1
$\begingroup$

On this excellent post, the following can be found:

... we're used to considering vectors as column vectors, and dual vectors as row vectors. So, when we write something like $$u^\top A v,$$ our notation suggests that $u^\top \in T^1(V)$ is a dual vector and that $v \in T_1(V)$ is a vector. This means that the bilinear map $V \times V^* \to \mathbb{R}$ given by $$(v, u^\top) \mapsto u^\top A v$$ is a type $(1,1)$-tensor.

Elements of $V^*$, or covectors, are linear functions or functionals. As such I can picture them as a matrix, because a matrix exerts a transformation on vectors (or elements of $V$). So it would have been less surprising if the writer had tried to make the claim that the matrix $A$ is an element of $V^*$.

Instead it is the row vector $u^\top$ that is a dual vector. Why $u^\top$ and not $A$?

Further, and it may be a notation problem, I thought that $T^0_1 V\equiv V^*$ in finite vector spaces, rather than $T^1(V).$


Updated question after the answer and comments at the time of the writing:

Can I interpret that since $V^*$ is the function

$v=\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}\in \underbrace{\quad V\quad}_{n\text{-dim. space}} \overbrace{{\Large{\longrightarrow}}}^{\Large\color{red}{V^*}} \underbrace{\quad \mathbb R\quad}_{1-\text{dim}}$

from the point of view of matching dimensions, it would make sense to picture it as row vector in the dot product:

$\underbrace{\color{red}{\begin{bmatrix}x^*_1,x^*_2,\cdots,x^*_d \end{bmatrix}}}_{[1\times d]}\cdot\underbrace{\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}}_{[d\times 1]}=[1\times 1]$

?

So what is the role of $A$? And, again, can I get some insight as to the $T$ script notation?

$\endgroup$
  • 1
    $\begingroup$ For the purposes of understanding row vectors as covectors, delete $A$ from the equations. Then $u^T$ is a matrix; it's a $1 \times n$ matrix, where $n = \dim V$. $\endgroup$ – Qiaochu Yuan Feb 25 '17 at 20:49
1
$\begingroup$

Note that $V^*$ is the space of all functions from $V \to \mathbb{R}$. Thus you are correct, you can represent elements of $V^*$ as matrices. These matrices are from a $n$ dimensional space to a $1$ dimensional space, and are thus $1 \times n$ matrices.

In your equation above, you can therefore think of $(v,u) \in V \otimes V^*$ mapping to $u(Av)$.

$\endgroup$
  • $\begingroup$ Can you explain a bit the role of $A$ regarding the dimensions, what $V\otimes V^*$ stands for, and the issues in the OP re: $T^i$ v $T^0_i$ notation? $\endgroup$ – Antoni Parellada Feb 25 '17 at 22:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.