My question has a negative answer.

**Lemma.** Suppose $X$ has the approximation property (AP), $Y$ is a subspace of $X$, and $X/Y$ fails the AP. Then there is a nuclear operator $T$ on $X$ s.t. $TX\subset Y$, $T^2=0$, and $tr(T)=1$.

Suppose you have $X$, $Y$, $T$ as in the lemma and $Y$ has the AP. Define $a:X\to Y$ to be $T$ considered as an operator into $Y$ and let $b:Y\to X$ be the inclusion map. Then $ba=T$ has trace one but $ab=0$.

Experts will see immediately that you can realize the situation in the previous paragraph by letting $Z$ be a James-Lindenstrauss space s.t. $Z^{**}/Z$ fails the AP while $Z^{**}$ and $Z$ have Schauder bases. More remarkable is that you can even have $X=\ell_p$ with $1<p<2$ and $Y$ isomorphic to $\ell_p$. This was proved by A. Szankowski a couple of years ago.

The lemma is easy: Since $X/Y$ fails the AP, by Grothendieck's classical characterization of the AP there is an absolutely summable sequence $f_n$ in $(X/Y)^*$ and a sequence $z_n$ in the open unit ball of $X/Y$ s.t. for all $z\in X/Y$, $\sum \langle f_n, z \rangle z_n=0$ but $\sum \langle f_n, z_n \rangle =1$ (that is, the trace of the zero operator on $X/Y$ is not well defined). Let $Q$ be the quotient mapping from $X$ onto $X/Y$ and get $x_n$ in the unit ball of $X$ s.t. $Qx_n=z_n$. Define a nuclear operator $T$ on $X$ by

$Tx = \sum Q^*f_n(x) x_n$.

$QT=0$ because $\sum \langle f_n, z \rangle z_n=0$ for all $z\in X/Y$ and hence $TX \subset Y$.

$T_{|Y} =$ because $Q^*$ ranges in the annihilator of $Y$ in $X^*$ and hence $T^2=0$.

Finally, $tr(T)= \sum \langle Q^*f_n, x_n \rangle =\sum \langle f_n, z_n \rangle =0$.

This construction raises more questions than it answers. For what Banach spaces $X$ and $Y$ is there an affirmative answer to the trace question? The only positive result I see is when one of the spaces is a Hilbert space and the other one is a weak Hilbert space in the sense of Pisier. The affirmative answer follows because Pisier proved that the Lidskii trace formula is valid for nuclear operators on a weak Hilbert space whose eigenvalues are absolutely summable (an old result due to Konig, Maurey, Retherford and me says that on any Banach space that is not isomorphic to a Hilbert space, there is a nuclear operator whose eigenvalues are not summable, so it is not clear that the trace question has an affirmative answer when $X$ and $Y$ are both weak Hilbert spaces).

ADDED 10/24/11: The paper of Szankowski I mentioned is

Szankowski A (2009)

Three-space problems for the approximation property.

J. Eur. Math. Soc., 11(2): 273-282.

Although obvious, I should have mentioned that from the negative answer to the question for $X=Y=\ell_p$, $1<p<2$, by duality you also get a negative answer for $X=Y=\ell_p$, $2<p<\infty$.

definesthe trace of an operator on a finite-dimensional space to be the sum of its eigenvalues, and only after that proves that it can be calculated by summing diagonal matrix entries. – Mark Meckes Sep 29 '11 at 13:47`$X^*$`

and $x_n$ in $X$ s.t. $\sup \|f_n\| <\infty$, $\sum \|x_n\| = 1$, $\sum f_n(x_n) = 1$, yet $\sum f_n(x) x_n = 0$ for all $x$ in $X$. – Bill Johnson Oct 6 '11 at 18:44