Mixing
show that $I, T, T^2, …, T^k$ are linear dependent
$begingroup$
I am learning linear algebra and new to it. I can not solve this problem. I think it has a trick that I don't know.
for T(a linear map), $T:Vrightarrow V$ and every $v$ in $V$ the $v, T(v), T^2(v), ..., T^k(v)$ are linearly dependent where $k$ is a natural number $leq dim(V)$.
show that $I, T, T^2, ..., T^k$ are linearly dependent
thanks in advance.
linear-algebra matrices linear-transformations independence
asked Dec 20 '18 at 22:08
Peyman mohseni kiasariPeyman mohseni kiasari
1089
$endgroup$
|
show 1 more comment
$begingroup$
I am learning linear algebra and new to it. I can not solve this problem. I think it has a trick that I don't know.
for T(a linear map), $T:Vrightarrow V$ and every $v$ in $V$ the $v, T(v), T^2(v), ..., T^k(v)$ are linearly dependent where $k$ is a natural number $leq dim(V)$.
show that $I, T, T^2, ..., T^k$ are linearly dependent
thanks in advance.
linear-algebra matrices linear-transformations independence
asked Dec 20 '18 at 22:08
Peyman mohseni kiasariPeyman mohseni kiasari
1089
$endgroup$
$begingroup$
What is T? A matrix?
$endgroup$
– H_1317
Dec 23 '18 at 23:36
$begingroup$
@H_1317 yes, T is actually "T:V→V". unfortunately, I've asked the question wrong! it must be actually "T(v)s are dependent prove Ts are dependent". I mean both "independent"s should be "dependent". when I saw guys were answering the question I didn't edit that and just accept the answer.
$endgroup$
– Peyman mohseni kiasari
Dec 23 '18 at 23:47
$begingroup$
Can u edit your question identifying what T is, and the proper wording for any other errors please? Also what vector space u believe T is in (a matrix space?) thanks
$endgroup$
– H_1317
Dec 23 '18 at 23:49
$begingroup$
@H_1317, sorry if I'm not so familiar but I think "T:V→V" is a linear map no? where T must be a matrix, am I wrong? T is a map from V to V where V can be any vector space I think.
$endgroup$
– Peyman mohseni kiasari
Dec 23 '18 at 23:57
1
$begingroup$
If $a_i T^i = 0$ is a linear combination of the $T^i,$ choose some $v$ for which the $T^i(v)$ are linearly independent vectors (you need to assume that only one such $v$ exists, not that every $v$ (which, by the way cannot be) satisfies it) and then conclude all scalars are zero.
$endgroup$
– Will M.
Dec 24 '18 at 7:55
|
show 1 more comment
$begingroup$
I am learning linear algebra and new to it. I can not solve this problem. I think it has a trick that I don't know.
for T(a linear map), $T:Vrightarrow V$ and every $v$ in $V$ the $v, T(v), T^2(v), ..., T^k(v)$ are linearly dependent where $k$ is a natural number $leq dim(V)$.
show that $I, T, T^2, ..., T^k$ are linearly dependent
thanks in advance.
linear-algebra matrices linear-transformations independence
asked Dec 20 '18 at 22:08
Peyman mohseni kiasariPeyman mohseni kiasari
1089
$endgroup$
I am learning linear algebra and new to it. I can not solve this problem. I think it has a trick that I don't know.
for T(a linear map), $T:Vrightarrow V$ and every $v$ in $V$ the $v, T(v), T^2(v), ..., T^k(v)$ are linearly dependent where $k$ is a natural number $leq dim(V)$.
show that $I, T, T^2, ..., T^k$ are linearly dependent
thanks in advance.
linear-algebra matrices linear-transformations independence
linear-algebra matrices linear-transformations independence
asked Dec 20 '18 at 22:08
Peyman mohseni kiasariPeyman mohseni kiasari
1089
asked Dec 20 '18 at 22:08
Peyman mohseni kiasariPeyman mohseni kiasari
1089
edited Jan 15 at 12:04
Peyman mohseni kiasari
asked Dec 20 '18 at 22:08
Peyman mohseni kiasariPeyman mohseni kiasari
1089
asked Dec 20 '18 at 22:08
Peyman mohseni kiasariPeyman mohseni kiasari
1089
asked Dec 20 '18 at 22:08
Peyman mohseni kiasariPeyman mohseni kiasari
1089
1089
$begingroup$
What is T? A matrix?
$endgroup$
– H_1317
Dec 23 '18 at 23:36
$begingroup$
@H_1317 yes, T is actually "T:V→V". unfortunately, I've asked the question wrong! it must be actually "T(v)s are dependent prove Ts are dependent". I mean both "independent"s should be "dependent". when I saw guys were answering the question I didn't edit that and just accept the answer.
