 Solution

For a self-adjoint C-star element $a$, the following are three equivalent characterizations of a Positive element of a C-star algebra:If $a=b^2$ with $b=b^{\ast }$, the spectral mapping theorem gives nonnegative spectrum. Conversely, the continuous functional calculus applied to $\sqrt{\lambda }$ on $\sigma (a)$ gives a self-adjoint C-star element $b$ with $b^2=a$. For real $\sigma (a)$,Thus the third condition means $\sigma (a)\subseteq [0,2t]$, while the first implies the second by taking the interval $[0,\Vert a\Vert ]$. The second implies the third with $t=\Vert a\Vert$ if $a\ne 0$; if $a=0$, any $t>0$ works. This proves every equivalence.

The positive cone of a C-star algebra contains zero and is closed under multiplication by nonnegative real scalars. To verify addition without assuming that positive elements commute, put $r=\Vert a\Vert$, $s=\Vert b\Vert$ for $a,b\ge 0$. ThenIf $r+s>0$, the third characterization makes $a+b$ positive; otherwise both are zero. Hence it is a convex cone. If $a$ and $-a$ are positive, $\sigma (a)=\{0\}$ and $a=0$ by the self-adjoint spectral norm formula; therefore the cone is pointed and defines a partial order on $A_{\mathrm{sa}}$ by $h\le k$ if $k-h\in A_{+}$. It is generating because the continuous functional calculus givesFor $a\ge 0$, the spectral mapping theorem yieldsFinally, let $a\in M_2(\mathbb{C})$ be Hermitian, and write $\tau =\mathrm{tr}a$, $d=\det a$. Its two eigenvalues are real. Both are nonnegative exactly when their sum $\tau$ and product $d$ are nonnegative. The corresponding sum and product for $I-a$ are $2-\tau$ and $1-\tau +d$. Therefore the necessary and sufficient conditions for a positive contraction areIndeed, nonnegative sum and product force two real numbers to be nonnegative, including the zero-product cases. Applying this to $a$ and $I-a$ places both eigenvalues in $[0,1]$.