Showing $\operatorname{tr}(T) = \sum_{i} v_i^* (T(v_i))$ for a linear map $T$.

Question

Let $V$ be a finite-dimensional vector space and $T:V \rightarrow V$ be a linear map. Let $\{v_i\}$ is a basis of $V$ and $\{v_i^*\}$ is a basis of $V^*$, dual space of $V$. Then I want to show \begin{align} \operatorname{tr}(T) = \sum_i v_i^* (T(v_i)). \end{align}

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I am familiar with the trace of a linear map as $\operatorname{tr}(T) = \langle v_i, T(v_i) \rangle$ and it seems the above formula is somehow the rigorous extension of this. But anyway it was something like given not proven.....

I know for the linear map $T$, $\operatorname{tr}(T) = \operatorname{tr}([T]_{\beta})$ but does not understand which step the dual basis comes out.

I found some posts from StackExchange

$T$ is a linear operator on a IPS $V$ which has a basis $\beta$. Prove that $A_{ij} = \langle T(v_j),v_i \rangle$

The trace formula of a linear map $T: V \to V$, $\operatorname{tr}(T) = \sum_k (Te_k,e_k)$

and for each case, the dual basis does not come out. Naively I guess there is a relation, but I want to know explicitly.

Answer 1

Write $T(v_i) = \sum_{j}k_{ij}v_j$. Note that $k_{ij} = v_{j}^{*}(T(v_i))$. In fact

$$v_{j}^{*}(T(v_i)) = v_{j}^{*}( \sum_{l}k_{il}v_l ) = \sum_{l}k_{il}v_{j}^{*}(v_l)= k_{ij} .$$

Then $T(v_i) = \sum_{j}v_{j}^{*}(T(v_i))v_j$. Now just write the $T$ matrix and take the Trace operator, you will see that it is exactly $\sum_{j}v_{j}^{*}(T(v_j))$.

Answer 2

The linear maps $T_{ij}=v_iv_j^*$ form a basis of $\mathrm {End}V$. The matrix representation of $T_{ij}$ with respect to the basis $v$ contains a $1$ at position $ij$ and is zero otherwise. Hence $\operatorname{tr}(T_{ij})=\partial_{ij}$.

Now the formula you want to show holds true for the $T_{ij}$ and hence by linearity for all $T\in\mathrm {End}V$.