Uniqueness of the Convex Combination of Positive-Definite Matrices I am trying to connect the matrices $X$ and $Y$ with a curve defined by the convex combination of $X X^T$ and $Y Y^T$.
If I define $Z Z^T = c(X X^T) + (1-c) (Y Y^T), \ c \in [0,1]$, it is true that $ZZ^T = YY^T$ at $c=0$, but it is not necessarily true that $Z = Y$.
My question is the following: if I enforce that $Z = X$ at $c=1$, would this imply that $Z = Y$ at $c=0$? What if I impose some condition on the product $XY^T$?
 A: I suppose that you want to find some $Z=Z(c)$ so that


*

*$Z(c)$ is a continuous path,

*$Z(0)=Y$ and $Z(1)=X$,

*$Z(c)Z(c)^T=c(XX^T)+(1-c)(YY^T)$ and the product is positive definite along the path.


(I) We consider the case where $X, Y$ are square matrices first. In this case, the above conditions can be simultaneously satisfied if and only if $\det(XY)>0$.
Proof. When $\det(XY)<0$, the determinants of $X$ and $Y$ have different signs. Therefore, if conditions 1, 2 are both satisfied, then $\det Z(c)$, being a continuous function in $c$, must vanish at some $c'\in[0,1]$. Hence $Z(c')$ is singular and $Z(c')Z(c')^T$ is not positive definite.
When $\det(XY)>0$, let $X=AU$ and $Y=BV$ be their polar decompositions. Then $\det UV^T>0$ and hence $UV^T=e^K$ for some skew-symmetric matrix $K$. Let $\widetilde{Z}(c)$ be the positive definite square root of $c(XX^T)+(1-c)(YY^T)$, so that $\widetilde{Z}(c)$ is continuous and $\widetilde{Z}(0)=B,\ \widetilde{Z}(1)=A$. Define $Z(c)=\widetilde{Z}(c)e^{cK}V$. Then conditions 1-3 are satisfied. $\square$

(II) Next, we consider the case where $X,Y$ are $m\times n$ with $m<n$ (i.e. they have more columns than rows). Given that $X,Y$ have full row ranks, it is always possible to find $Z(c)$ that satisfies conditions 1-3.

Proof.
Let $X_1$ be a full-rank $(n-m)\times n$ matrix whose rows are orthogonal to the rows of $X$. Let $\mathbf X=\pmatrix{X\\ X_1}$. Note that, by negating the bottom row of $X_1$ if necessary, we may always pick an $X_1$ such that in the polar decomposition $\mathbf X=AU$, we have $U\in SO(n)$. Perform the analogous construction for $Y$ and call the resulting matrix $\mathbf Y$. Using the result of part I, we can construct a continuous path $\mathbf Z(c)$ that starts at $\mathbf Y$, ends at $\mathbf X$ and $\mathbf Z(c)\mathbf Z(c)^T$ is a convex combination of $\mathbf X\mathbf X^T$ and $\mathbf Y\mathbf Y^T$.
Note that $\mathbf X\mathbf X^T=\pmatrix{XX^T\\ &X_1X_1^T}$ and similarly for $\mathbf Y\mathbf Y^T$. It follows that their convex combination must be block diagonal too. So, if we partition $\mathbf Z(c)$ as $\pmatrix{Z(c)\\ Z_1(c)}$, where $Z(c)$ is $m\times n$, we must have $Z(c)Z(c)^T=cXX^T+(1-c)YY^T$ for every $c$. Therefore all the three conditions are now satisfied by $Z(c)$.