$endgroup$
– Peyman mohseni kiasari
Dec 23 '18 at 23:47
$begingroup$
Can u edit your question identifying what T is, and the proper wording for any other errors please? Also what vector space u believe T is in (a matrix space?) thanks
$endgroup$
– H_1317
Dec 23 '18 at 23:49
$begingroup$
@H_1317, sorry if I'm not so familiar but I think "T:V→V" is a linear map no? where T must be a matrix, am I wrong? T is a map from V to V where V can be any vector space I think.
$endgroup$
– Peyman mohseni kiasari
Dec 23 '18 at 23:57
1
$begingroup$
If $a_i T^i = 0$ is a linear combination of the $T^i,$ choose some $v$ for which the $T^i(v)$ are linearly independent vectors (you need to assume that only one such $v$ exists, not that every $v$ (which, by the way cannot be) satisfies it) and then conclude all scalars are zero.
$endgroup$
– Will M.
Dec 24 '18 at 7:55
|
show 1 more comment
$begingroup$
What is T? A matrix?
$endgroup$
– H_1317
Dec 23 '18 at 23:36
$begingroup$
@H_1317 yes, T is actually "T:V→V". unfortunately, I've asked the question wrong! it must be actually "T(v)s are dependent prove Ts are dependent". I mean both "independent"s should be "dependent". when I saw guys were answering the question I didn't edit that and just accept the answer.
$endgroup$
– Peyman mohseni kiasari
Dec 23 '18 at 23:47
$begingroup$
Can u edit your question identifying what T is, and the proper wording for any other errors please? Also what vector space u believe T is in (a matrix space?) thanks
$endgroup$
– H_1317
Dec 23 '18 at 23:49
$begingroup$
@H_1317, sorry if I'm not so familiar but I think "T:V→V" is a linear map no? where T must be a matrix, am I wrong? T is a map from V to V where V can be any vector space I think.
$endgroup$
– Peyman mohseni kiasari
Dec 23 '18 at 23:57
1
$begingroup$
If $a_i T^i = 0$ is a linear combination of the $T^i,$ choose some $v$ for which the $T^i(v)$ are linearly independent vectors (you need to assume that only one such $v$ exists, not that every $v$ (which, by the way cannot be) satisfies it) and then conclude all scalars are zero.
$endgroup$
– Will M.
Dec 24 '18 at 7:55
$begingroup$
What is T? A matrix?
$endgroup$
– H_1317
Dec 23 '18 at 23:36
$begingroup$
What is T? A matrix?
$endgroup$
– H_1317
Dec 23 '18 at 23:36
$begingroup$
@H_1317 yes, T is actually "T:V→V". unfortunately, I've asked the question wrong! it must be actually "T(v)s are dependent prove Ts are dependent". I mean both "independent"s should be "dependent". when I saw guys were answering the question I didn't edit that and just accept the answer.
$endgroup$
– Peyman mohseni kiasari
Dec 23 '18 at 23:47
$begingroup$
@H_1317 yes, T is actually "T:V→V". unfortunately, I've asked the question wrong! it must be actually "T(v)s are dependent prove Ts are dependent". I mean both "independent"s should be "dependent". when I saw guys were answering the question I didn't edit that and just accept the answer.
$endgroup$
– Peyman mohseni kiasari
Dec 23 '18 at 23:47
$begingroup$
Can u edit your question identifying what T is, and the proper wording for any other errors please? Also what vector space u believe T is in (a matrix space?) thanks
$endgroup$
– H_1317
Dec 23 '18 at 23:49
$begingroup$
Can u edit your question identifying what T is, and the proper wording for any other errors please? Also what vector space u believe T is in (a matrix space?) thanks
$endgroup$
– H_1317
Dec 23 '18 at 23:49
$begingroup$
@H_1317, sorry if I'm not so familiar but I think "T:V→V" is a linear map no? where T must be a matrix, am I wrong? T is a map from V to V where V can be any vector space I think.
$endgroup$
– Peyman mohseni kiasari
Dec 23 '18 at 23:57
$begingroup$
@H_1317, sorry if I'm not so familiar but I think "T:V→V" is a linear map no? where T must be a matrix, am I wrong? T is a map from V to V where V can be any vector space I think.
$endgroup$
– Peyman mohseni kiasari
Dec 23 '18 at 23:57
1
1
$begingroup$
If $a_i T^i = 0$ is a linear combination of the $T^i,$ choose some $v$ for which the $T^i(v)$ are linearly independent vectors (you need to assume that only one such $v$ exists, not that every $v$ (which, by the way cannot be) satisfies it) and then conclude all scalars are zero.
