Analogy of covectors
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Given that the bases of a vector and their duals satisfy:
$$epsilon^ie_j=delta^i_j$$
in which $epsilon$ and $e$ are the dual bases and bases respectively, and $delta^i_j$ is the Kronecker symbol, and that the definition of the dual vector space is:
$$V^*=varphi:V longrightarrow mathbbR$$
with $V$ being a vector space. Is it "correct" to think of the covector and vector as analogous to row and column vectors respectively ? I came to this conclusion due to the fact that the matrix product of a row vector and a column vector is a scalar, and $varphi$ also takes vectors to numbers, but I have a feeling that I'm still missing something.
Thanks in advance!
P/s: I also find this way of thinking of vectors and covectors very nice since row and column vectors are sort of doppelgängers, hence the covectors and dual spaces
multilinear-algebra
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up vote
1
down vote
favorite
Given that the bases of a vector and their duals satisfy:
$$epsilon^ie_j=delta^i_j$$
in which $epsilon$ and $e$ are the dual bases and bases respectively, and $delta^i_j$ is the Kronecker symbol, and that the definition of the dual vector space is:
$$V^*=varphi:V longrightarrow mathbbR$$
with $V$ being a vector space. Is it "correct" to think of the covector and vector as analogous to row and column vectors respectively ? I came to this conclusion due to the fact that the matrix product of a row vector and a column vector is a scalar, and $varphi$ also takes vectors to numbers, but I have a feeling that I'm still missing something.
Thanks in advance!
P/s: I also find this way of thinking of vectors and covectors very nice since row and column vectors are sort of doppelgängers, hence the covectors and dual spaces
multilinear-algebra
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given that the bases of a vector and their duals satisfy:
$$epsilon^ie_j=delta^i_j$$
in which $epsilon$ and $e$ are the dual bases and bases respectively, and $delta^i_j$ is the Kronecker symbol, and that the definition of the dual vector space is:
$$V^*=varphi:V longrightarrow mathbbR$$
with $V$ being a vector space. Is it "correct" to think of the covector and vector as analogous to row and column vectors respectively ? I came to this conclusion due to the fact that the matrix product of a row vector and a column vector is a scalar, and $varphi$ also takes vectors to numbers, but I have a feeling that I'm still missing something.
Thanks in advance!
P/s: I also find this way of thinking of vectors and covectors very nice since row and column vectors are sort of doppelgängers, hence the covectors and dual spaces
multilinear-algebra
Given that the bases of a vector and their duals satisfy:
$$epsilon^ie_j=delta^i_j$$
in which $epsilon$ and $e$ are the dual bases and bases respectively, and $delta^i_j$ is the Kronecker symbol, and that the definition of the dual vector space is:
$$V^*=varphi:V longrightarrow mathbbR$$
with $V$ being a vector space. Is it "correct" to think of the covector and vector as analogous to row and column vectors respectively ? I came to this conclusion due to the fact that the matrix product of a row vector and a column vector is a scalar, and $varphi$ also takes vectors to numbers, but I have a feeling that I'm still missing something.
Thanks in advance!
P/s: I also find this way of thinking of vectors and covectors very nice since row and column vectors are sort of doppelgängers, hence the covectors and dual spaces
multilinear-algebra
edited Aug 21 at 16:39
asked Apr 3 at 17:02
user518704
336
336
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1 Answer
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It is partially correct, or at least meaningful. Indeed, given $v in V$ and $w in V^*$, if you choose dual bases, you have that $w(v)=w^Tv$ (check this). But note that I am considering all vectors as vertical and then transposing $w$, since in principle $V^*$ is not a special vector space, so that it is not very precise to write its vectors in a different way.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It is partially correct, or at least meaningful. Indeed, given $v in V$ and $w in V^*$, if you choose dual bases, you have that $w(v)=w^Tv$ (check this). But note that I am considering all vectors as vertical and then transposing $w$, since in principle $V^*$ is not a special vector space, so that it is not very precise to write its vectors in a different way.
add a comment |Â
up vote
2
down vote
accepted
It is partially correct, or at least meaningful. Indeed, given $v in V$ and $w in V^*$, if you choose dual bases, you have that $w(v)=w^Tv$ (check this). But note that I am considering all vectors as vertical and then transposing $w$, since in principle $V^*$ is not a special vector space, so that it is not very precise to write its vectors in a different way.
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It is partially correct, or at least meaningful. Indeed, given $v in V$ and $w in V^*$, if you choose dual bases, you have that $w(v)=w^Tv$ (check this). But note that I am considering all vectors as vertical and then transposing $w$, since in principle $V^*$ is not a special vector space, so that it is not very precise to write its vectors in a different way.
It is partially correct, or at least meaningful. Indeed, given $v in V$ and $w in V^*$, if you choose dual bases, you have that $w(v)=w^Tv$ (check this). But note that I am considering all vectors as vertical and then transposing $w$, since in principle $V^*$ is not a special vector space, so that it is not very precise to write its vectors in a different way.
answered Apr 3 at 17:40
57Jimmy
3,132421
3,132421
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