Matrix as tensor exercise with answer Can I get an explanation for the following exercise with answer in the book Linear Algebra via Exterior Products by S. Winitzki?

Exercise 1 - Matrices as Tensors. Now suppose you have a matrix $A_{jk}$ that specifies the linear operator $\hat A$ in a basis $\{\mathbf e_k\}.$ Which tensor $A\in V\otimes V^*$ corresponds to this operator?
  Answer:
$$A=\sum_{j,k=1}^n\,A_{jk}\,\mathbf e_j\otimes \mathbf e_k^*.$$

A tensor of the form $\mathbf e_j \otimes\mathbf e_k^*$ is presumably a $(1,1)$-tensor, and should take a covector and a vector. But if it represents a matrix, then it is only taking in a vector...
Say the matrix is $$\mathbf A=\begin{bmatrix}6&7&1\\-2&0&10\end{bmatrix}.$$ Then, it would just need a vector to implement the linear transformation. For example, $\small\begin{bmatrix}4\\3\\1\end{bmatrix}.$
Aren't the columns of the matrix $\mathbf A$ expressed in the the basis $\mathbf e_k^*$ as in
$$\mathbf A=\begin{bmatrix}6 \mathbf e_{11}^*&7\mathbf e_{12}^*&1\mathbf e_{13}^*\\-2\mathbf e_{21}^*&0\mathbf e_{22}^*&10\mathbf e_{23}^*\end{bmatrix}$$
in some fashion? What goes with $\mathbf e_j$?

One of the basic insights of multilinear algebra is that "taking a vector" is in some sense equivalent to "giving you a covector" (because taking a vector is what that covector does), and conversely "taking a covector" is the same as "giving you a vector".
So we can view an $(1,1)$-tensor either as something that takes in a vector and a covector (and gives you a scalar back), or as something that takes a vector and gives you another vector back. Not coincidentally, these are also two things you can do with a matrix: Given $A$ you can either take a column $v$ and a row $z^T$ and compute the scalar $z^T A v$, or you can take just the column $v$ and compute the column $Av$.
In particular, $w\otimes u^*$ can represent the operation that


*

*takes in a vector $v$.

*computes a scalar contracting $u^*$ and $v$.

*returns that scalar times $w$.


If you also give it a covector $z^*$, you can contract that with the result of the above operation and get a scalar. This gives the same scalar as if you contracted the input vector and covector separately:
$$ \langle z^*, \langle u^*,v\rangle w \rangle = \langle u^*, v\rangle\langle z^*, w\rangle  $$
by (right) linearity of the inner product.
