An orthogonal projection is an Hermitian matrix $P$ such that $P^2=P$. Denote $U^*$ the conjugate transpose of a matrix $U$.

It can be easily shown that for two projections $P_1$ and $P_2$, there exists a unitary $U$ such that both $UP_1 U^*$ and $UP_2U^*$ are block diagonal with blocks of size one or two (And both resulting matrices have the same block structure).

My question is whether this block decomposition of projections can be generalised, for more than two projections: Given orthogonal projections $P_1, P_2, ..., P_k$, Is there a unitary $U$ such that for each $i$, $UP_i U^*$ is block diagonal with blocks of size at most $k$? (The resulting matrices must have the same block structure)

Two weaker questions are:

Is there a bound on the size of the blocks in function of $k$ only, i.e. in function of the number of projectors independently of their dimensions?

If a block decomposition is not possible, then what about decomposing the projectors into $k$-diagonal matrices? (All entries of the matrix are zero except (possible) for the diagonal and the $k$-upper and $k$-lower diagonals)

I would deeply appreciate any help or reference on how to handle these problems.

Best regards,



2 Answers 2


If $P$ is a projection then $I-2P$ is a reflection. Two reflections generate the dihedral group and all irreducible representations of the dihedral group have dimension at most two. This explains your observation about two projections.

But the alternating group $Alt(n)$ can be generated by three involutions when $n\ge9$ (a result of Nuzhin). Take the usual permutation representation of $Alt(n)$ with degree $n$. If $R$ is the permutation matrix representing an involution, then $R$ is symmetric and $(I-R)^2 =2(I-R)$. So $\frac12(I-R)$ is a projection.

This permutation representation has one invariant subspace of degree one (the span of the constant vectors) and the orthogonal complement to this is irreducible with dimension $n-1$. This shows that the size of your blocks is not bounded by the number of projections.

  • $\begingroup$ Thank you very much for the nice answer. Do you have some indication on the more general case, where the matrices are just required to be k-diagonal? $\endgroup$ May 31, 2010 at 6:40

Each projection gives a decomposition of the vector space into two subspaces. So with two projections you have four subspaces. The four subspace problem is tame. For $m$ projections you have $2m$ subspaces and the $n$-subspace problem is known to wild for $n\ge 5$.

This point of view has not taken into account the requirement that pairs of subspaces have zero intersection. Once this has been taken into account you can explain your observation about two idempotents. For more than two idempotents I expect this is a wild problem.


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