Showing that $\text{tr}(u \otimes v) = u \cdot v$ I am currently working on the following problem from P. Chadwick's Continuum Mechanics and was wondering if my appraoch is correct (and if it is, if it is rigorous enough):

Let $u,v \in \mathbb{R}^3$. Show that the trace of a $(1,1)$ tensor $T = u \otimes v$ gives the scaler $u \cdot v \in \mathbb{R},$ i.e., $\text{tr}(u \otimes v) = u \cdot v$.

A few things to note before my approach is that in my course we are only considering second-order tensors, $\cdot$ is the usual dot product on $ \mathbb{R}^3$ ( i.e., $u \cdot v = u_1v_1+u_2v_2+u_3v_3$), and $u \otimes v := \begin{bmatrix}
    u_1 v_1 & u_1 v_2 & u_1 v_3 \\
    u_2 v_1 & u_2 v_2 & u_2 v_3 \\
    u_3 v_1 & u_3 v_2 & u_3 v_3 \\
  \end{bmatrix}.$ (our version of the tensor product: the outer product).
I feel like this problem is really easy, but my textbook's solution is using the scaler triple product to show this. Here is what I have written so far:
Since we are given what $u \cdot v$ is, we can expand this expression as $u_1v_1+u_2v_2+u_3v_3$. We also know that each entry of the matrix $u \otimes v$ is of the form $u_{i}v_{j}$, thus since the trace of the matrix $u \otimes v$ is the sum of the diagonal elements we have that $$\text{tr}(u \otimes v) = u_1v_1+u_2v_2+u_3v_3=u \cdot v.$$ Is this enough for this proof? Or is there a more rigorous way of showing this?
 A: The trace of a type $(1,1)$ tensor $T^i_j\varepsilon^j\otimes e_i$ is by definition the contraction $T^i_i$. I interpret the question as "Show that $T^i_i$ is a scalar". This is also true by definition since the quantities obtained by contraction of a type $(r,s)$-tensor constitute the components of a tensor of type $(r-1,s-1)$. But we can at least verify it for clarity.
To verify that the trace is a scalar we can use the trick "If it looks like a tensor, swims like a tensor, and quacks like a tensor, then it is a tensor."
The components of a type $(1,1)$-tensor transform like
$$\bar{T}^u_w=\frac{\partial \bar{x}^u}{\partial x^i}\frac{\partial x^j}{\partial \bar{x}^w}T^i_j$$
The trace $\bar{T}^u_u$ is therefore (let $w=u$)
$$\bar{T}^u_u=\frac{\partial \bar{x}^u}{\partial x^i}\frac{\partial x^j}{\partial \bar{x}^u}T^i_j=\delta^j_iT^i_j=T^i_i$$
So the trace is indeed a $(0,0)$-tensor, ie a scalar. Sometimes referred to as an invariant (it is invariant with respect to a change of basis). Notice that this concept of a scalar might be different from what you are used to in linear algebra where pretty much any real number is considered a "scalar".
Now with the components $T^i_j=u^iv_j$ in 3 dimensions obviously $T^k_k=u^1v_1+u^2 v_2+u^3v_3$ which artificially might be interpreted as $\mathbf{u}\cdot \mathbf{v}=g_{ij}u^iv^j$.
A: @ContraKinta's answer using tensor transformation properties and contraction is excellent. I shall try to present an alternate approach sticking only to linear algebra that attempts to formalize OP's initial proof.
The outer product of two vectors $u$ and $v$ in $\mathbb{R^n}$ is
$$ u \otimes v = uv^T $$
The Trace of a matrix $A_{m \times m}$ is defined as,
$$ \text{Tr}(A) = \sum_{i=1}^{m} a_{ii} $$
For us to take the trace of the outer product $ \text{Tr}(uv^T) $, $uv^T$ must be a square matrix which is only possible if $u$ and $v$ are both of the same dimension, which is the case here as both $u, v \in \mathbb{R}^n$.
Thus, $uv^T$ is a square matrix with dimensions $n \times n$ and its trace is,
$$ \text{Tr}(uv^T) = \sum_{i=1}^{n} (uv^T)_{ii} \tag{1} $$
By definition of the outer product,
$$ (uv^T)_{ij} = u_{i}v_{j}$$
We require the elements where $i = j$,
$$ (uv^T)_{ii} = u_{i}v_{i} \tag{2} $$
$ (2) $ in $ (1) $,
$$ \text{Tr}(uv^T) = \sum_{i=1}^{n} u_{i}v_{i} $$
Which by definition is the dot product. Therefore,
$$ \boxed{\text{Tr}(uv^T) = u \cdot v} $$
