Proof regarding the positive definite matrix

Question

Assume $y,s\in \mathbb{R}^n$ are such that $y^Ts > 0$.

Can we prove that there exists a symmetric and positive definite matrix $M \in \mathbb{R}^{n\times n}$ such that $$Ms= y$$

In particular, there exists a non-singular matrix $W \in \mathbb{R}^{n\times n}$ such that $WW^Ts = y$.

Answer 1

Let's first consider the case in which every entry except the first entry $s_1$ of $s$ is equal to zero. In this case we know that the first column of $M$ must be obtained by dividing the entries of $y$ by $s_1$. Choose the remaining entries of $M$ as follows: the first row of $M$ shall be the transpose of its first column; the remaining diagonal entries of $M$ shall be very, very large and positive; the remaining off diagonal entries of $M$ shall be zero. Clearly $M$ is symmetric. The 1,1 entry of $M$ is positive because of the condition $y^T s > 0$. And so Sylvester's Criterion tells us that $M$ is positive definite provided that the remaining diagonal entries are sufficiently large.

For the general case, note that the set consisting only of $s$ divided by its norm can be extended to an orthonormal basis of $\mathbb R^n$. So there's an orthogonal matrix $U$ with the property that $$ Us = \left[ \begin{array}{c} ||s|| \\ 0 \\ 0 \\ \vdots \\ 0 \end{array}\right]$$ Choose a symmetric positive definite matrix $M'$ such that $M' (Us) = Uy$. Then put $M=U^T M' U$.

Answer 2

Here is a simple idea that is messy to describe: The essence of the idea is that it can be reduced down to a two dimensional problem (the span of $s,y$) and that any pair of vectors in $\mathbb{R}^2$ that have positive inner product can be rotated so that both are strictly in the first quadrant.

Note that we can assume that $s,y$ are unit vectors.

First, deal with a simple case; if $s=y$, then the identity works, so we can suppose $s \neq y$.

Use Gram Schmidt starting with the vectors $s, y$ to produce an orthonormal basis $b_1,...,b_n$. Note that $\operatorname{sp} \{s,y\} = \operatorname{sp} \{b_1, b_2\} $.

Let $B = \begin{bmatrix} b_1 &\cdots & b_n \end{bmatrix}$ and note that $B^T B = I$.

Let $\tilde{s}, \tilde{y} \in \mathbb{R}^2$ be the coordinates of $s,y$ in terms of $b_1,b_2$, and note that $\langle \tilde{s}, \tilde{y} \rangle_{\mathbb{R}^2} = \langle s,y \rangle >0$.

Now find a rotation $R$ such that both $R\tilde{s}, R\tilde{y}$ lie in the interior of the first quadrant (all components are strictly positive). Let $\tilde{M} = \operatorname{diag} ({ [R\tilde{y}]_1 \over [R\tilde{s}]_1}, { [R\tilde{y}]_2 \over [R\tilde{s}]_2})$ and note that $\tilde{M}$ is positive definite and $\tilde{M} R\tilde{s} = R\tilde{y}$, or $(R^T \tilde{M} R ) \tilde{s} = \tilde{y}$.

Now let $M = B \begin{bmatrix} R^T \tilde{M} R & 0 \\ 0 & I\end{bmatrix} B^T$ and note that $M$ is positive definite.

Answer 3

Suppose $y$ and $s$ are linearly independent, otherwise an appropriate scalar matrix will do. Let $V$ be any matrix whose first two columns are respectively $y$ and $s$ and whose remaining columns are orthogonal to $y$ and $s$. Let $P$ be any positive definite matrix of the form $\pmatrix{y^Ts&\|s\|^2\\ \|s\|^2&\ast}\oplus C$. Then $M=VP^{-1}V^T$ is positive definite and $Ms=y$.