Mixing
Easiest way to prove that the operator is zero
$begingroup$
If an operator $T in L(V)$ satisfies $Tv = 0$ for every $v in V$, then $T = 0$.
By the way, $V$ is a finite dimensional vector space.
Though, I am pretty certain that the statement is true, but I haven't proved it. Is there any way to prove that statement in an easy way?
(What I thought was, if we can represent that T in a basis of V, then if we put vectors $(1,0,...0), (0,1,...0), ... (0,0,...1)$, it is easy to show that all the entries of the matrix is zero which makes the operator $T$ an zero operator. However, I just wonder is there any shorter way to prove it)
linear-algebra matrices
asked Mar 19 '14 at 2:31
JaebumJaebum
1063
$endgroup$
add a comment |
$begingroup$
If an operator $T in L(V)$ satisfies $Tv = 0$ for every $v in V$, then $T = 0$.
By the way, $V$ is a finite dimensional vector space.
Though, I am pretty certain that the statement is true, but I haven't proved it. Is there any way to prove that statement in an easy way?
(What I thought was, if we can represent that T in a basis of V, then if we put vectors $(1,0,...0), (0,1,...0), ... (0,0,...1)$, it is easy to show that all the entries of the matrix is zero which makes the operator $T$ an zero operator. However, I just wonder is there any shorter way to prove it)
linear-algebra matrices
asked Mar 19 '14 at 2:31
JaebumJaebum
1063
$endgroup$
$begingroup$
How about using Contradiction? It works on the same idea but backwards. Assume that the operator is nonzero and design a $v$ such that $Tvneq0$?
$endgroup$
– Nitish
Mar 19 '14 at 2:36
$begingroup$
@Nitish I thought about that but I dont know how to come up a vector that $Tv neq 0$.
$endgroup$
– Jaebum
Mar 19 '14 at 2:39
$begingroup$
Is it a CIRCULAR question ?.
$endgroup$
– Felix Marin
Mar 19 '14 at 2:43
1
$begingroup$
What is the definition of two operators being equal? Check your textbook.
$endgroup$
– Omnomnomnom
Mar 19 '14 at 2:45
add a comment |
$begingroup$
If an operator $T in L(V)$ satisfies $Tv = 0$ for every $v in V$, then $T = 0$.
By the way, $V$ is a finite dimensional vector space.
Though, I am pretty certain that the statement is true, but I haven't proved it. Is there any way to prove that statement in an easy way?
(What I thought was, if we can represent that T in a basis of V, then if we put vectors $(1,0,...0), (0,1,...0), ... (0,0,...1)$, it is easy to show that all the entries of the matrix is zero which makes the operator $T$ an zero operator. However, I just wonder is there any shorter way to prove it)
linear-algebra matrices
asked Mar 19 '14 at 2:31
JaebumJaebum
1063
$endgroup$
If an operator $T in L(V)$ satisfies $Tv = 0$ for every $v in V$, then $T = 0$.
By the way, $V$ is a finite dimensional vector space.
Though, I am pretty certain that the statement is true, but I haven't proved it. Is there any way to prove that statement in an easy way?
(What I thought was, if we can represent that T in a basis of V, then if we put vectors $(1,0,...0), (0,1,...0), ... (0,0,...1)$, it is easy to show that all the entries of the matrix is zero which makes the operator $T$ an zero operator. However, I just wonder is there any shorter way to prove it)
linear-algebra matrices
linear-algebra matrices
asked Mar 19 '14 at 2:31
JaebumJaebum
1063
asked Mar 19 '14 at 2:31
JaebumJaebum
1063
asked Mar 19 '14 at 2:31
JaebumJaebum
1063
asked Mar 19 '14 at 2:31
JaebumJaebum
1063
asked Mar 19 '14 at 2:31
JaebumJaebum
1063
1063
$begingroup$
How about using Contradiction? It works on the same idea but backwards. Assume that the operator is nonzero and design a $v$ such that $Tvneq0$?
