Linear independence and basis within linear transformations
$begingroup$
I am new to linear algebra, and without a teacher. I just want to check my understanding based on the following question:
Let ${v_1,…,v_k}$ be a subset in the linear space V and $T:V to V$ a linear transformation.
- If ${Tv_1,…,Tv_k}$ is independent, then ${v_1,…,v_k}$ is independent
- If ${Tv_1,…,Tv_k}$ spans V, then ${v_1,…,v_k}$ is a basis of $V$.
My answer was "yes" to no. 1 (by definition), and "no" to 2. (because we do not know if ${v_1,…,v_k}$ is linearly independent)...
Am I on the right path?
Thank you!
linear-algebra linear-transformations
edited Jan 1 at 18:38
Hans Hüttel
3,3572921
asked Jan 1 at 18:32
daltadalta
1508
$endgroup$
add a comment |
$begingroup$
I am new to linear algebra, and without a teacher. I just want to check my understanding based on the following question:
Let ${v_1,…,v_k}$ be a subset in the linear space V and $T:V to V$ a linear transformation.
- If ${Tv_1,…,Tv_k}$ is independent, then ${v_1,…,v_k}$ is independent
- If ${Tv_1,…,Tv_k}$ spans V, then ${v_1,…,v_k}$ is a basis of $V$.
My answer was "yes" to no. 1 (by definition), and "no" to 2. (because we do not know if ${v_1,…,v_k}$ is linearly independent)...
Am I on the right path?
Thank you!
linear-algebra linear-transformations
edited Jan 1 at 18:38
Hans Hüttel
3,3572921
asked Jan 1 at 18:32
daltadalta
1508
$endgroup$
add a comment |
$begingroup$
I am new to linear algebra, and without a teacher. I just want to check my understanding based on the following question:
Let ${v_1,…,v_k}$ be a subset in the linear space V and $T:V to V$ a linear transformation.
- If ${Tv_1,…,Tv_k}$ is independent, then ${v_1,…,v_k}$ is independent
- If ${Tv_1,…,Tv_k}$ spans V, then ${v_1,…,v_k}$ is a basis of $V$.
My answer was "yes" to no. 1 (by definition), and "no" to 2. (because we do not know if ${v_1,…,v_k}$ is linearly independent)...
Am I on the right path?
Thank you!
linear-algebra linear-transformations
edited Jan 1 at 18:38
Hans Hüttel
3,3572921
asked Jan 1 at 18:32
daltadalta
1508
$endgroup$
I am new to linear algebra, and without a teacher. I just want to check my understanding based on the following question:
Let ${v_1,…,v_k}$ be a subset in the linear space V and $T:V to V$ a linear transformation.
- If ${Tv_1,…,Tv_k}$ is independent, then ${v_1,…,v_k}$ is independent
- If ${Tv_1,…,Tv_k}$ spans V, then ${v_1,…,v_k}$ is a basis of $V$.
My answer was "yes" to no. 1 (by definition), and "no" to 2. (because we do not know if ${v_1,…,v_k}$ is linearly independent)...
Am I on the right path?
Thank you!
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Jan 1 at 18:38
Hans Hüttel
3,3572921
asked Jan 1 at 18:32
daltadalta
1508
edited Jan 1 at 18:38
Hans Hüttel
3,3572921
asked Jan 1 at 18:32
daltadalta
1508
edited Jan 1 at 18:38
Hans Hüttel
3,3572921
edited Jan 1 at 18:38
Hans Hüttel
3,3572921
edited Jan 1 at 18:38
Hans Hüttel
3,3572921
3,3572921
asked Jan 1 at 18:32
daltadalta
1508
asked Jan 1 at 18:32
daltadalta
1508
asked Jan 1 at 18:32
daltadalta
1508
1508
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
- It is true, but not “by definition”. If $alpha_1,ldots,alpha_k$ are such that $alpha_1v_1+cdots+alpha_kv_k=0$, thenbegin{align}0&=T(0)\&=T(alpha_1v_1+cdots+alpha_kv_k)\&=alpha_1T(v_1)+cdots+alpha_kT(v_k).end{align}Since ${T(v_1),ldots,T(v_k}$ is linearly independent, the $alpha_i$'s are all equal to $0$.
