Proving $aa^*\geq0$ in a $C^*$ algebra

Question

I tried to construct a proof of above statement by using matrices. It goes like this :- Let $a$ be any arbitrary element in a $C^*$ algebra. Consider the matrix $\begin{pmatrix}0 & a\\a^* & 0 \end{pmatrix}$. This is a self adjoint operator. Hence its spectrum is in $\mathbb{R}$. The square of this matrix is $\begin{pmatrix} aa^* &0\\0 & a^*a\end{pmatrix}$. Being square of a self adjoint operator, its spectrum lies in $[0,\infty)$. Hence the spectrum of $aa^*$ is also contained in $[0,\infty)$. I have a doubt that whether this proof is correct, but could not point the mistake. Please help.

Answer 1

Writing $$ M := \begin{bmatrix} aa^* & 0 \\ 0 & a^*a \end{bmatrix} \in M_2(A), $$ the only thing missing (as far as I can tell - hopefully someone will set me straight if I've missed something) is a proof that $ \sigma(aa^*) \subseteq \sigma (M)$, and that shouldn't be too difficult. Given $\lambda \in \mathbb{C}$, it is easy to see that if $$ \begin{bmatrix} aa^* - \lambda 1 & 0 \\ 0 & a^*a - \lambda 1 \end{bmatrix} \begin{bmatrix} b & c \\ d & e \end{bmatrix} = \begin{bmatrix} b & c \\ d & e \end{bmatrix} \begin{bmatrix} aa^* - \lambda 1 & 0 \\ 0 & a^*a - \lambda 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, $$ for $b,c,d,e \in A$, then $(aa^* - \lambda 1)b = b(aa^* - \lambda 1) = 1$. So if $aa^* - \lambda 1$ is not invertible in $\widetilde{A}$ (the unitization of $A$), then $M - \lambda I$ is not invertible in $M_{2}(\widetilde{A})$, and hence $\sigma(aa^*) \subseteq \sigma(M)$. This proves the result.