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?

  • 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

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 $u(Av)$.

  • $\begingroup$ 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

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