Simplifying a matrix equality

Question

I want to know whether there exist any $Y$ and $Z$ satisfying this matrix equality: $$ \sum_{i=1}^N X_iaa^TX_i^T = Yaa^TZ $$ where $X_i$'s are matrices with appropriate dimensions and $a$ is a vector with appropriate dimension.

Answer 1

It depends on the matrices.

For example, if $ a a^t=\begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix}$, then $$ \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0\\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix} \begin{pmatrix} 0 & 0\\ 0 & 2 \end{pmatrix}= \begin{pmatrix} 1 & 0\\ 0 & 4 \end{pmatrix} $$ has rank $2$, while $Ya a^t Z$ has rank at most $1$. On the other hand, if $a=0$, then you can find such $Y,Z$.

Answer 2

Usually not, because the rank of the RHS is at most 1 but the LHS can have higher ranks. In particular, when the underlying field is real, each summand on the LHS is a positive semidefinite matrix of rank $\le1$. Hence the LHS has rank $\le1$ if and only if all vectors $X_ka$ are multiples of some common vector $v$.