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Linear Algebra - Proof on Linear Transformations and Independence
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Let $T: BbbR^n rightarrow BbbR^n$ be an invertible linear transformation. Prove that if $ vecv_1,vecv_2,...,vecv_n$ is a linearly independent set of $BbbR^n$ if and only if $T( vecv_1),T( vecv_2),...,T( vecv_k)$ is a linearly independent set of $BbbR^n$.
So I know that for a set of vectors to be linearly independent then they cannot be written as a linear combination of one another. So can I say something along the lines of $t_1 v_1 + t_2 v_2 + ... + t_k v_k = 0$ then taking the transformations the set I get is $T( t_1 vecv_1),T( t_2 vecv_2),...,T( t_k vecv_k)$ which given the definition of linear independence can be written as $ t_1 T( vecv_1) + t_2 T( vecv_2) +,..., + t_k T( vecv_k) = 0$. So then is that it?
linear-algebra proof-verification linear-transformations
edited 4 hours ago
Chinnapparaj R
3,818724
asked 4 hours ago
FundementalJTheorem
373
add a comment |Â
up vote
1
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Let $T: BbbR^n rightarrow BbbR^n$ be an invertible linear transformation. Prove that if $ vecv_1,vecv_2,...,vecv_n$ is a linearly independent set of $BbbR^n$ if and only if $T( vecv_1),T( vecv_2),...,T( vecv_k)$ is a linearly independent set of $BbbR^n$.
So I know that for a set of vectors to be linearly independent then they cannot be written as a linear combination of one another. So can I say something along the lines of $t_1 v_1 + t_2 v_2 + ... + t_k v_k = 0$ then taking the transformations the set I get is $T( t_1 vecv_1),T( t_2 vecv_2),...,T( t_k vecv_k)$ which given the definition of linear independence can be written as $ t_1 T( vecv_1) + t_2 T( vecv_2) +,..., + t_k T( vecv_k) = 0$. So then is that it?
linear-algebra proof-verification linear-transformations
edited 4 hours ago
Chinnapparaj R
3,818724
asked 4 hours ago
FundementalJTheorem
373
Yes. You have the right line of thinking. Be careful:you've only shown that a linear combination in the domain implies a linear combination in the range. You may want to remark why none of the images T(v$_i$) can be 0 in this situation. Otherwise, a non-trivial solution may map to a trivial solution if certain images are 0.
– Joel Pereira
4 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $T: BbbR^n rightarrow BbbR^n$ be an invertible linear transformation. Prove that if $ vecv_1,vecv_2,...,vecv_n$ is a linearly independent set of $BbbR^n$ if and only if $T( vecv_1),T( vecv_2),...,T( vecv_k)$ is a linearly independent set of $BbbR^n$.
So I know that for a set of vectors to be linearly independent then they cannot be written as a linear combination of one another. So can I say something along the lines of $t_1 v_1 + t_2 v_2 + ... + t_k v_k = 0$ then taking the transformations the set I get is $T( t_1 vecv_1),T( t_2 vecv_2),...,T( t_k vecv_k)$ which given the definition of linear independence can be written as $ t_1 T( vecv_1) + t_2 T( vecv_2) +,..., + t_k T( vecv_k) = 0$. So then is that it?
linear-algebra proof-verification linear-transformations
edited 4 hours ago
Chinnapparaj R
3,818724
asked 4 hours ago
FundementalJTheorem
373
Let $T: BbbR^n rightarrow BbbR^n$ be an invertible linear transformation. Prove that if $ vecv_1,vecv_2,...,vecv_n$ is a linearly independent set of $BbbR^n$ if and only if $T( vecv_1),T( vecv_2),...,T( vecv_k)$ is a linearly independent set of $BbbR^n$.
So I know that for a set of vectors to be linearly independent then they cannot be written as a linear combination of one another. So can I say something along the lines of $t_1 v_1 + t_2 v_2 + ... + t_k v_k = 0$ then taking the transformations the set I get is $T( t_1 vecv_1),T( t_2 vecv_2),...,T( t_k vecv_k)$ which given the definition of linear independence can be written as $ t_1 T( vecv_1) + t_2 T( vecv_2) +,..., + t_k T( vecv_k) = 0$. So then is that it?
linear-algebra proof-verification linear-transformations
linear-algebra proof-verification linear-transformations
edited 4 hours ago
Chinnapparaj R
3,818724
asked 4 hours ago
FundementalJTheorem
373
edited 4 hours ago
Chinnapparaj R
3,818724
asked 4 hours ago
FundementalJTheorem
373
edited 4 hours ago
Chinnapparaj R
3,818724
edited 4 hours ago
Chinnapparaj R
3,818724
edited 4 hours ago
Chinnapparaj R
3,818724
3,818724
asked 4 hours ago
FundementalJTheorem
373
asked 4 hours ago
FundementalJTheorem
373
asked 4 hours ago
FundementalJTheorem
373
373
Yes. You have the right line of thinking. Be careful:you've only shown that a linear combination in the domain implies a linear combination in the range. You may want to remark why none of the images T(v$_i$) can be 0 in this situation. Otherwise, a non-trivial solution may map to a trivial solution if certain images are 0.
