lambda {bf v}_1(t)+mu {bf v}_2(t)={bf 0}qquadforall> t .tag{1}$$
                Inspecting the first component of $(1)$ we obtain the condition $lambda e^t+mu=0$. This condition cannot be satisfied for all $t$ with fixed numbers $lambda$, $mu$, other than $lambda=mu=0$. It follows that the two given vectors are linearly independent in $X$.






                share|cite|improve this answer












                Linear dependence of ${bf v}_1$ and ${bf v}_2$ in the vector space $X$ of functions ${bf x}:>tmapstobigl(x_1(t),x_2(t)bigr)$ means that there are fixed numbers $lambda$, $mu in{mathbb R}$, not both zero, such that $$lambda {bf v}_1(t)+mu {bf v}_2(t)={bf 0}qquadforall> t .tag{1}$$
                Inspecting the first component of $(1)$ we obtain the condition $lambda e^t+mu=0$. This condition cannot be satisfied for all $t$ with fixed numbers $lambda$, $mu$, other than $lambda=mu=0$. It follows that the two given vectors are linearly independent in $X$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 9 at 20:27









                Christian Blatter

                172k7112325




                172k7112325






























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