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 ?

  • $\begingroup$ Sorry for cross product term in higher dimension. I meant $u \neq \alpha v$ for any scalar $\alpha$. I have edited the question now. $\endgroup$
    – user402940
    Dec 31 '16 at 8:18
  • $\begingroup$ The only positive definite projection matrix is the identity matrix. $\endgroup$ Dec 31 '16 at 10:53

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.

  • $\begingroup$ So, can we say that, given any two vectors $u$ and $v$ as defined above, and a matrix $A \in \mathbb{R}^{n\times n}$ such that $v^TAu=0$, we can't have $A$ to be symmetric positive definite matrix but rather symmetric psd. $\endgroup$
    – user402940
    Dec 31 '16 at 11:58
  • $\begingroup$ No, for example $(0 \ \ 1)\pmatrix{2&1\\1&1}\pmatrix{1\\-1}=0$ with $\pmatrix{2&1\\1&1}$ pd (not psd). $\endgroup$
    – Jean Marie
    Dec 31 '16 at 12:07
  • $\begingroup$ So as per your example, $A$ is not projecting $(1 -1)^T$ on $v^\perp$ where $v=(0 1)^T$, but rather some other kind of operator. Can we find such a $A$ (pd matrix) given $u$ and $v$ ? $\endgroup$
    – user402940
    Dec 31 '16 at 12:26

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