Mixing
A special construction in $mathbb{R}^n$
$begingroup$
I have a question from my professor' notes. We defined $Lambda^k(V)$ as the set of all $k$-Tensor' forms (multilinear transformations), $omega$, which fulfuill $omega(v_1,...,v_i,v_j,...,v_k)=-omega(v_1,...,v_j,v_i,...,v_k)$.
Let $v_1,...,v_{n-1}inmathbb{R}^n$. Let $varphi:mathbb{R}^ntomathbb{R}$ defined by $varphi(omega)=operatorname{det}left(begin{smallmatrix}
v_1\
vdots\
v_{n-1}\
omega
end{smallmatrix}right)$. We observe that $varphiinLambda^1(mathbb{R}^n)$, thus, there exsists a unique $zinmathbb{R}^n$ such that
$$\ left langle z,omega right rangle=varphi(omega)=operatorname{det}left(begin{smallmatrix} v_1\ vdots \ v_{n-1}\ omega end{smallmatrix}right)
$$
My question is why is there a uniuqe $z$ as above?
I know that $varphi(omega)$ is some number in $mathbb{R}$, so I have a lot of options to choose $z$. For example, if $varphi(omega)=3$ and $omega=(2,5,-24)$ and $z=(z_1,z_2,z_3)$, I can find a lot of options such that $2z_1+5z_2-24z_3=3$, so how is $z$ unique?
multivariable-calculus determinant tensor-products tensors multilinear-algebra
asked Jan 26 at 17:41
J. DoeJ. Doe
13713
$endgroup$
add a comment |
$begingroup$
I have a question from my professor' notes. We defined $Lambda^k(V)$ as the set of all $k$-Tensor' forms (multilinear transformations), $omega$, which fulfuill $omega(v_1,...,v_i,v_j,...,v_k)=-omega(v_1,...,v_j,v_i,...,v_k)$.
Let $v_1,...,v_{n-1}inmathbb{R}^n$. Let $varphi:mathbb{R}^ntomathbb{R}$ defined by $varphi(omega)=operatorname{det}left(begin{smallmatrix}
v_1\
vdots\
v_{n-1}\
omega
end{smallmatrix}right)$. We observe that $varphiinLambda^1(mathbb{R}^n)$, thus, there exsists a unique $zinmathbb{R}^n$ such that
$$\ left langle z,omega right rangle=varphi(omega)=operatorname{det}left(begin{smallmatrix} v_1\ vdots \ v_{n-1}\ omega end{smallmatrix}right)
$$
My question is why is there a uniuqe $z$ as above?
I know that $varphi(omega)$ is some number in $mathbb{R}$, so I have a lot of options to choose $z$. For example, if $varphi(omega)=3$ and $omega=(2,5,-24)$ and $z=(z_1,z_2,z_3)$, I can find a lot of options such that $2z_1+5z_2-24z_3=3$, so how is $z$ unique?
multivariable-calculus determinant tensor-products tensors multilinear-algebra
asked Jan 26 at 17:41
J. DoeJ. Doe
13713
$endgroup$
1
$begingroup$
That is $phi$ defined as function of $w$. So your formula should hold for any $w$. In that sense, $z$ is unique.
$endgroup$
– i707107
Jan 26 at 17:57
$begingroup$
Every element of $Lambda^1(Bbb R^n)$ is given by a linear transformation from $Bbb R^n$ to $Bbb R$. Every such is dot product with some (unique) vector in $Bbb R^n$. (Think about the matrix representation of the linear transformation.)
$endgroup$
– Ted Shifrin
Jan 26 at 22:31
add a comment |
$begingroup$
I have a question from my professor' notes. We defined $Lambda^k(V)$ as the set of all $k$-Tensor' forms (multilinear transformations), $omega$, which fulfuill $omega(v_1,...,v_i,v_j,...,v_k)=-omega(v_1,...,v_j,v_i,...,v_k)$.
Let $v_1,...,v_{n-1}inmathbb{R}^n$. Let $varphi:mathbb{R}^ntomathbb{R}$ defined by $varphi(omega)=operatorname{det}left(begin{smallmatrix}
v_1\
vdots\
v_{n-1}\
omega
end{smallmatrix}right)$. We observe that $varphiinLambda^1(mathbb{R}^n)$, thus, there exsists a unique $zinmathbb{R}^n$ such that
$$\ left langle z,omega right rangle=varphi(omega)=operatorname{det}left(begin{smallmatrix} v_1\ vdots \ v_{n-1}\ omega end{smallmatrix}right)
$$
My question is why is there a uniuqe $z$ as above?