$endgroup$
– Will M.
Dec 24 '18 at 7:55
$begingroup$
If $a_i T^i = 0$ is a linear combination of the $T^i,$ choose some $v$ for which the $T^i(v)$ are linearly independent vectors (you need to assume that only one such $v$ exists, not that every $v$ (which, by the way cannot be) satisfies it) and then conclude all scalars are zero.
$endgroup$
– Will M.
Dec 24 '18 at 7:55
|
show 1 more comment
2 Answers
2
active
oldest
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$begingroup$
Suppose $a_1T+...+a_kT^k=0$. Let $xneq 0$, $a_1T(x)+...+a_kT^k(x)=0$, this implies that $a_1=..=a_k=0$ since $T(x),...T^k(x)$ are linearly independent. It results that $T,....T^k$ is linearly independent.
Remark, you have to suppose for every $vneq 0$, $T(v),...,T^k(v)$ is linearly independent since it is not true for $v=0$.
answered Dec 20 '18 at 22:12
Tsemo AristideTsemo Aristide
58.3k11445
$endgroup$
add a comment |
$begingroup$
Hint :
Suppose that $T,T^2,dots, T^k$ are not linearly independent. Then, there exists $lambda_1, lambda_2, dots, lambda_k in mathbb R$ not all zero, such that :
$$lambda_1T + lambda_2T^2 + cdots + lambda_kT^k = 0$$
answered Dec 20 '18 at 22:13
RebellosRebellos
14.8k31248
$endgroup$
$begingroup$
such that what!?
$endgroup$
– Peyman mohseni kiasari
Dec 20 '18 at 22:13
$begingroup$
@Peyman Fixed. Thanks.
$endgroup$
– Rebellos
Dec 20 '18 at 22:14
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose $a_1T+...+a_kT^k=0$. Let $xneq 0$, $a_1T(x)+...+a_kT^k(x)=0$, this implies that $a_1=..=a_k=0$ since $T(x),...T^k(x)$ are linearly independent. It results that $T,....T^k$ is linearly independent.
Remark, you have to suppose for every $vneq 0$, $T(v),...,T^k(v)$ is linearly independent since it is not true for $v=0$.
answered Dec 20 '18 at 22:12
Tsemo AristideTsemo Aristide
58.3k11445
$endgroup$
add a comment |
$begingroup$
Suppose $a_1T+...+a_kT^k=0$. Let $xneq 0$, $a_1T(x)+...+a_kT^k(x)=0$, this implies that $a_1=..=a_k=0$ since $T(x),...T^k(x)$ are linearly independent. It results that $T,....T^k$ is linearly independent.
Remark, you have to suppose for every $vneq 0$, $T(v),...,T^k(v)$ is linearly independent since it is not true for $v=0$.
answered Dec 20 '18 at 22:12
Tsemo AristideTsemo Aristide
58.3k11445
$endgroup$
add a comment |
$begingroup$
Suppose $a_1T+...+a_kT^k=0$. Let $xneq 0$, $a_1T(x)+...+a_kT^k(x)=0$, this implies that $a_1=..=a_k=0$ since $T(x),...T^k(x)$ are linearly independent. It results that $T,....T^k$ is linearly independent.
Remark, you have to suppose for every $vneq 0$, $T(v),...,T^k(v)$ is linearly independent since it is not true for $v=0$.
answered Dec 20 '18 at 22:12
Tsemo AristideTsemo Aristide
58.3k11445
$endgroup$
Suppose $a_1T+...+a_kT^k=0$. Let $xneq 0$, $a_1T(x)+...+a_kT^k(x)=0$, this implies that $a_1=..=a_k=0$ since $T(x),...T^k(x)$ are linearly independent. It results that $T,....T^k$ is linearly independent.
Remark, you have to suppose for every $vneq 0$, $T(v),...,T^k(v)$ is linearly independent since it is not true for $v=0$.
answered Dec 20 '18 at 22:12
Tsemo AristideTsemo Aristide
58.3k11445
answered Dec 20 '18 at 22:12
Tsemo AristideTsemo Aristide
58.3k11445
answered Dec 20 '18 at 22:12
Tsemo AristideTsemo Aristide
58.3k11445
answered Dec 20 '18 at 22:12
Tsemo AristideTsemo Aristide
58.3k11445
58.3k11445
add a comment |
add a comment |
$begingroup$
Hint :
Suppose that $T,T^2,dots, T^k$ are not linearly independent. Then, there exists $lambda_1, lambda_2, dots, lambda_k in mathbb R$ not all zero, such that :
$$lambda_1T + lambda_2T^2 + cdots + lambda_kT^k = 0$$
answered Dec 20 '18 at 22:13
RebellosRebellos
14.8k31248
$endgroup$
$begingroup$
such that what!?
