# Is $\left(x^TAx\right)^{-1}Axx^T=I$?

Let $A$ be a real $n\times n$ matrix that is symmetric and positive. Let $x\in\mathbb{R}^n$.

If $I$ is the $n\times n$ identity matrix is the following true? and if so why? $$\frac{1}{\left(x^TAx\right)}Axx^T=I$$.

I know that $\left(x^TAx\right)$ is simply a scalar and that $xx^T$ is an $n\times n$ matrix, but how do I know this must be true? (if it is)

• The rank of $AB$ must be bounded by the minimum of the ranks of $A$ and $B$, for any two matrices $A$ and $B$. Since $xx^{T}$ has rank one, can the equality possibly hold? – RideTheWavelet Oct 7 '17 at 22:33

This cannot be true for $n \geq 2$. If it was, $\frac{1}{x^TAx}xx^T$ would be the inverse of the matrix $A$. But you can easily check that $xx^T$ is not an invertible matrix : actually, it is of rank exactly $1$ if $x \not = 0$, and $0$ if $x = 0$.

The assertion

$\frac{1}{\left(x^TAx\right)}Axx^T=I \tag 1$

looks false to me. Take, for example $y \ne 0$ such that

$x^Ty = 0; \tag 2$

then

$\frac{1}{\left(x^TAx\right)}Axx^Ty = 0, \tag 3$

but

$Iy = y; \tag 4$

the left and right sides disagree when applied to $y$.