Is my understanding of an annihilator correct?
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This is how I understand the annihilator now, but I feel like it might be incorrect.
So for some $U subset V$, the annihilator of $U$ is all of the linear functionals $t(v)$ in $V'$, such that $t(u)=0$ for $u in U$.
So in other words, I understand it as the following: You pick the subspace U that you want to annihilate, then the annihilator of U is simply the set of linear functions that map each vector in U to zero?
Notation: $V'$ is the set of all linear maps that map vectors in $V$ from $V$ to $F$
linear-algebra vector-spaces
asked Mar 5 '14 at 7:50
user123429user123429
14911
$endgroup$
add a comment |
$begingroup$
This is how I understand the annihilator now, but I feel like it might be incorrect.
So for some $U subset V$, the annihilator of $U$ is all of the linear functionals $t(v)$ in $V'$, such that $t(u)=0$ for $u in U$.
So in other words, I understand it as the following: You pick the subspace U that you want to annihilate, then the annihilator of U is simply the set of linear functions that map each vector in U to zero?
Notation: $V'$ is the set of all linear maps that map vectors in $V$ from $V$ to $F$
linear-algebra vector-spaces
asked Mar 5 '14 at 7:50
user123429user123429
14911
$endgroup$
2
$begingroup$
Yes. The annihilator of $U$ can be viewed as a substitute for the more familiar orthogonal complement of $U$, in the case where $V$ is not an inner product space. If $V$ is an inner product space, then $V'$ can be naturally identified with $V$, and the annihilator of $U$ is identified with the orthogonal complement of $U$. Using the notation $t(u) = langle t, u rangle$, the annihilator of $U$ consists of all $t in V'$ such that $langle t, u rangle = 0$ for all $u in U$.
$endgroup$
– littleO
Mar 5 '14 at 7:54
add a comment |
$begingroup$
This is how I understand the annihilator now, but I feel like it might be incorrect.
So for some $U subset V$, the annihilator of $U$ is all of the linear functionals $t(v)$ in $V'$, such that $t(u)=0$ for $u in U$.
So in other words, I understand it as the following: You pick the subspace U that you want to annihilate, then the annihilator of U is simply the set of linear functions that map each vector in U to zero?
Notation: $V'$ is the set of all linear maps that map vectors in $V$ from $V$ to $F$
linear-algebra vector-spaces
asked Mar 5 '14 at 7:50
user123429user123429
14911
$endgroup$
This is how I understand the annihilator now, but I feel like it might be incorrect.
So for some $U subset V$, the annihilator of $U$ is all of the linear functionals $t(v)$ in $V'$, such that $t(u)=0$ for $u in U$.
So in other words, I understand it as the following: You pick the subspace U that you want to annihilate, then the annihilator of U is simply the set of linear functions that map each vector in U to zero?
Notation: $V'$ is the set of all linear maps that map vectors in $V$ from $V$ to $F$
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Mar 5 '14 at 7:50
user123429user123429
14911
asked Mar 5 '14 at 7:50
user123429user123429
14911
asked Mar 5 '14 at 7:50
user123429user123429
14911
asked Mar 5 '14 at 7:50
user123429user123429
14911
asked Mar 5 '14 at 7:50
user123429user123429
14911
14911
2
$begingroup$
Yes. The annihilator of $U$ can be viewed as a substitute for the more familiar orthogonal complement of $U$, in the case where $V$ is not an inner product space. If $V$ is an inner product space, then $V'$ can be naturally identified with $V$, and the annihilator of $U$ is identified with the orthogonal complement of $U$. Using the notation $t(u) = langle t, u rangle$, the annihilator of $U$ consists of all $t in V'$ such that $langle t, u rangle = 0$ for all $u in U$.
$endgroup$
– littleO
Mar 5 '14 at 7:54
add a comment |
2
$begingroup$
Yes. The annihilator of $U$ can be viewed as a substitute for the more familiar orthogonal complement of $U$, in the case where $V$ is not an inner product space. If $V$ is an inner product space, then $V'$ can be naturally identified with $V$, and the annihilator of $U$ is identified with the orthogonal complement of $U$. Using the notation $t(u) = langle t, u rangle$, the annihilator of $U$ consists of all $t in V'$ such that $langle t, u rangle = 0$ for all $u in U$.
