Define the trace and deteraminant of a linear transformation in a basis-free manner

linear algebramatrices

Trace and determinant of a matrix can be shown to be invariant under basis transformation. This should suggest that they can be defined for a linear transformation in a basis-free manner. How does one do that?

P.S. I'm really looking for a definition which can illustrate why the trace and determinant of matrices are invariant under basis transformations.

Thanks very much!

Best Answer

$\newcommand{\ab}[1]{\langle #1 \rangle}$ $\newcommand{\mr}{\mathscr}$ $\newcommand{\mc}{\mathcal}$ $\newcommand{\bw}{\bigwedge}$ $\newcommand{\tr}{\text{trace}}$

Below I give the definitions using exterior powers. I also prove some theorems so that the reader can gain some familiarity if this approach is new to him/her.

Trace

Let $V$ be an $n$-dimensional vector space over a field $F$ and $T$ be a linear operator on $V$. Define a map $f:V^n\to \bw^n V$ as $$ f(v_1 , \ldots, v_n)=\sum_{i=1}^n v_1\wedge \cdots \wedge v_{i-1}\wedge Tv_i\wedge v_{i+1}\wedge \cdots \wedge v_n $$ Then $f$ is an alternating multilinear map. Therefore, we get a unique linear map $\theta:\bw^n V\to \bw^n V$ such that the following diagram commutes:

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Since $\dim(\bw^n V)=1$, there is a unique $c\in F$ such that $\theta(v_1 \wedge \cdots \wedge v_n)=c(v_1 \wedge \cdots \wedge v_n)$ for all $v_1 , \ldots, v_n\in V$.

Definition. This unique element $c\in F$ is termed as the trace of $T$ and is written as $\tr(T)$.

Theorem 1. Let $V$ be a finite dimensional vector space and $T$ be a linear operator on $V$. Let $\ab{\cdot, \cdot}$ be a non-degenerate symmetric bilinear form on $V$ which affords an orthonormal basis $\mc B=(e_1 , \ldots, e_n)$ on $V$. Then $$ \text{trace}(T) = \sum_{i=1}^n\ab{Te_i, e_i} $$

Proof. Immediate from the definition of trace. $\blacksquare$

Going back to the definition of trace, more generally, given an $n$ tuple $(v_1, \ldots, v_n)$ of vectors in $V$ and an increasing $k$-tuple $I=(i_1, \ldots , i_k)$ of integers between $1$ and $n$, write $v_{I, j}$ to denote $Tv_j$ if $j$ appears in $I$ and simply $v_j$ if $j$ does not appear in $I$. Further write $v_I$ to denote $v_{I, 1}\wedge \cdots \wedge v_{I, n}$. Define $f_k:V^n\to \Lambda^n V$ as $$ f_k(v_1, \ldots, v_n)= \sum_{I \text{ an increasing }k\text{-tuple}}v_I $$ Then $f_k$ is an alternating multilinear map and this induces a unique linear map $\Lambda^n V\to \Lambda^n V$. Again, this linear map is multiplication by a constant which we call the $k$-th-trace of $T$ and denote it as $\text{trace}_k(T)$. Now we define determinants and the reader should convince oneself that the determinant is the $n$-th trace.

Determinants

Definition. Let $V$ be a finite dimensional vector space. Writing $\dim V=n$, and knowing that $\dim(\bw^n V)=1$, we deduce that there is a unique constant $c\in F$ such that $$ \textstyle{\bw^n T(v_1 \wedge \cdots \wedge v_n)} =c\cdot(v_1 \wedge \cdots \wedge v_n) $$ for all $v_1 , \ldots, v_n\in V$. This constant $c$ is termed as the determinant of $T$ and is written as $\det T$. Since $\bw^n V$ has dimension $1$, the determinant is well-defined.

Theorem 2. Let $T$ and $S$ be linear operators on $V$. Then $\det(TS)=(\det T)(\det S)$.

Proof. Say $\dim V=n$. We have $$ \begin{array}{rcl} (\det(TS))v_1\wedge \cdots \wedge v_n &=& T(Sv_1)\wedge \cdots \wedge T(Sv_n)\\ \\ &=& (\det T)Sv_1 \wedge \cdots \wedge Sv_n\\ \\ &=& (\det T)(\det S) v_1\wedge \cdots\wedge v_n \end{array} $$ Since this is true for all $v_1 , \ldots, v_n\in V$, we must have $\det(TS)=(\det T)(\det S)$. $\blacksquare$

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