$endgroup$
– Nitish
Mar 19 '14 at 2:36
$begingroup$
@Nitish I thought about that but I dont know how to come up a vector that $Tv neq 0$.
$endgroup$
– Jaebum
Mar 19 '14 at 2:39
$begingroup$
Is it a CIRCULAR question ?.
$endgroup$
– Felix Marin
Mar 19 '14 at 2:43
1
$begingroup$
What is the definition of two operators being equal? Check your textbook.
$endgroup$
– Omnomnomnom
Mar 19 '14 at 2:45
add a comment |
$begingroup$
How about using Contradiction? It works on the same idea but backwards. Assume that the operator is nonzero and design a $v$ such that $Tvneq0$?
$endgroup$
– Nitish
Mar 19 '14 at 2:36
$begingroup$
@Nitish I thought about that but I dont know how to come up a vector that $Tv neq 0$.
$endgroup$
– Jaebum
Mar 19 '14 at 2:39
$begingroup$
Is it a CIRCULAR question ?.
$endgroup$
– Felix Marin
Mar 19 '14 at 2:43
1
$begingroup$
What is the definition of two operators being equal? Check your textbook.
$endgroup$
– Omnomnomnom
Mar 19 '14 at 2:45
$begingroup$
How about using Contradiction? It works on the same idea but backwards. Assume that the operator is nonzero and design a $v$ such that $Tvneq0$?
$endgroup$
– Nitish
Mar 19 '14 at 2:36
$begingroup$
How about using Contradiction? It works on the same idea but backwards. Assume that the operator is nonzero and design a $v$ such that $Tvneq0$?
$endgroup$
– Nitish
Mar 19 '14 at 2:36
$begingroup$
@Nitish I thought about that but I dont know how to come up a vector that $Tv neq 0$.
$endgroup$
– Jaebum
Mar 19 '14 at 2:39
$begingroup$
@Nitish I thought about that but I dont know how to come up a vector that $Tv neq 0$.
$endgroup$
– Jaebum
Mar 19 '14 at 2:39
$begingroup$
Is it a CIRCULAR question ?.
$endgroup$
– Felix Marin
Mar 19 '14 at 2:43
$begingroup$
Is it a CIRCULAR question ?.
$endgroup$
– Felix Marin
Mar 19 '14 at 2:43
1
1
$begingroup$
What is the definition of two operators being equal? Check your textbook.
$endgroup$
– Omnomnomnom
Mar 19 '14 at 2:45
$begingroup$
What is the definition of two operators being equal? Check your textbook.
$endgroup$
– Omnomnomnom
Mar 19 '14 at 2:45
add a comment |
1 Answer
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$begingroup$
"$T=0$" means "$Tv=0$ for all $vin V$". There is nothing to prove.
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
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$begingroup$
"$T=0$" means "$Tv=0$ for all $vin V$". There is nothing to prove.
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
"$T=0$" means "$Tv=0$ for all $vin V$". There is nothing to prove.
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
"$T=0$" means "$Tv=0$ for all $vin V$". There is nothing to prove.
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
"$T=0$" means "$Tv=0$ for all $vin V$". There is nothing to prove.
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
127k1182183
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
127k1182183
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
127k1182183
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
127k1182183
127k1182183
add a comment |
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$begingroup$
How about using Contradiction? It works on the same idea but backwards. Assume that the operator is nonzero and design a $v$ such that $Tvneq0$?
$endgroup$
– Nitish
Mar 19 '14 at 2:36
$begingroup$
@Nitish I thought about that but I dont know how to come up a vector that $Tv neq 0$.
$endgroup$
– Jaebum
Mar 19 '14 at 2:39
$begingroup$
Is it a CIRCULAR question ?.
$endgroup$
– Felix Marin
Mar 19 '14 at 2:43
1
$begingroup$
What is the definition of two operators being equal? Check your textbook.
$endgroup$
– Omnomnomnom
Mar 19 '14 at 2:45