$operatornamerangeT_1subsetoperatornamerangeT_2$ if and only if there exists $SinmathcalL(V,V)$ such that $T_1 = T_2S$

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Is the Following argument correct?




Suppose $V$ is finite-dimensional and $T_1,T_2inmathcalL(V,W)$.
Prove that $operatornamerangeT_1subsetoperatornamerangeT_2$ if
and only if there exists $SinmathcalL(V,V)$ such that $T_1 =
T_2S$.




Proof. Since $V$ is finite-dimensional we may therefore invoke a basis $v_1,v_2,dots,v_n$ for $V$.



$(Rightarrow)$. Assume that $operatornamerangeT_1subsetoperatornamerangeT_2$, then for each $T_1v_j$ there exists a $T_2u_j$ such that $T_1v_j=T_2u_j$. Now using theorem $textbf3.5$ we may define a linear map $T:Vto V$ as follows.
$$Sv_j = u_j$$
Given the above definition, it is evident that the images of the basis vectors under $T_2S$ and $T_1$ are the same implying that $T_1=T_2S$



$(Leftarrow)$. Conversely assume $T_1=T_2S$ and let $Tv_1inoperatornamerangeT_1$ and $Sv_1=u$, then $T_1v=T_2S = T_2u$ implying $T_1vinoperatornamerangeT_2$.



$blacksquare$



Note:



  • Theroem $textbf3.5$ says that a linear transformation is uniquely determined by its action on a basis.



linear-algebra proof-verification




share|cite|improve this question



















  • Seems correct to me!
    – Davide Morgante
    Jul 22 at 10:49










  • It's fine, other than a typo.
    – José Carlos Santos
    Jul 22 at 10:50














up vote
1
down vote

favorite












Is the Following argument correct?




Suppose $V$ is finite-dimensional and $T_1,T_2inmathcalL(V,W)$.
Prove that $operatornamerangeT_1subsetoperatornamerangeT_2$ if
and only if there exists $SinmathcalL(V,V)$ such that $T_1 =
T_2S$.




Proof. Since $V$ is finite-dimensional we may therefore invoke a basis $v_1,v_2,dots,v_n$ for $V$.



$(Rightarrow)$. Assume that $operatornamerangeT_1subsetoperatornamerangeT_2$, then for each $T_1v_j$ there exists a $T_2u_j$ such that $T_1v_j=T_2u_j$. Now using theorem $textbf3.5$ we may define a linear map $T:Vto V$ as follows.
$$Sv_j = u_j$$
Given the above definition, it is evident that the images of the basis vectors under $T_2S$ and $T_1$ are the same implying that $T_1=T_2S$



$(Leftarrow)$. Conversely assume $T_1=T_2S$ and let $Tv_1inoperatornamerangeT_1$ and $Sv_1=u$, then $T_1v=T_2S = T_2u$ implying $T_1vinoperatornamerangeT_2$.



$blacksquare$



Note:



  • Theroem $textbf3.5$ says that a linear transformation is uniquely determined by its action on a basis.



linear-algebra proof-verification




share|cite|improve this question



















  • Seems correct to me!
    – Davide Morgante
    Jul 22 at 10:49










  • It's fine, other than a typo.
    – José Carlos Santos
    Jul 22 at 10:50












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Is the Following argument correct?




Suppose $V$ is finite-dimensional and $T_1,T_2inmathcalL(V,W)$.
Prove that $operatornamerangeT_1subsetoperatornamerangeT_2$ if
and only if there exists $SinmathcalL(V,V)$ such that $T_1 =
T_2S$.




Proof. Since $V$ is finite-dimensional we may therefore invoke a basis $v_1,v_2,dots,v_n$ for $V$.



$(Rightarrow)$. Assume that $operatornamerangeT_1subsetoperatornamerangeT_2$, then for each $T_1v_j$ there exists a $T_2u_j$ such that $T_1v_j=T_2u_j$. Now using theorem $textbf3.5$ we may define a linear map $T:Vto V$ as follows.
$$Sv_j = u_j$$
Given the above definition, it is evident that the images of the basis vectors under $T_2S$ and $T_1$ are the same implying that $T_1=T_2S$



$(Leftarrow)$. Conversely assume $T_1=T_2S$ and let $Tv_1inoperatornamerangeT_1$ and $Sv_1=u$, then $T_1v=T_2S = T_2u$ implying $T_1vinoperatornamerangeT_2$.



$blacksquare$



Note:



  • Theroem $textbf3.5$ says that a linear transformation is uniquely determined by its action on a basis.



linear-algebra proof-verification




share|cite|improve this question











Is the Following argument correct?




Suppose $V$ is finite-dimensional and $T_1,T_2inmathcalL(V,W)$.
Prove that $operatornamerangeT_1subsetoperatornamerangeT_2$ if
and only if there exists $SinmathcalL(V,V)$ such that $T_1 =
T_2S$.




Proof. Since $V$ is finite-dimensional we may therefore invoke a basis $v_1,v_2,dots,v_n$ for $V$.



$(Rightarrow)$. Assume that $operatornamerangeT_1subsetoperatornamerangeT_2$, then for each $T_1v_j$ there exists a $T_2u_j$ such that $T_1v_j=T_2u_j$. Now using theorem $textbf3.5$ we may define a linear map $T:Vto V$ as follows.
$$Sv_j = u_j$$
Given the above definition, it is evident that the images of the basis vectors under $T_2S$ and $T_1$ are the same implying that $T_1=T_2S$



$(Leftarrow)$. Conversely assume $T_1=T_2S$ and let $Tv_1inoperatornamerangeT_1$ and $Sv_1=u$, then $T_1v=T_2S = T_2u$ implying $T_1vinoperatornamerangeT_2$.



$blacksquare$



Note:



  • Theroem $textbf3.5$ says that a linear transformation is uniquely determined by its action on a basis.




linear-algebra proof-verification





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asked Jul 22 at 10:44









Atif Farooq

2,7092824




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  • Seems correct to me!
    – Davide Morgante
    Jul 22 at 10:49










  • It's fine, other than a typo.
    – José Carlos Santos
    Jul 22 at 10:50
















  • Seems correct to me!
    – Davide Morgante
    Jul 22 at 10:49










  • It's fine, other than a typo.
    – José Carlos Santos
    Jul 22 at 10:50















Seems correct to me!
– Davide Morgante
Jul 22 at 10:49




Seems correct to me!
– Davide Morgante
Jul 22 at 10:49












It's fine, other than a typo.
– José Carlos Santos
Jul 22 at 10:50




It's fine, other than a typo.
– José Carlos Santos
Jul 22 at 10:50















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