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The rank-nullity theorem states that for $U, V$ finite dimensional vector spaces and $T:U \to V$ a linear map $$\dim(U) = \dim(im(T)) + \dim(ker(T)) $$ Let $M \subset B(H) $ be a finite von Neumann algebra with a trace $tr$, let $p \in M$ be a projection and $T: pH \to H$ a linear map such that $im(T)=q_1H$ and $ker(T)=q_2H$ with $q_1, q_2 \in M$ projections.

Question: Is it true that $tr(p) = tr(q_1) + tr(q_2)$ ?

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  • $\begingroup$ If necessary we can assume that $M$ is a factor and $T = \tilde{T}p\vert_{pH}$ with $\tilde{T} \in M$. $\endgroup$
    – Sebastien Palcoux
    Commented Aug 4, 2015 at 6:41
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    $\begingroup$ Without the extra assumption in your comment, it seems to me you could just take something like $H = L^2([0,1])$, $M = L^\infty([0,1]) \subset B(H)$, $\mathrm{tr} = \int dx$, $p=1$, $q_1 = \chi_{[0.5]}$, $q_2=0$ and $T \in B(H)$ given by $(T \xi)(x) = \begin{cases} \xi(2x) & \text{ if } 0 \leq x \leq .5 \\ 0 & \text{ if } .5 \leq x \leq 1 \end{cases}$. $\endgroup$
    – Michael
    Commented Aug 4, 2015 at 7:34
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    $\begingroup$ With the extra assumption this is just as easily seen to be true since then the projections onto the ranges of $\tilde Tp$ and $(\tilde T p)^*$ are equivalent in $M$ (just consider the partial isometry in the polar decomposition of $\tilde T p$). $\endgroup$
    – Jesse Peterson
    Commented Aug 4, 2015 at 17:00
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    $\begingroup$ You just need a dimension function which assigns equal dimensions to equivalent projections. You don't need factoriality. In the abelian case $M = L^\infty(X, \mu)$ this just says $\mu( E \cup F ) = \mu(E) + \mu(F)$ for disjoint measurable sets $E$ and $F$. $\endgroup$
    – Jesse Peterson
    Commented Aug 5, 2015 at 21:23
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    $\begingroup$ You still need that $T$ is contained in $M$. In Michael's example the operator $T$ is not contained in $M$. $\endgroup$
    – Jesse Peterson
    Commented Aug 6, 2015 at 18:59

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