– Joel Pereira
4 hours ago
add a comment |Â
Yes. You have the right line of thinking. Be careful:you've only shown that a linear combination in the domain implies a linear combination in the range. You may want to remark why none of the images T(v$_i$) can be 0 in this situation. Otherwise, a non-trivial solution may map to a trivial solution if certain images are 0.
– Joel Pereira
4 hours ago
Yes. You have the right line of thinking. Be careful:you've only shown that a linear combination in the domain implies a linear combination in the range. You may want to remark why none of the images T(v$_i$) can be 0 in this situation. Otherwise, a non-trivial solution may map to a trivial solution if certain images are 0.
– Joel Pereira
4 hours ago
Yes. You have the right line of thinking. Be careful:you've only shown that a linear combination in the domain implies a linear combination in the range. You may want to remark why none of the images T(v$_i$) can be 0 in this situation. Otherwise, a non-trivial solution may map to a trivial solution if certain images are 0.
– Joel Pereira
4 hours ago
add a comment |Â
1 Answer
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For $Longrightarrow$ : You are on the right way!
$t_1T(v_1)+t_2T(v_2)+cdots+t_nT(v_n)=0 $ implies $$T(t_1v_1+cdots+t_nv_n)=0=T(0)$$ and so by one-one of $T$, $$t_1v_1+cdots+t_nv_n=0$$
and by independence of $v_i$, $$t_1=t_2=cdots=t_n=0$$
For $Longleftarrow$ : Let $alpha_1v_1+alpha_2v_2+cdots+alpha_nv_n=0$ and apply $T$ on both sides and use independence of $T(v_i)$'s to get the result!
answered 4 hours ago
Chinnapparaj R
3,818724
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
For $Longrightarrow$ : You are on the right way!
$t_1T(v_1)+t_2T(v_2)+cdots+t_nT(v_n)=0 $ implies $$T(t_1v_1+cdots+t_nv_n)=0=T(0)$$ and so by one-one of $T$, $$t_1v_1+cdots+t_nv_n=0$$
and by independence of $v_i$, $$t_1=t_2=cdots=t_n=0$$
For $Longleftarrow$ : Let $alpha_1v_1+alpha_2v_2+cdots+alpha_nv_n=0$ and apply $T$ on both sides and use independence of $T(v_i)$'s to get the result!
answered 4 hours ago
Chinnapparaj R
3,818724
add a comment |Â
up vote
3
down vote
For $Longrightarrow$ : You are on the right way!
$t_1T(v_1)+t_2T(v_2)+cdots+t_nT(v_n)=0 $ implies $$T(t_1v_1+cdots+t_nv_n)=0=T(0)$$ and so by one-one of $T$, $$t_1v_1+cdots+t_nv_n=0$$
and by independence of $v_i$, $$t_1=t_2=cdots=t_n=0$$
For $Longleftarrow$ : Let $alpha_1v_1+alpha_2v_2+cdots+alpha_nv_n=0$ and apply $T$ on both sides and use independence of $T(v_i)$'s to get the result!
answered 4 hours ago
Chinnapparaj R
3,818724
add a comment |Â
up vote
3
down vote
up vote
3
down vote
For $Longrightarrow$ : You are on the right way!
$t_1T(v_1)+t_2T(v_2)+cdots+t_nT(v_n)=0 $ implies $$T(t_1v_1+cdots+t_nv_n)=0=T(0)$$ and so by one-one of $T$, $$t_1v_1+cdots+t_nv_n=0$$
and by independence of $v_i$, $$t_1=t_2=cdots=t_n=0$$
For $Longleftarrow$ : Let $alpha_1v_1+alpha_2v_2+cdots+alpha_nv_n=0$ and apply $T$ on both sides and use independence of $T(v_i)$'s to get the result!
answered 4 hours ago
Chinnapparaj R
3,818724
For $Longrightarrow$ : You are on the right way!
$t_1T(v_1)+t_2T(v_2)+cdots+t_nT(v_n)=0 $ implies $$T(t_1v_1+cdots+t_nv_n)=0=T(0)$$ and so by one-one of $T$, $$t_1v_1+cdots+t_nv_n=0$$
and by independence of $v_i$, $$t_1=t_2=cdots=t_n=0$$
For $Longleftarrow$ : Let $alpha_1v_1+alpha_2v_2+cdots+alpha_nv_n=0$ and apply $T$ on both sides and use independence of $T(v_i)$'s to get the result!
answered 4 hours ago
Chinnapparaj R
3,818724
edited 4 hours ago
answered 4 hours ago
Chinnapparaj R
3,818724
answered 4 hours ago
Chinnapparaj R
3,818724
answered 4 hours ago
Chinnapparaj R
3,818724
3,818724
add a comment |Â
add a comment |Â
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Yes. You have the right line of thinking. Be careful:you've only shown that a linear combination in the domain implies a linear combination in the range. You may want to remark why none of the images T(v$_i$) can be 0 in this situation. Otherwise, a non-trivial solution may map to a trivial solution if certain images are 0.
– Joel Pereira
4 hours ago