# Polar decomposition of invertible elements in a unital C$^{*}$-algebra.

If $A$ is a unital C$^{*}$-algebra and $a$ is invertible, then

1. $a = u|a|$ for a unique unitary element $u$ of $A$.
2. If $\| a \| = \| a^{-1} \| = 1$, what can you say about $|a|$?

I don’t know how to start!

• You can just say that $A$ is a C$^{*}$-algebra. – Haskell Curry Mar 3 '13 at 3:15
• Set $u:=a/|a|$ and show that $u$ is a unitary element of $A$. The rest follows from Haskell Curry's answer. – user56706 Mar 3 '13 at 3:30
• @VahidShirbisheh do you mean put $u=a|a|^{-1}$ ?? – user61965 Mar 3 '13 at 3:41
• @edab12: Yes, your expression is more correct. – user56706 Mar 3 '13 at 4:03
• @edab12: As $|a|$ must be invertible, we have $u = a |a|^{-1}$. Hence, $u$ is uniquely determined. To show that $u$ is unitary, simply observe that $$u^{\ast} u = |a|^{-1} a^{\ast} a |a|^{-1} = |a|^{-1} |a|^{2} |a|^{-1} = \mathbf{1}_{A}$$ and $$u u^{\ast} = a |a|^{-1} |a|^{-1} a^{\ast} = a |a|^{-2} a^{\ast} = a a^{-1} (a^{*})^{-1} a^{\ast} = \mathbf{1}_{A}.$$ – Haskell Curry Mar 3 '13 at 7:09

We will make use of the following basic result from the theory of Banach algebras.

Theorem 1 Let $A$ be a Banach algebra. If $x \in A$ is invertible, then $0 \notin {\sigma_{A}}(x)$ and $${\sigma_{A}}(x^{-1}) = \left\{ \frac{1}{\lambda} ~ \Bigg| ~ \lambda \in {\sigma_{A}}(x) \right\}.$$

We also need the following theorem from the theory of C$^{*}$-algebras.

Theorem 2 Let $A$ be a C$^{*}$-algebra. Then for any normal element $x$ of $A$, we have $${r_{A}}(x) = \| x \|_{A},$$ where ${r_{A}}(x)$ denotes the spectral radius of $x$.

Let $A$ be a C$^{*}$-algebra. Let $a \in A$ be invertible, and suppose that $\| a \|_{A} = \| a^{-1} \|_{A} = 1$. Then \begin{align} {\sigma_{A}}(|a|) &= {\sigma_{A}} \left( \sqrt{a^{*} a} \right) \\ &= \sqrt{{\sigma_{A}}(a^{*} a)} \quad (\text{By the Spectral Mapping Theorem applied to $a^{*} a$.}) \\ &= \{ 1 \}. \quad (\clubsuit) \end{align}

Proof of $(\clubsuit)$: Firstly, observe that ${\sigma_{A}}(a^{*} a) \subseteq [0,\infty)$. Next, using the C$^{*}$-condition and the fact that involution in a C$^{*}$-algebra is an isometry, we have $$\| a^{*} a \|_{A} = \| (a^{*}) (a^{*})^{*} \|_{A} = \| a^{*} \|_{A}^{2} = \| a \|_{A}^{2} = 1$$ and \begin{align} \| (a^{*} a)^{-1} \|_{A} &= \| a^{-1} (a^{*})^{-1} \|_{A} \\ &= \| a^{-1} (a^{-1})^{*} \|_{A} \\ &= \| a^{-1} \|_{A}^{2} \\ &= 1. \end{align} By Theorems $1$ and $2$, we therefore obtain ${\sigma_{A}}(a^{*} a) = \{ 1 \}$.

Let us now work with the Continuous Functional Calculus corresponding to the self-adjoint element $a^{*} a$. As ${\sigma_{A}}(a^{*} a) = \{ 1 \}$, applying both the constant function $1$ and the identity function to $a^{*} a$ yields $$\mathbf{1}_{A} = a^{*} a.$$ Therefore, $|a| := \sqrt{a^{*} a} = \mathbf{1}_{A}$ and $a = u|a| = u$.

Conclusion: If $a \in A$ is invertible and $\| a \|_{A} = \| a^{-1} \|_{A} = 1$, then $|a| = \mathbf{1}_{A}$ and $a$ is unitary.

Note: In the argument above, we first proved that $a^{*} a = \mathbf{1}_{A}$, then used polar decomposition to deduce that $a$ is unitary and hence $a a^{*} = \mathbf{1}_{A}$. However, we could have proven directly that $a$ is unitary by performing the argument with $a^{*}$ instead of $a$, as the given conditions already imply that $a^{*}$ is invertible and that $\| a^{*} \|_{A} = \| (a^{*})^{-1} \|_{A} = 1$. Once again, we have to use the fact that involution in a C$^{*}$-algebra is an isometry.

• Thank you very much, but does this mean that $|a|$ must be the identity of $A$ ? – user61965 Mar 3 '13 at 3:28
• in Gerard Murphy's : $C^{*}$ Algebras and Operator Theory, page 73 , the second condition holds iff $a$ is unitary ! – user61965 Mar 3 '13 at 3:39
• @edab12: Yes, $|a|$ must be the identity. – Haskell Curry Mar 3 '13 at 7:01
• Yes, I didn't use the polar decomposition to prove that $a$ is unitary, Thank you for ur help! – user61965 Mar 3 '13 at 7:28