Unitary element in $M_2(A)$

Question

Let $A$ be a unital C*-algebra and let $a∈A$ with $\|a\|≤1.$

1. Justify that $$af(a^* a) =f(aa^*)a$$ for every continuous function $f: [0,1]→ℂ.$
2. Use this to show that$$v:=\begin{pmatrix}a& (1−aa^*)^{1/2}\\ −(1−a^* a)^{1/2}& a^*\end{pmatrix}$$is a unitary element in $M_2(A).$
3. Why is $w:=\begin{pmatrix}a&(1−aa^*)^{1/2}\\0&0\end{pmatrix}$ a partial isometry?

How might one show 1., and is my theory correct for 2.?

Let's calculate v*v:

v* v = [a*, (1 - a* a)^(1/2), -(1 - aa*)^(1/2), a] * [a, (1 - aa*)^(1/2), -(1 - a* a)^(1/2), a*] = [a* a + (1 - a* a), a*(1 - aa*)^(1/2) - (1 - a* a)^(1/2)a, -(1 - aa*)^(1/2)a* - (1 - a* a)^(1/2)a, aa* + a(1 - a* a)a*]

Notice that i have the term (1 - a^*a)^(1/2) in the second and third entries. To simplify the calculation, so i use the given property: af(a^a) = f(aa^)a for every continuous function f: [0,1] -> ℂ.

Setting f(t) = (1 - t)^(1/2), we have:

a(1 - a* a)^(1/2) = (1 - aa*)^(1/2)a.

Now let's substitute this into the expression for v* v:

v* v = [a* a + (1 - a* a), a*(1 - aa*)^(1/2) - (1 - a* a)^(1/2)a, -(1 - aa*)^(1/2)a* - (1 - a* a)^(1/2)a, aa* + a(1 - a* a)a*] = [I, 0, 0, I]

Thus, we have v* v = vv* = I, which means that v is a unitary element in M_2(A).

here V is in column bracket

Answer 1

1. A commented, this equality is straightforward for $f(x)=x^n.$ The property then extends by linearity to polynomials and by density to continuous functions on $[0,1]$ (which contains the spectra of $aa^*$ and $a^*a,$ so that both sides of the equality make sense).
2. Your proof looks fine but should stay in your post, not in an answer, and should be formatted correctly, as commented also.
3. Check that $ww^*=\begin{pmatrix}1&0\\0&0\end{pmatrix}.$ So $ww^*$ is a projection hence $w$ is a partial isometry.