What is the n-fold composition of a linear map “T”, if n is zero?
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I am doing an assignment on Groups of Linear Transformations. Here is a defnition with regards to n-fold compositions:
"Let $T in mathcal{L}(V)$ denote the n-fold compositions of $T$ with itself as $T^{n}$. For instance, the two-fold composition, $Tcirc T$ is denoted $T^{2}$. Let $T^{-n}$ denote the n-fold composition of $T^{-1}$."
So, with the above definition in mind, $T^{1}$ is just T, but what would $T^{0}$ yeild?
linear-algebra group-theory linear-transformations function-and-relation-composition
asked Dec 1 '18 at 5:42
JaigusJaigus
2218
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add a comment |
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I am doing an assignment on Groups of Linear Transformations. Here is a defnition with regards to n-fold compositions:
"Let $T in mathcal{L}(V)$ denote the n-fold compositions of $T$ with itself as $T^{n}$. For instance, the two-fold composition, $Tcirc T$ is denoted $T^{2}$. Let $T^{-n}$ denote the n-fold composition of $T^{-1}$."
So, with the above definition in mind, $T^{1}$ is just T, but what would $T^{0}$ yeild?
linear-algebra group-theory linear-transformations function-and-relation-composition
asked Dec 1 '18 at 5:42
JaigusJaigus
2218
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1
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Identity linear function.
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– Will M.
Dec 1 '18 at 5:46
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@WillM. Thanks but can you explain how/why? Or is this just the way it is defined?
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– Jaigus
Dec 1 '18 at 5:50
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If $T$ were invertible and you wanted the relation $T^{p+q}=T^p circ T^q$ to hold for all integers $p$ and $q,$ then $T^0$ has to be the identity.
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– Will M.
Dec 1 '18 at 5:54
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@WillM. Thanks. If you write this as an answer I can accept it...
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– Jaigus
Dec 1 '18 at 5:55
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You are welcome. I am OK like this. I'm lazy to write it down, :P
$endgroup$
– Will M.
Dec 1 '18 at 5:59
add a comment |
$begingroup$
I am doing an assignment on Groups of Linear Transformations. Here is a defnition with regards to n-fold compositions:
"Let $T in mathcal{L}(V)$ denote the n-fold compositions of $T$ with itself as $T^{n}$. For instance, the two-fold composition, $Tcirc T$ is denoted $T^{2}$. Let $T^{-n}$ denote the n-fold composition of $T^{-1}$."
So, with the above definition in mind, $T^{1}$ is just T, but what would $T^{0}$ yeild?
linear-algebra group-theory linear-transformations function-and-relation-composition
asked Dec 1 '18 at 5:42
JaigusJaigus
2218
$endgroup$
I am doing an assignment on Groups of Linear Transformations. Here is a defnition with regards to n-fold compositions:
"Let $T in mathcal{L}(V)$ denote the n-fold compositions of $T$ with itself as $T^{n}$. For instance, the two-fold composition, $Tcirc T$ is denoted $T^{2}$. Let $T^{-n}$ denote the n-fold composition of $T^{-1}$."
So, with the above definition in mind, $T^{1}$ is just T, but what would $T^{0}$ yeild?
linear-algebra group-theory linear-transformations function-and-relation-composition
linear-algebra group-theory linear-transformations function-and-relation-composition
asked Dec 1 '18 at 5:42
JaigusJaigus
2218
asked Dec 1 '18 at 5:42
JaigusJaigus
2218
asked Dec 1 '18 at 5:42
JaigusJaigus
2218
asked Dec 1 '18 at 5:42
JaigusJaigus
2218
asked Dec 1 '18 at 5:42
JaigusJaigus
2218
2218
1
$begingroup$
Identity linear function.
$endgroup$
– Will M.
Dec 1 '18 at 5:46
$begingroup$
@WillM. Thanks but can you explain how/why? Or is this just the way it is defined?
