On the norm of a quadratic form associated with a matrix product Let $x,z\in\mathbb{R}^n$ be a couple of arbitrary vectors and let $A\neq A^T\in\mathbb{R} ^{n \times n}$ be a square matrix such that $z^TAz>0$ for all $z\neq 0$. Consider the product $z^T\frac{xx^T}{x^Tx}Az$. Note that $\|\frac{xx^T}{x^Tx}\|=1$ for all $x\neq 0$.
Is the inequality $z^T\frac{xx^T}{x^Tx}Az\leq z^TAz$ valid?
 A: No. For example, take $A=\begin{pmatrix}2&-1\\-1&1\end{pmatrix}+\epsilon P$ with $P$ anti-symmetric, $x=(1,0)$, $z=(0,1)$. Then $$z^\top(A-xx^\top A)z=(0,1)\begin{pmatrix}2&-1\\0&0\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}-\epsilon(z^\top xx^\top Pz)<0$$
A: $\boxed{\text{No, the inequality is not valid.}}$
The matrix $X=\frac{x\,x^T}{x^Tx}$ can be considered just as $X=x\,x^T$ with $\|x\| = 1$. We should consider $3$ vectors:

*

*$z$ ; so that $z^Tz>0$

*$a=A\,z$ ; so that $z^Ta > 0$

*$y = X\,z$ ; so that  $0\leq y^Ty\leq z^Tz$
We want to verify if $y^Ta \leq z^Ta$ is always satisfied.
Vector $a$ is $z$ rotated and stretched, vector $y$ is $z$ projected at direction $x$.
If $x$ is aligned to $a$ or to $z$, those expressions will be equal. However, there are cases in which the considered inequality will not be satisfied, e.g., when $x$ is between $a$ and $z$. This is shown below.

For general vectors such that $z$ and $a$ are not aligned, and $x$ is between $a$ and $z$:

*

*$cos(\theta) = \frac{z^Ta}{\|z\|\,\|a\|} \in \;]-1,\;1[$

*$x = \gamma\,a + (1-\gamma)\,z$
$$\gamma\in]0,1[ \;\;\Rightarrow\; \frac{z^Tx\,x^T\,a}{x^Tx} > z^Ta$$
$$\left[\gamma\,z^Ta + (1-\gamma)\,z^Tz\right]\left[\gamma\,a^Ta + (1-\gamma)\,z^Ta\right] > \left[\gamma\,a + (1-\gamma)\,z\right]^T\left[\gamma\,a + (1-\gamma)\,z\right]\,z^Ta$$
$$(1-\gamma)\,\gamma\,z^Tz\;a^Ta > (1-\gamma)\,\gamma\,z^Ta\;z^Ta$$
$$z^Tz\;a^Ta = \|z\|^2\,\|a\|^2 > \|z\|^2\,\|a\|^2\,cos(\theta)^2 =  z^Ta\;z^Ta$$
$$cos(\theta)^2 < 1$$

Numerical example:
$$A = \begin{bmatrix}1&2\\0&1
\end{bmatrix} \;\;;\;\;z = \begin{bmatrix}1\\1
\end{bmatrix} \;\;;\;\; x = \frac{1}{\sqrt{5}}\begin{bmatrix}2\\1
\end{bmatrix}$$
$$a = \begin{bmatrix}3\\1
\end{bmatrix} \;\;;\;\;y = \begin{bmatrix}6/5\\3/5
\end{bmatrix}$$
$$y^Ta = 4.2>4=z^Ta$$
