Given two vectors $u$ and $v$ in $\mathbb{R}^n$ such that $u^Tv \neq 0$ and $u \neq \alpha v$ for any scalar $\alpha$. Let $v^\perp$ be a subspace to which $v$ is orthogonal. A orthogonal basis $[s_1, s_2,\ldots,s_{n-1}]$ for such a subspace can be generated by some standard approach. I want a symmetric positive definite matrix $A \in \mathbb{R}^{n\times n}$ such that $A$ projects vector $u$ on $v^\perp$ or span($s_1,s_2,\ldots,s_{n-1}$) i.e. $Au \in v^\perp$ or $v^TAu=0$.
If $d \in v^\perp$ be a vector along orthogonal projection of $u$ on $v^\perp$ then $A=\frac{dd^T}{d^Td}$ (symmetric). In other words, can we find such a $d$ for which $\frac{dd^T}{d^Td}$ matrix is positive definite ?
Best Answer
We assume $n \geq 2$.
$A:=\frac{dd^T}{d^Td}$ is a rank-one matrix. Explanation: every $c \in d^{\perp}$ is such that $\frac{dd^T}{d^Td}c=\frac{dd^Tc}{d^Td}=\frac{d0}{d^Td}=0$. Thus $c$ is an eigenvector associated to eigenvalue $0$. In other terms, subspace $d^{\perp}$ (which is $n-1$ dimensional) is included in the kernel of $A$; it is immediate that it is in fact the kernel of $A$. From there, using the rank-nullity theorem, we can conclude to the "rank 1" property of $A$.
Thus 0 is an eigenvalue of $A$ with multiplicity $n-1$. Due to relationship $Ad=d$, the other eigenvalue is 1.
Thus $A$ is only a symmetrical semi-definite positive matrix.