- It is false, but you should provide a counterexample.
answered Jan 1 at 18:39
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Your answers are correct but your reasoning is invalid.
As a rule of thumb, if you want to justify "yes" answers, you (usually) need a conventional proof using theorems, definitions, and so forth. But if you want to justify a "no" answer you (usually) need to exhibit a specific counter-example.
So, in your case for #2, you need to give a vector space $V$, a set ${v_1,cdots,v_k}$ which is not a basis for $V$, and a linear map $T:Vto V$ such that ${Tv_1,cdots,Tv_k}$ spans $V$.
The easiest way to do this is to let $T$ be the identity map on $V$ and let ${v_1,cdots,v_k}$ be any overcomplete set.
For #1, it is true but not by definition. If ${v_1,cdots v_k}$ were not independent we would have a nontrivial linear combination $sum_{i=1}^ka_iv_i=0$ to which we could apply $T$, violating independence of ${Tv_1,cdots,Tv_k}$.
answered Jan 1 at 18:39
Ben WBen W
2,321615
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
- It is true, but not “by definition”. If $alpha_1,ldots,alpha_k$ are such that $alpha_1v_1+cdots+alpha_kv_k=0$, thenbegin{align}0&=T(0)\&=T(alpha_1v_1+cdots+alpha_kv_k)\&=alpha_1T(v_1)+cdots+alpha_kT(v_k).end{align}Since ${T(v_1),ldots,T(v_k}$ is linearly independent, the $alpha_i$'s are all equal to $0$.
- It is false, but you should provide a counterexample.
answered Jan 1 at 18:39
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
- It is true, but not “by definition”. If $alpha_1,ldots,alpha_k$ are such that $alpha_1v_1+cdots+alpha_kv_k=0$, thenbegin{align}0&=T(0)\&=T(alpha_1v_1+cdots+alpha_kv_k)\&=alpha_1T(v_1)+cdots+alpha_kT(v_k).end{align}Since ${T(v_1),ldots,T(v_k}$ is linearly independent, the $alpha_i$'s are all equal to $0$.
- It is false, but you should provide a counterexample.
answered Jan 1 at 18:39
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
- It is true, but not “by definition”. If $alpha_1,ldots,alpha_k$ are such that $alpha_1v_1+cdots+alpha_kv_k=0$, thenbegin{align}0&=T(0)\&=T(alpha_1v_1+cdots+alpha_kv_k)\&=alpha_1T(v_1)+cdots+alpha_kT(v_k).end{align}Since ${T(v_1),ldots,T(v_k}$ is linearly independent, the $alpha_i$'s are all equal to $0$.
- It is false, but you should provide a counterexample.
answered Jan 1 at 18:39
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
- It is true, but not “by definition”. If $alpha_1,ldots,alpha_k$ are such that $alpha_1v_1+cdots+alpha_kv_k=0$, thenbegin{align}0&=T(0)\&=T(alpha_1v_1+cdots+alpha_kv_k)\&=alpha_1T(v_1)+cdots+alpha_kT(v_k).end{align}Since ${T(v_1),ldots,T(v_k}$ is linearly independent, the $alpha_i$'s are all equal to $0$.
- It is false, but you should provide a counterexample.
answered Jan 1 at 18:39
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 1 at 18:39
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 1 at 18:39
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 1 at 18:39
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
add a comment |
add a comment |
$begingroup$
Your answers are correct but your reasoning is invalid.
As a rule of thumb, if you want to justify "yes" answers, you (usually) need a conventional proof using theorems, definitions, and so forth. But if you want to justify a "no" answer you (usually) need to exhibit a specific counter-example.
