Cauchy-Schwarz Inequality in a $C^∗$-algebra

Question

I am trying to prove the following lemma:

Let $\omega$ be a positive linear functional over the $^*$-algebra $\mathfrak{A}$. It follows that \begin{align*} \text{(a)}\quad & \omega(A^*B)=\overline{\omega(B^*A)}, \\ \text{(b)}\quad & |\omega(A^*B)|^2 \le \omega(A^*A)\omega(B^*B) \quad \forall A,B \in > \mathfrak{A}. \\ \end{align*}

The proof is pretty straightforward. In essence, it is similar to this.

For $A,B \in \mathfrak{A}$ and $\lambda \in \mathbb{C},$ positivity of $\omega$ implies that

$$ \omega((\lambda A + B)^*(\lambda A + B)) \ge 0 $$

Please note that this is a partial ordering in the complex plane: $\text{For } a,b \in \mathbb{C},$ $a \le b \iff \mathcal{Re}(a-b) \le 0 \ \land \ \mathcal{Im}(a-b) = 0 \,.$

By linearity this becomes

$$ |\lambda|^2 \omega(A^*A) + \overline{\lambda}\omega(A^*B) + \lambda \omega(B^*A) + \omega(B^*B) \ge 0$$

From here, (a) follows immediately. To prove (b), we note that the following matrix is self-adjoint.

$$ Q_{AB}(\lambda) := \begin{bmatrix} |\lambda|^2 \omega(A^*A) & \lambda \omega(B^*A) \\ \overline{\lambda}\omega(A^*B) & \omega(B^*B) \\ \end{bmatrix}$$

Claim (b) follows straightforwardly from the non-negativity of $\mathrm{det}(Q_{AB}(\lambda)).$

So my question boils down to something altogether different. If a Hermitian matrix has non-negative diagonal entries, does it also have only non-negative eigenvalues?

If not, how should I end the above proof?

$$\begin{pmatrix} 1 & -2 \ -2 & 1 \end{pmatrix} $$ has an eigenvalue of $-1$ — Ofek Gillon, May 06 '17 at 08:45
@OfekGillon That's right. So how is it that the necessary, and sufficient, conditions for the positivity of the quadratic form in $\lambda$ are exactly the two conditions of the lemma? — Nanashi No Gombe, May 06 '17 at 08:51
@MartinArgerami How can we prove claim (b) from the non-negativity of the quadratic form in $\lambda$? — Nanashi No Gombe, May 06 '17 at 15:21
Ok. I have written a proof. The matrix thing makes no sense to me, but I never paid much attention to quadratic forms per se. — Martin Argerami, May 06 '17 at 15:30

score 2 · Accepted Answer · answered May 06 '17 at 15:29

2

It is best to write the quadratic equation like this: $$\tag{1} |\lambda|^2 \omega(A^*A) + 2\text{Re}\, \lambda \omega(B^*A) + \omega(B^*B) \ge 0. $$ It is important to notice that we are free to choose $\lambda$, and $A$ and $B$. Fix $A,B$. Let $\gamma\in\mathbb T$ such that $\gamma\,\omega(B^*A)=|\omega(B^*A)|$. Given $t\in \mathbb R$, put $\lambda=t\gamma$. Then $(1)$ becomes $$\tag{2} t^2 \omega(A^*A) + 2t|\omega(B^*A)| + \omega(B^*B) \ge 0 $$ As we can do this for any $t\in\mathbb R$ and this is a real quadratic, for this to be always non-negative we need "$b^2\leq 4ac"$, i.e., $$ 4|\omega(B^*A)|^2\leq 4|\omega(A^*A)|\,|\omega(B^*B)|. $$

answered May 06 '17 at 15:29

Martin Argerami

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What is $\mathbb T$? Moreover, don't you need "$b^2 -4ac \ge 0$" for real solutions? – Nanashi No Gombe May 06 '17 at 15:40
$\mathbb T$ is the unit circle. If $b^2-4ac>0$, the quadratic has two roots and so it cannot be always positive. – Martin Argerami May 06 '17 at 16:27
But isn't it worse if the roots are not even real? Sorry, I must be missing something basic. – Nanashi No Gombe May 06 '17 at 16:30
I suggest you draw a parabola and think about the roots. – Martin Argerami May 06 '17 at 17:11
Got it. Thanks. – Nanashi No Gombe May 07 '17 at 09:54
Glad I could help. – Martin Argerami May 07 '17 at 11:07

Cauchy-Schwarz Inequality in a $C^∗$-algebra

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