$endgroup$
– Peyman mohseni kiasari
Dec 20 '18 at 22:13
$begingroup$
@Peyman Fixed. Thanks.
$endgroup$
– Rebellos
Dec 20 '18 at 22:14
add a comment |
$begingroup$
Hint :
Suppose that $T,T^2,dots, T^k$ are not linearly independent. Then, there exists $lambda_1, lambda_2, dots, lambda_k in mathbb R$ not all zero, such that :
$$lambda_1T + lambda_2T^2 + cdots + lambda_kT^k = 0$$
answered Dec 20 '18 at 22:13
RebellosRebellos
14.8k31248
$endgroup$
$begingroup$
such that what!?
$endgroup$
– Peyman mohseni kiasari
Dec 20 '18 at 22:13
$begingroup$
@Peyman Fixed. Thanks.
$endgroup$
– Rebellos
Dec 20 '18 at 22:14
add a comment |
$begingroup$
Hint :
Suppose that $T,T^2,dots, T^k$ are not linearly independent. Then, there exists $lambda_1, lambda_2, dots, lambda_k in mathbb R$ not all zero, such that :
$$lambda_1T + lambda_2T^2 + cdots + lambda_kT^k = 0$$
answered Dec 20 '18 at 22:13
RebellosRebellos
14.8k31248
$endgroup$
Hint :
Suppose that $T,T^2,dots, T^k$ are not linearly independent. Then, there exists $lambda_1, lambda_2, dots, lambda_k in mathbb R$ not all zero, such that :
$$lambda_1T + lambda_2T^2 + cdots + lambda_kT^k = 0$$
answered Dec 20 '18 at 22:13
RebellosRebellos
14.8k31248
edited Dec 20 '18 at 22:13
answered Dec 20 '18 at 22:13
RebellosRebellos
14.8k31248
answered Dec 20 '18 at 22:13
RebellosRebellos
14.8k31248
answered Dec 20 '18 at 22:13
RebellosRebellos
14.8k31248
14.8k31248
$begingroup$
such that what!?
$endgroup$
– Peyman mohseni kiasari
Dec 20 '18 at 22:13
$begingroup$
@Peyman Fixed. Thanks.
$endgroup$
– Rebellos
Dec 20 '18 at 22:14
add a comment |
$begingroup$
such that what!?
$endgroup$
– Peyman mohseni kiasari
Dec 20 '18 at 22:13
$begingroup$
@Peyman Fixed. Thanks.
$endgroup$
– Rebellos
Dec 20 '18 at 22:14
$begingroup$
such that what!?
$endgroup$
– Peyman mohseni kiasari
Dec 20 '18 at 22:13
$begingroup$
such that what!?
$endgroup$
– Peyman mohseni kiasari
Dec 20 '18 at 22:13
$begingroup$
@Peyman Fixed. Thanks.
$endgroup$
– Rebellos
Dec 20 '18 at 22:14
$begingroup$
@Peyman Fixed. Thanks.
$endgroup$
– Rebellos
Dec 20 '18 at 22:14
add a comment |
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$begingroup$
What is T? A matrix?
$endgroup$
– H_1317
Dec 23 '18 at 23:36
$begingroup$
@H_1317 yes, T is actually "T:V→V". unfortunately, I've asked the question wrong! it must be actually "T(v)s are dependent prove Ts are dependent". I mean both "independent"s should be "dependent". when I saw guys were answering the question I didn't edit that and just accept the answer.
$endgroup$
– Peyman mohseni kiasari
Dec 23 '18 at 23:47
$begingroup$
Can u edit your question identifying what T is, and the proper wording for any other errors please? Also what vector space u believe T is in (a matrix space?) thanks
$endgroup$
– H_1317
Dec 23 '18 at 23:49
$begingroup$
@H_1317, sorry if I'm not so familiar but I think "T:V→V" is a linear map no? where T must be a matrix, am I wrong? T is a map from V to V where V can be any vector space I think.
$endgroup$
– Peyman mohseni kiasari
Dec 23 '18 at 23:57
1
$begingroup$
If $a_i T^i = 0$ is a linear combination of the $T^i,$ choose some $v$ for which the $T^i(v)$ are linearly independent vectors (you need to assume that only one such $v$ exists, not that every $v$ (which, by the way cannot be) satisfies it) and then conclude all scalars are zero.
$endgroup$
– Will M.
Dec 24 '18 at 7:55