So, in your case for #2, you need to give a vector space $V$, a set ${v_1,cdots,v_k}$ which is not a basis for $V$, and a linear map $T:Vto V$ such that ${Tv_1,cdots,Tv_k}$ spans $V$.
The easiest way to do this is to let $T$ be the identity map on $V$ and let ${v_1,cdots,v_k}$ be any overcomplete set.
For #1, it is true but not by definition. If ${v_1,cdots v_k}$ were not independent we would have a nontrivial linear combination $sum_{i=1}^ka_iv_i=0$ to which we could apply $T$, violating independence of ${Tv_1,cdots,Tv_k}$.
answered Jan 1 at 18:39
Ben WBen W
2,321615
$endgroup$
add a comment |
$begingroup$
Your answers are correct but your reasoning is invalid.
As a rule of thumb, if you want to justify "yes" answers, you (usually) need a conventional proof using theorems, definitions, and so forth. But if you want to justify a "no" answer you (usually) need to exhibit a specific counter-example.
So, in your case for #2, you need to give a vector space $V$, a set ${v_1,cdots,v_k}$ which is not a basis for $V$, and a linear map $T:Vto V$ such that ${Tv_1,cdots,Tv_k}$ spans $V$.
The easiest way to do this is to let $T$ be the identity map on $V$ and let ${v_1,cdots,v_k}$ be any overcomplete set.
For #1, it is true but not by definition. If ${v_1,cdots v_k}$ were not independent we would have a nontrivial linear combination $sum_{i=1}^ka_iv_i=0$ to which we could apply $T$, violating independence of ${Tv_1,cdots,Tv_k}$.
answered Jan 1 at 18:39
Ben WBen W
2,321615
$endgroup$
add a comment |
$begingroup$
Your answers are correct but your reasoning is invalid.
As a rule of thumb, if you want to justify "yes" answers, you (usually) need a conventional proof using theorems, definitions, and so forth. But if you want to justify a "no" answer you (usually) need to exhibit a specific counter-example.
So, in your case for #2, you need to give a vector space $V$, a set ${v_1,cdots,v_k}$ which is not a basis for $V$, and a linear map $T:Vto V$ such that ${Tv_1,cdots,Tv_k}$ spans $V$.
The easiest way to do this is to let $T$ be the identity map on $V$ and let ${v_1,cdots,v_k}$ be any overcomplete set.
For #1, it is true but not by definition. If ${v_1,cdots v_k}$ were not independent we would have a nontrivial linear combination $sum_{i=1}^ka_iv_i=0$ to which we could apply $T$, violating independence of ${Tv_1,cdots,Tv_k}$.
answered Jan 1 at 18:39
Ben WBen W
2,321615
$endgroup$
Your answers are correct but your reasoning is invalid.
As a rule of thumb, if you want to justify "yes" answers, you (usually) need a conventional proof using theorems, definitions, and so forth. But if you want to justify a "no" answer you (usually) need to exhibit a specific counter-example.
So, in your case for #2, you need to give a vector space $V$, a set ${v_1,cdots,v_k}$ which is not a basis for $V$, and a linear map $T:Vto V$ such that ${Tv_1,cdots,Tv_k}$ spans $V$.
The easiest way to do this is to let $T$ be the identity map on $V$ and let ${v_1,cdots,v_k}$ be any overcomplete set.
For #1, it is true but not by definition. If ${v_1,cdots v_k}$ were not independent we would have a nontrivial linear combination $sum_{i=1}^ka_iv_i=0$ to which we could apply $T$, violating independence of ${Tv_1,cdots,Tv_k}$.
answered Jan 1 at 18:39
Ben WBen W
2,321615
edited Jan 1 at 18:45
answered Jan 1 at 18:39
Ben WBen W
2,321615
answered Jan 1 at 18:39
Ben WBen W
2,321615
answered Jan 1 at 18:39
Ben WBen W
2,321615
2,321615
add a comment |
add a comment |
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