Kim: Exercise 3.1.R (Tensors)

This exercise is from D. Kim, A rough guide to linear algebra, see p. 62, [link](https://stanford.edu/~dkim04/writings/LinAlg.pdf), accessed on 31 July 2025.

Some useful isomorphisms

Suppose $V$ is a finite-dimensional vector space with basis $B_V = \{e_1, \ldots, e_n\}$ and let the dual basis be $B_{V^*}=\{e_1^*, \ldots, e_n^*\}$.

Here we identify ${\rm Hom}(V, V)$ with $V^* \otimes V$. Each $f\in {\rm Hom}(V, V)$ corresponds to the following element of the tensor product space $V^* \otimes V$

$$\hat{f} = \sum_{i=1}^{n} e_i^* \otimes f(e_i).$$

Conversely, every element of $V^* \otimes V$,

$$q = \sum_{i=1}^{k}a_i \otimes v_i,$$

can be seen as an endomorphism

$$\check{q}(\xi) = \sum_{i=1}^{k} a_i(\xi) v_i,$$

for $\xi \in V$.

Now we can do a little tango: ${\rm End}(V) \overset{\hat{\cdot}}{\to} V^* \otimes V \overset{\check{\cdot}}{\to} {\rm End}(V)$

$$f(\xi) = \check{\hat{f}}(\xi) = \sum_{i=1}^{n} e_i^*(\xi)f(e_i).$$

Hereafter, with some abuse of notation we will identify $f$ with $\hat{f}$ and $\check{\hat{f}}$ and write just $f$.

Compositions

Suppose $f, g\in {\rm End}(V)$ with $g(u) = \sum_{j=1}^{n} e_j^*(u)g(e_j)$ and $f(\xi)=\sum_{i=1}^{n}e_i^*(\xi) f(e_i)$. Then,

$$(g\circ f)(\xi) = \sum_{j=1}^{n} e_j^*\left(\sum_{i=1}^{n}e_i^*(\xi) f(e_i)\right)g(e_j) = \sum_{i, j} e_i^*(\xi)e_j^*(f(e_i))g(e_j).$$

Solution

Treating $f\in {\rm Hom}(V,V)$ as an element of $V^*\otimes V$, $f = \sum_i e_i^* \otimes f(e_i)$, and, likewise, $\hat{g} = \sum_i e_i^* \otimes g(e_i)$. Taking a $\otimes$ we have

$$\begin{aligned} f \otimes g {}={}& \left(\sum_i e_i^* \otimes f(e_i)\right)\left(\sum_i e_i^* \otimes g(e_i)\right) \\ {}={}& \sum_{i}\sum_{j} e_i^* \otimes f(e_i) \otimes e_j^* \otimes g(e_j) \end{aligned}$$

Therefore,

$$\begin{aligned} (\mathrm{id} \otimes \mathrm{tr} \otimes \mathrm{id})(f\otimes g) {}={}& \sum_{i}\sum_{j} e_i^* \otimes \underbrace{\mathrm{tr}(f(e_i) \otimes e_j^*)}_{\in k} \otimes g(e_j) \\ {}={}& \sum_{i}\sum_{j} e_j^*(f(e_i)) \cdot (e_i^* \otimes g(e_j)) \end{aligned}$$

Here we used the fact that $k \otimes V \cong V$. Next, $e_i^* \otimes g(e_j)$ is an element of $V^* \otimes V$, so we can treat it as an element of ${\rm End}(V)$. We can write

$$\begin{aligned} (\mathrm{id} \otimes \mathrm{tr} \otimes \mathrm{id})(f\otimes g)(\xi) {}={}& \sum_{i}\sum_{j} e_j^*(f(e_i)) e_i(\xi)g(e_j) = (g\circ f)(\xi). \end{aligned}$$

We have shown that $(\mathrm{id} \otimes \mathrm{tr} \otimes \mathrm{id})(f\otimes g) = g\circ f$. $\Box$