📂Linear Algebra

Linear Transformation Trace

Definition

Let $V$ be a $n$-dimensional vector space. Let $f : V \to V$ be a linear transformation. Let $B = \left\{ e_{i} \right\}$ be a basis of $V$. Let $n \times n$ matrix $A$ be the matrix representation of $f$ with respect to $B$.

$$A = [f]_{B}$$

Since $f(e_{i}) \in V$, we represent it as $f(e_{i}) = \sum f_{j}(e_{i})e_{j}$. Then,

$$A = \begin{bmatrix} f_{1}(e_{1}) & f_{2}(e_{1}) & \cdots & f_{n}(e_{1}) \\ f_{1}(e_{2}) & f_{2}(e_{2}) & \cdots & f_{n}(e_{2}) \\ \vdots & \vdots & \ddots & \vdots \\ f_{1}(e_{n}) & f_{2}(e_{n}) & \cdots & f_{n}(e_{n}) \end{bmatrix}$$

Define the trace of the linear transformation $f$, $\tr f$, as follows:

$$\tr f := \tr(A) = \sum_{i} f_{i}(e_{i})$$

Here, $\tr(A)$ is the trace of matrix $A$.

Explanation

Since there exists a corresponding matrix for a linear transformation, the trace of the linear transformation can be naturally defined.

Basis Invariance

• The trace of a linear transformation does not depend on the choice of basis.

By the definition, it might seem that the matrix $A$ depends on the basis, so choosing a different basis $B^{\prime}$ to obtain matrix $A^{\prime}$ would change the value of $\tr f$. However, the two matrices $A$ and $A^{\prime}$ are similar. Since the trace is invariant under similarity, $\tr f$ is well defined irrespective of the choice of basis for $V$.

Inner Product Representation

• Use the Einstein notation.

$\tr f$ can be expressed using the given metric. Let’s say metric $g$ is given. Then,

$$g(f(e_{i}), e_{k}) = g( f_{j}(e_{i})e_{j}, e_{k} ) = f_{j}(e_{i}) g( e_{j}, e_{k} ) = f_{j}(e_{i})g_{jk}$$

Multiply both sides by $g^{kl}$ and sum over index $k$,

$$g^{kl} g(f(e_{i}), e_{k}) = f_{j}(e_{i})g_{jk}g^{kl} = f_{j}(e_{i})\delta_{j}^{l} = f_{l}(e_{i})$$

Therefore, we obtain the following.

$$\tr f = f_{i}(e_{i}) = g(f(e_{i}), e_{k})g^{ki}$$