$endgroup$
– Jaigus
Dec 1 '18 at 5:50
$begingroup$
If $T$ were invertible and you wanted the relation $T^{p+q}=T^p circ T^q$ to hold for all integers $p$ and $q,$ then $T^0$ has to be the identity.
$endgroup$
– Will M.
Dec 1 '18 at 5:54
$begingroup$
@WillM. Thanks. If you write this as an answer I can accept it...
$endgroup$
– Jaigus
Dec 1 '18 at 5:55
$begingroup$
You are welcome. I am OK like this. I'm lazy to write it down, :P
$endgroup$
– Will M.
Dec 1 '18 at 5:59
add a comment |
1
$begingroup$
Identity linear function.
$endgroup$
– Will M.
Dec 1 '18 at 5:46
$begingroup$
@WillM. Thanks but can you explain how/why? Or is this just the way it is defined?
$endgroup$
– Jaigus
Dec 1 '18 at 5:50
$begingroup$
If $T$ were invertible and you wanted the relation $T^{p+q}=T^p circ T^q$ to hold for all integers $p$ and $q,$ then $T^0$ has to be the identity.
$endgroup$
– Will M.
Dec 1 '18 at 5:54
$begingroup$
@WillM. Thanks. If you write this as an answer I can accept it...
$endgroup$
– Jaigus
Dec 1 '18 at 5:55
$begingroup$
You are welcome. I am OK like this. I'm lazy to write it down, :P
$endgroup$
– Will M.
Dec 1 '18 at 5:59
1
1
$begingroup$
Identity linear function.
$endgroup$
– Will M.
Dec 1 '18 at 5:46
$begingroup$
Identity linear function.
$endgroup$
– Will M.
Dec 1 '18 at 5:46
$begingroup$
@WillM. Thanks but can you explain how/why? Or is this just the way it is defined?
$endgroup$
– Jaigus
Dec 1 '18 at 5:50
$begingroup$
@WillM. Thanks but can you explain how/why? Or is this just the way it is defined?
$endgroup$
– Jaigus
Dec 1 '18 at 5:50
$begingroup$
If $T$ were invertible and you wanted the relation $T^{p+q}=T^p circ T^q$ to hold for all integers $p$ and $q,$ then $T^0$ has to be the identity.
$endgroup$
– Will M.
Dec 1 '18 at 5:54
$begingroup$
If $T$ were invertible and you wanted the relation $T^{p+q}=T^p circ T^q$ to hold for all integers $p$ and $q,$ then $T^0$ has to be the identity.
$endgroup$
– Will M.
Dec 1 '18 at 5:54
$begingroup$
@WillM. Thanks. If you write this as an answer I can accept it...
$endgroup$
– Jaigus
Dec 1 '18 at 5:55
$begingroup$
@WillM. Thanks. If you write this as an answer I can accept it...
$endgroup$
– Jaigus
Dec 1 '18 at 5:55
$begingroup$
You are welcome. I am OK like this. I'm lazy to write it down, :P
$endgroup$
– Will M.
Dec 1 '18 at 5:59
$begingroup$
You are welcome. I am OK like this. I'm lazy to write it down, :P
$endgroup$
– Will M.
Dec 1 '18 at 5:59
add a comment |
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1
$begingroup$
Identity linear function.
$endgroup$
– Will M.
Dec 1 '18 at 5:46
$begingroup$
@WillM. Thanks but can you explain how/why? Or is this just the way it is defined?
$endgroup$
– Jaigus
Dec 1 '18 at 5:50
$begingroup$
If $T$ were invertible and you wanted the relation $T^{p+q}=T^p circ T^q$ to hold for all integers $p$ and $q,$ then $T^0$ has to be the identity.
$endgroup$
– Will M.
Dec 1 '18 at 5:54
$begingroup$
@WillM. Thanks. If you write this as an answer I can accept it...
$endgroup$
– Jaigus
Dec 1 '18 at 5:55
$begingroup$
You are welcome. I am OK like this. I'm lazy to write it down, :P
$endgroup$
– Will M.
Dec 1 '18 at 5:59