# is every element in a C* algebra a product of normal elements?

I have the following question and since I am not an expert on C*-algebras, I thought I ask it here:

I know that in general the sum and product of normal elements need not be normal. It is even true that every element in a C*-algebra is the sum of two normal elements. What I do not know is if that is true for multiplication, too, i.e. if every element in a C*-algebra is a product of normal elements.

I think I know that every invertible element can be written as a product of a unitary and a positive element but this decomposition does not work for non-invertible elements.

My standard example is the right shift operator on the l^2(N)-space. This is a non-normal operator, which is not invertible and I have no idea how to write it as a product of two normal operators.

I would be very grateful if anyone could help.

Greetings, Tom.

Since the question was asked wrt C$*$-algebras, I guess there is room for a general remark. Suppose that $xy = 1$ where $x$ is a product of normal elements, say $v_1v_2\cdots v_n$. Then $v_1$ is normal and has a right inverse; therefore it is (two-sided) invertible. Thus $v_2\dots v_n y$ is invertible, so $v_2$ has a right inverse, and thus is invertible. By induction, we obtain that all the $v_i$ are two-sided invertible, so that $x$ is too.

The upshot of this argument is that a product of normal elements that is one-sided invertible, is actually invertible. In particular, the right shift is not a product of normal elements---although of course, the index argument above is decisive in this case.

If every element of a C$*$-algebra were a product of normals, then it would be directly finite (all one-sided inverses are two-sided). Whether this is sufficient is unclear (as is the question as to whether the product of normals property extends to matrix rings over the original). But if the C*-algebra has one in the stable range, then every element is an invertible times a positive, so is a product of (three) normals.

And for von Neumann algebras, if of finite type, then every element is a product of two normals, as the isometry in the polar decomposition argument (below) can be modified to be a unitary in that case. If not of finite type, the property fails.

• Interesting addition. I don't see that it suggests a classification of the operators on $\ell_2$ that are products of normal operators. Apr 4, 2016 at 22:44

The shift in $\ell^2$ cannot be a product of normal operators, since its Fredholm index is nonzero, and a normal operator cannot have nonzero Fredholm index.

• Is there a classification of operators on $\ell_2$ that are products of normal operators? Apr 4, 2016 at 16:06
• Thm. 2.24 of P.Y. Wu's "The operator factorization problems" proves that if H is infinite-dim., then $T \in B(H)$ is a product of finitely many normal operators iff it is a product of 3, iff $nullity\ T = nullity\ T^*$ or the range of $T$ is not closed. This also coincides with the set of prod's of $\geq 6$ hermitian operators, and the set of products of $\geq 18$ non-neg. operators. Wu also has a paper with the name "products of normal operators" from 1988 that includes the proof. Terry Quinn has some ext's of Wu's results to direct integrals of B(H) and prod's of decomp. positive operators. May 3, 2022 at 20:45

Recall the polar decomposition: Let $T$ be any operator on a Hilbert space, let $|T|=\sqrt{T^*T}$. Then there exists an isometry $U$ from the image of $|T|$ to the image of $T$ such that $$T=U|T|.$$ Now, whenever $U$ may be extended to a unitary on $H$, we are done. This is equivalent to saying that the image of $|T|$ and the image of $T$ have the same codimension. A non-zero Fredholm-index is an obstacle to that.