Solution for Linear Transformation question seems wrong.












0












$begingroup$


I'm studying with a Linear Algebra book which presents the following question:



Be $A:Erightarrow F$ a linear transformation. If the vectors $Av_1,ldots,Av_n in F$ are LI, then prove that $v_1,ldots,v_n in E$ are also linear.



Below is the official solution:



Assume $c_1v_1+ldots+c_nv_n=0$. Applying $A$ (which is linear), yields $c_1Av_1+ldots+c_nAv_n=0$. Since $Av_1,ldots,Av_n$ are LI, then $c_1=c_2=ldots=c_n=0$. Thus, $v_1,ldots,v_n$ are LI.



It seems to me that this solution does not take me anywhere. It already assumed a LI set $v_1,ldots,v_n$ and applied $A$ to it, to conclude that the same set is LI. But what if $A$ is singular?



I think it's more reasonable to linearly combine $Av_1,ldots,Av_n$:



$$
c_1Av_1+ldots+c_nAv_n=0
$$



Thus,



$$
Aleft(c_1v_1+ldots+c_nv_nright)=0
$$



And then to conclude that, iff $A$ is nonsingular, then $v_1,ldots,v_n$ is LI. Otherwise, we can't be sure.



Am I missing something?



Thanks in advance.






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$












  • $begingroup$
    How singula nature of $A$ affects the linear independency?The proof seems fine to me
    $endgroup$
    – Tom.
    Dec 10 '18 at 12:56


















0












$begingroup$


I'm studying with a Linear Algebra book which presents the following question:



Be $A:Erightarrow F$ a linear transformation. If the vectors $Av_1,ldots,Av_n in F$ are LI, then prove that $v_1,ldots,v_n in E$ are also linear.



Below is the official solution:



Assume $c_1v_1+ldots+c_nv_n=0$. Applying $A$ (which is linear), yields $c_1Av_1+ldots+c_nAv_n=0$. Since $Av_1,ldots,Av_n$ are LI, then $c_1=c_2=ldots=c_n=0$. Thus, $v_1,ldots,v_n$ are LI.



It seems to me that this solution does not take me anywhere. It already assumed a LI set $v_1,ldots,v_n$ and applied $A$ to it, to conclude that the same set is LI. But what if $A$ is singular?



I think it's more reasonable to linearly combine $Av_1,ldots,Av_n$:



$$
c_1Av_1+ldots+c_nAv_n=0
$$



Thus,



$$
Aleft(c_1v_1+ldots+c_nv_nright)=0
$$



And then to conclude that, iff $A$ is nonsingular, then $v_1,ldots,v_n$ is LI. Otherwise, we can't be sure.



Am I missing something?



Thanks in advance.






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$












  • $begingroup$
    How singula nature of $A$ affects the linear independency?The proof seems fine to me
    $endgroup$
    – Tom.
    Dec 10 '18 at 12:56
















0












0








0





$begingroup$


I'm studying with a Linear Algebra book which presents the following question:



Be $A:Erightarrow F$ a linear transformation. If the vectors $Av_1,ldots,Av_n in F$ are LI, then prove that $v_1,ldots,v_n in E$ are also linear.



Below is the official solution:



Assume $c_1v_1+ldots+c_nv_n=0$. Applying $A$ (which is linear), yields $c_1Av_1+ldots+c_nAv_n=0$. Since $Av_1,ldots,Av_n$ are LI, then $c_1=c_2=ldots=c_n=0$. Thus, $v_1,ldots,v_n$ are LI.



It seems to me that this solution does not take me anywhere. It already assumed a LI set $v_1,ldots,v_n$ and applied $A$ to it, to conclude that the same set is LI. But what if $A$ is singular?



I think it's more reasonable to linearly combine $Av_1,ldots,Av_n$:



$$
c_1Av_1+ldots+c_nAv_n=0
$$



Thus,



$$
Aleft(c_1v_1+ldots+c_nv_nright)=0
$$



And then to conclude that, iff $A$ is nonsingular, then $v_1,ldots,v_n$ is LI. Otherwise, we can't be sure.



Am I missing something?



Thanks in advance.






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$




I'm studying with a Linear Algebra book which presents the following question:



Be $A:Erightarrow F$ a linear transformation. If the vectors $Av_1,ldots,Av_n in F$ are LI, then prove that $v_1,ldots,v_n in E$ are also linear.



Below is the official solution:



Assume $c_1v_1+ldots+c_nv_n=0$. Applying $A$ (which is linear), yields $c_1Av_1+ldots+c_nAv_n=0$. Since $Av_1,ldots,Av_n$ are LI, then $c_1=c_2=ldots=c_n=0$. Thus, $v_1,ldots,v_n$ are LI.



It seems to me that this solution does not take me anywhere. It already assumed a LI set $v_1,ldots,v_n$ and applied $A$ to it, to conclude that the same set is LI. But what if $A$ is singular?



