# Let $A \in M_n(\mathbb{C})$ and $x$ a complex unit vector. Show $(x^*Ax)(x^*A^*x) \leq x^*AA^*x$.

Let $$A \in M_n(\mathbb{C})$$ and $$x$$ a complex unit vector. Show $$$$(x^*Ax)(x^*A^*x) \leq x^*AA^*x. \;\;[1]$$$$

Solution attempt: We know that $$AA^*$$ is positive semi-definite so it is Hermitian and its eigenvalues are nonnegative. Thus, it has a basis $$v_1, \ldots, v_n$$ of eigenvectors respectively associated with the nonnegative eigenvalues $$\lambda_1,\ldots,\lambda_{n-1}.$$

Thus, if $$x= c_1 v_1 + \ldots + c_nv_n$$, the right hand side of [1] becomes $$\lambda_1 |c_1|^2 +\ldots+\lambda_n|c_n|^2.$$

I no longer know what to do with the left hand side of [1].

Define matrix $$M\in\mathbb{C}^{n\times 2}$$ with $$M=\begin{bmatrix}x & A^*x\end{bmatrix}$$ and note that $$M^*M = \begin{bmatrix}x^*x & x^*A^*x\\ x^*Ax & x^*AA^*x\end{bmatrix}\,.$$ Now, we have $$0 \leq |\det(M)|^2 = \det(M^*M) = (x^*x)(x^*AA^*x)-(x^*Ax)(x^*A^*x)\,,$$ from where the inequality follows because $$x$$ is a unit vector.

• [+1] I have given an equivalent proof. May 11, 2019 at 6:52

Expanding the LHS of $$\|x^{\ast}(\lambda A+ A^{\ast})x\|^2 \geq 0$$

one gets quadratic expression in $$\lambda$$ :

$$(x^{\ast}Ax)^2\lambda^2+(x^{\ast}Ax+x^{\ast}A^{\ast}x)\lambda+(x^{\ast}A^{\ast}x)^2$$

with real coefficients that is is always $$\geq 0$$.

This is equivalent to the fact that its discriminant is $$\leq 0$$, giving the looked for inequality.