$endgroup$
– littleO
Mar 5 '14 at 7:54
2
2
$begingroup$
Yes. The annihilator of $U$ can be viewed as a substitute for the more familiar orthogonal complement of $U$, in the case where $V$ is not an inner product space. If $V$ is an inner product space, then $V'$ can be naturally identified with $V$, and the annihilator of $U$ is identified with the orthogonal complement of $U$. Using the notation $t(u) = langle t, u rangle$, the annihilator of $U$ consists of all $t in V'$ such that $langle t, u rangle = 0$ for all $u in U$.
$endgroup$
– littleO
Mar 5 '14 at 7:54
$begingroup$
Yes. The annihilator of $U$ can be viewed as a substitute for the more familiar orthogonal complement of $U$, in the case where $V$ is not an inner product space. If $V$ is an inner product space, then $V'$ can be naturally identified with $V$, and the annihilator of $U$ is identified with the orthogonal complement of $U$. Using the notation $t(u) = langle t, u rangle$, the annihilator of $U$ consists of all $t in V'$ such that $langle t, u rangle = 0$ for all $u in U$.
$endgroup$
– littleO
Mar 5 '14 at 7:54
add a comment |
1 Answer
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$begingroup$
From the comments above by @littleO.
Yes. The annihilator of $U$ can be viewed as a substitute for the more familiar orthogonal complement of $U$, in the case where $V$ is not an inner product space. If $V$ is an inner product space, then $V'$ can be naturally identified with $V$, and the annihilator of $U$ is identified with the orthogonal complement of $U$. Using the notation $t(u)=langle t,urangle$, the annihilator of $U$ consists of all $t in V'$ such that $langle t,u rangle =0$ for all $u in U$.
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add a comment |
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$begingroup$
From the comments above by @littleO.
Yes. The annihilator of $U$ can be viewed as a substitute for the more familiar orthogonal complement of $U$, in the case where $V$ is not an inner product space. If $V$ is an inner product space, then $V'$ can be naturally identified with $V$, and the annihilator of $U$ is identified with the orthogonal complement of $U$. Using the notation $t(u)=langle t,urangle$, the annihilator of $U$ consists of all $t in V'$ such that $langle t,u rangle =0$ for all $u in U$.
$endgroup$
add a comment |
$begingroup$
From the comments above by @littleO.
Yes. The annihilator of $U$ can be viewed as a substitute for the more familiar orthogonal complement of $U$, in the case where $V$ is not an inner product space. If $V$ is an inner product space, then $V'$ can be naturally identified with $V$, and the annihilator of $U$ is identified with the orthogonal complement of $U$. Using the notation $t(u)=langle t,urangle$, the annihilator of $U$ consists of all $t in V'$ such that $langle t,u rangle =0$ for all $u in U$.
$endgroup$
add a comment |
$begingroup$
From the comments above by @littleO.
Yes. The annihilator of $U$ can be viewed as a substitute for the more familiar orthogonal complement of $U$, in the case where $V$ is not an inner product space. If $V$ is an inner product space, then $V'$ can be naturally identified with $V$, and the annihilator of $U$ is identified with the orthogonal complement of $U$. Using the notation $t(u)=langle t,urangle$, the annihilator of $U$ consists of all $t in V'$ such that $langle t,u rangle =0$ for all $u in U$.
$endgroup$
From the comments above by @littleO.
Yes. The annihilator of $U$ can be viewed as a substitute for the more familiar orthogonal complement of $U$, in the case where $V$ is not an inner product space. If $V$ is an inner product space, then $V'$ can be naturally identified with $V$, and the annihilator of $U$ is identified with the orthogonal complement of $U$. Using the notation $t(u)=langle t,urangle$, the annihilator of $U$ consists of all $t in V'$ such that $langle t,u rangle =0$ for all $u in U$.
answered Dec 6 '18 at 13:58
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Brahadeesh
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$begingroup$
Yes. The annihilator of $U$ can be viewed as a substitute for the more familiar orthogonal complement of $U$, in the case where $V$ is not an inner product space. If $V$ is an inner product space, then $V'$ can be naturally identified with $V$, and the annihilator of $U$ is identified with the orthogonal complement of $U$. Using the notation $t(u) = langle t, u rangle$, the annihilator of $U$ consists of all $t in V'$ such that $langle t, u rangle = 0$ for all $u in U$.
$endgroup$
– littleO
Mar 5 '14 at 7:54