I think it's more reasonable to linearly combine $Av_1,ldots,Av_n$:



$$
c_1Av_1+ldots+c_nAv_n=0
$$



Thus,



$$
Aleft(c_1v_1+ldots+c_nv_nright)=0
$$



And then to conclude that, iff $A$ is nonsingular, then $v_1,ldots,v_n$ is LI. Otherwise, we can't be sure.



Am I missing something?



Thanks in advance.






linear-algebra linear-transformations




linear-algebra linear-transformations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 12:50









César PachecoCésar Pacheco

81




81












  • $begingroup$
    How singula nature of $A$ affects the linear independency?The proof seems fine to me
    $endgroup$
    – Tom.
    Dec 10 '18 at 12:56




















  • $begingroup$
    How singula nature of $A$ affects the linear independency?The proof seems fine to me
    $endgroup$
    – Tom.
    Dec 10 '18 at 12:56


















$begingroup$
How singula nature of $A$ affects the linear independency?The proof seems fine to me
$endgroup$
– Tom.
Dec 10 '18 at 12:56






$begingroup$
How singula nature of $A$ affects the linear independency?The proof seems fine to me
$endgroup$
– Tom.
Dec 10 '18 at 12:56












1 Answer
1






active

oldest

votes


















1












$begingroup$

We want to check if $v_i$ are linearly independent, hence we form the equation that we use to check linearly independence.



$$sum_{i=1}^n c_i v_i = 0$$



At this moment, I still do not know if all the $c_i$ must be $0$.



Now, let's multiply by $A$.



$$sum_{i=1}^n c_i (Av_i) = 0$$



Now, I know that all the $c_i$'s are zero since we are told that ${Av_1, ldots, Av_n}$ is linearly independent. Hence ${ v_1, ldots, v_n}$ is linearly independent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot. Your explanation helped me see the problem with my reasoning.
    $endgroup$
    – César Pacheco
    Dec 10 '18 at 13:16











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1 Answer
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1 Answer
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active

oldest

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1












$begingroup$

We want to check if $v_i$ are linearly independent, hence we form the equation that we use to check linearly independence.



$$sum_{i=1}^n c_i v_i = 0$$



At this moment, I still do not know if all the $c_i$ must be $0$.



Now, let's multiply by $A$.



$$sum_{i=1}^n c_i (Av_i) = 0$$



Now, I know that all the $c_i$'s are zero since we are told that ${Av_1, ldots, Av_n}$ is linearly independent. Hence ${ v_1, ldots, v_n}$ is linearly independent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot. Your explanation helped me see the problem with my reasoning.
    $endgroup$
    – César Pacheco
    Dec 10 '18 at 13:16
















1












$begingroup$

We want to check if $v_i$ are linearly independent, hence we form the equation that we use to check linearly independence.



$$sum_{i=1}^n c_i v_i = 0$$



At this moment, I still do not know if all the $c_i$ must be $0$.



Now, let's multiply by $A$.



$$sum_{i=1}^n c_i (Av_i) = 0$$



Now, I know that all the $c_i$'s are zero since we are told that ${Av_1, ldots, Av_n}$ is linearly independent. Hence ${ v_1, ldots, v_n}$ is linearly independent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot. Your explanation helped me see the problem with my reasoning.
    $endgroup$
    – César Pacheco
    Dec 10 '18 at 13:16














1












1








1





$begingroup$

We want to check if $v_i$ are linearly independent, hence we form the equation that we use to check linearly independence.



$$sum_{i=1}^n c_i v_i = 0$$



At this moment, I still do not know if all the $c_i$ must be $0$.



Now, let's multiply by $A$.



$$sum_{i=1}^n c_i (Av_i) = 0$$



Now, I know that all the $c_i$'s are zero since we are told that ${Av_1, ldots, Av_n}$ is linearly independent. Hence ${ v_1, ldots, v_n}$ is linearly independent.






share|cite|improve this answer









$endgroup$



We want to check if $v_i$ are linearly independent, hence we form the equation that we use to check linearly independence.



$$sum_{i=1}^n c_i v_i = 0$$



At this moment, I still do not know if all the $c_i$ must be $0$.



Now, let's multiply by $A$.



$$sum_{i=1}^n c_i (Av_i) = 0$$



Now, I know that all the $c_i$'s are zero since we are told that ${Av_1, ldots, Av_n}$ is linearly independent. Hence ${ v_1, ldots, v_n}$ is linearly independent.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 12:58









Siong Thye GohSiong Thye Goh

101k1466117




101k1466117












  • $begingroup$
    Thanks a lot. Your explanation helped me see the problem with my reasoning.
    $endgroup$
    – César Pacheco
    Dec 10 '18 at 13:16


















  • $begingroup$
    Thanks a lot. Your explanation helped me see the problem with my reasoning.
    $endgroup$
    – César Pacheco
    Dec 10 '18 at 13:16
















$begingroup$
Thanks a lot. Your explanation helped me see the problem with my reasoning.
$endgroup$
– César Pacheco
Dec 10 '18 at 13:16




$begingroup$
Thanks a lot. Your explanation helped me see the problem with my reasoning.
$endgroup$
– César Pacheco
Dec 10 '18 at 13:16


















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