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Is it true, that if $T$ and $S$ are selfadjoint and positive operators in a finite-dimensional Hilbert space, such that $T^2=S$, that then $\text{rank}T=\text{rank}S$ ?

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up vote 7 down vote accepted

This is true more generally if $T$ is diagonalizable (recall that if $T$ is self-adjoint, it is diagonalizable).

If $T$ is diagonalisable, there is an invertible matrix $P$ and a diagonal matrix $D$ such that $T=PDP^{-1}$, the rank of $T$ is the number (counted with multiplicities) of the non-zero eigenvalues of $T$. Now since $T^2=PDP^{-1}PDP^{-1}=PD^2P^{-1}$, the rank of $T^2$ is the number of the non-zero eigenvalues of $D^2$, which is the same as the number of non-zero eigenvalues of $D$.

The ranks of $T$ and $T^2$ are thus equal.

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Since we are in a finite dimensional vector space, we can diagonalize $T$: there is a hermitian matrix $P$ and a diagonal matrix $D$ such that $T=P^*DP$. The rank of $T$ is the same as the rank of $D$, which is the same as the rank of $D^2$ (the number of non-zero elements of the diagonal), and $S=P^*D^2P$, which proves that $S$ and $T$ have the same rank.

The fact that the operators are self-adjoint is necessary here: if $T^2=0$, $T\neq 0$ and $S=0$, it won't work. Positiveness is not necessary here.

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Whenever $T$ is a matrix which has the same rank as its square, $\mathrm{i}T$ also is. However it is not possible that $T$ and $\mathrm{i}T$ are both self-adjoined unless $T=0$. Thus your claim that self-adjointness is necessary for this property is disproven. – celtschk Jul 30 '12 at 11:04

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