I know that $varphi(omega)$ is some number in $mathbb{R}$, so I have a lot of options to choose $z$. For example, if $varphi(omega)=3$ and $omega=(2,5,-24)$ and $z=(z_1,z_2,z_3)$, I can find a lot of options such that $2z_1+5z_2-24z_3=3$, so how is $z$ unique?
multivariable-calculus determinant tensor-products tensors multilinear-algebra
asked Jan 26 at 17:41
J. DoeJ. Doe
13713
$endgroup$
I have a question from my professor' notes. We defined $Lambda^k(V)$ as the set of all $k$-Tensor' forms (multilinear transformations), $omega$, which fulfuill $omega(v_1,...,v_i,v_j,...,v_k)=-omega(v_1,...,v_j,v_i,...,v_k)$.
Let $v_1,...,v_{n-1}inmathbb{R}^n$. Let $varphi:mathbb{R}^ntomathbb{R}$ defined by $varphi(omega)=operatorname{det}left(begin{smallmatrix}
v_1\
vdots\
v_{n-1}\
omega
end{smallmatrix}right)$. We observe that $varphiinLambda^1(mathbb{R}^n)$, thus, there exsists a unique $zinmathbb{R}^n$ such that
$$\ left langle z,omega right rangle=varphi(omega)=operatorname{det}left(begin{smallmatrix} v_1\ vdots \ v_{n-1}\ omega end{smallmatrix}right)
$$
My question is why is there a uniuqe $z$ as above?
I know that $varphi(omega)$ is some number in $mathbb{R}$, so I have a lot of options to choose $z$. For example, if $varphi(omega)=3$ and $omega=(2,5,-24)$ and $z=(z_1,z_2,z_3)$, I can find a lot of options such that $2z_1+5z_2-24z_3=3$, so how is $z$ unique?
multivariable-calculus determinant tensor-products tensors multilinear-algebra
multivariable-calculus determinant tensor-products tensors multilinear-algebra
asked Jan 26 at 17:41
J. DoeJ. Doe
13713
asked Jan 26 at 17:41
J. DoeJ. Doe
13713
edited Jan 26 at 17:49
J. Doe
asked Jan 26 at 17:41
J. DoeJ. Doe
13713
asked Jan 26 at 17:41
J. DoeJ. Doe
13713
asked Jan 26 at 17:41
J. DoeJ. Doe
13713
13713
1
$begingroup$
That is $phi$ defined as function of $w$. So your formula should hold for any $w$. In that sense, $z$ is unique.
$endgroup$
– i707107
Jan 26 at 17:57
$begingroup$
Every element of $Lambda^1(Bbb R^n)$ is given by a linear transformation from $Bbb R^n$ to $Bbb R$. Every such is dot product with some (unique) vector in $Bbb R^n$. (Think about the matrix representation of the linear transformation.)
$endgroup$
– Ted Shifrin
Jan 26 at 22:31
add a comment |
1
$begingroup$
That is $phi$ defined as function of $w$. So your formula should hold for any $w$. In that sense, $z$ is unique.
$endgroup$
– i707107
Jan 26 at 17:57
$begingroup$
Every element of $Lambda^1(Bbb R^n)$ is given by a linear transformation from $Bbb R^n$ to $Bbb R$. Every such is dot product with some (unique) vector in $Bbb R^n$. (Think about the matrix representation of the linear transformation.)
$endgroup$
– Ted Shifrin
Jan 26 at 22:31
1
1
$begingroup$
That is $phi$ defined as function of $w$. So your formula should hold for any $w$. In that sense, $z$ is unique.
$endgroup$
– i707107
Jan 26 at 17:57
$begingroup$
That is $phi$ defined as function of $w$. So your formula should hold for any $w$. In that sense, $z$ is unique.
$endgroup$
– i707107
Jan 26 at 17:57
$begingroup$
Every element of $Lambda^1(Bbb R^n)$ is given by a linear transformation from $Bbb R^n$ to $Bbb R$. Every such is dot product with some (unique) vector in $Bbb R^n$. (Think about the matrix representation of the linear transformation.)
$endgroup$
– Ted Shifrin
Jan 26 at 22:31
$begingroup$
Every element of $Lambda^1(Bbb R^n)$ is given by a linear transformation from $Bbb R^n$ to $Bbb R$. Every such is dot product with some (unique) vector in $Bbb R^n$. (Think about the matrix representation of the linear transformation.)
$endgroup$
– Ted Shifrin
Jan 26 at 22:31
add a comment |
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1
$begingroup$
That is $phi$ defined as function of $w$. So your formula should hold for any $w$. In that sense, $z$ is unique.
$endgroup$
– i707107
Jan 26 at 17:57
$begingroup$
Every element of $Lambda^1(Bbb R^n)$ is given by a linear transformation from $Bbb R^n$ to $Bbb R$. Every such is dot product with some (unique) vector in $Bbb R^n$. (Think about the matrix representation of the linear transformation.)
$endgroup$
– Ted Shifrin
Jan 26 at 22:31