# Generalizing trace from positive to trace class operators

This is a question about generalizing trace defined for positive operators to trace class operators, in the terminology below, in the context of functional analysis.

Let $T \in B(H)$ (i.e. $T$ is a bounded linear operator on a Hilbert space $H$). Say $T$ is positive (denoted $T \geq 0$) if $T$ is self-adjoint and $\langle Tx, x\rangle \geq 0$ for all $x$ (where $\langle\cdot ,\cdot \rangle$ is the inner product on $H$). Define $$\mathrm{tr}(T)=\sum_j \left<Te_j, e_j \right>$$

the trace of $T$ for $T$ positive, where $e_j$ is an orthonormal basis for $H$. This definition is independent of the choice of basis. Let $B_0(H)$ be the compact operators in $B(H)$. Define

$B^1(H) = \mathrm{span} \{ T \in B_0(H) : T \geq 0, \mathrm{tr}\,T<\infty \}$

the trace class operators. Why is it that for $T \in B^1(H)$, $T=\sum_0^3 i^k T_k$ for some $T_k \geq 0$ (where $i \in \mathbb{C}$), and why is the extension of trace to $B^1(H)$ defined as

$\mathrm{tr}\,T = \sum_0^3 i^k \mathrm{tr} T_k$

well-defined? (This is from Pedersen, Analysis Now.)

• Please give some context when you ask questions instead of just tossing symbols out and expecting people to immediately know which of the 20 fields of mathematics that might use the same symbols is the one you are referring to. After a little reading and back-and-forth, I have finally deduced that $H$ is a Hilbert space and $B(H)$ is the set of bounded operators, but at this point I'm ticked off enough at having to deduce this, I'm not really interested in the rest. – Paul Sinclair Nov 23 '17 at 0:41

Because any $T\in B(H)$ can be written as $T=\sum_{k=0}^3i^k T_k$ with $T_k\geq0$.
Maybe it helps to see this first in the complex numbers: given $z\in\mathbb C$, you are used to write it $a+ib$ with $a,b\in\mathbb R$. We can write $a=a_0-a_2$, $b=a_1-a_3$, with $a_0,a_1,a_2,a_3\geq0$, and then $$z=a+ib=a_0+ia_1-a_2-ia_3=\sum_{k=0}^3i^ka_k.$$ In the case of $T$, we can take $$\text{Re}\,T=\frac{T+T^*}2,\ \ \ \ \ \text{Im}\,T=\frac{T-T^*}{2i},$$ and then $T=\text{Re}\,T+i\,\text{Im}\,T$, with $\text{Re},\text{Im}\,T$ selfadjoint. As any selfadjoint operator can be written as the difference of two positive operators, we can write $$\text{Re}\,T=T_0-T_2,\ \ \ \ \text{Im}\,T=T_1-T_3$$ with $T_0,T_1,T_2,T_3$ positive. Then $$T=\sum_{k=0}^3i^kT_k.$$