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$X^{T}$ is the transpose of $X$. $A$ is a $n$ x $n$ matrix and $B$ is a $m$ x $m$ matrix, $m$ > $n$, both of them are known, $A$ is positive definitive and $B$ is symmetric. I would like to find $X$.

george7
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    The equality is contradictory. The rank of $XAX^T$ is at most $n$ while the rank of $B$ is $m$. – Git Gud Jun 02 '15 at 09:40
  • Thank you very much for your answer. What you say is correct, however I would like to find $X$, when approximating $B$ with $XAX^{T}$... – george7 Jun 02 '15 at 12:01
  • However, if we consider that only $A$ is positive definite, how could this be solved? Thank you in advance. – george7 Jun 02 '15 at 12:05
  • @GitGud I don't understand why the equality is contradictory? If we take $\mathbf{A} = \mathbf{I}{3\times 3}$, where $\mathbf{I}{n \times n}$ is the $n\times n$ identity matrix, and we have: $$\mathbf{B} = \begin{pmatrix}3 & 3 & 3 & 3 \ 3 & 3 & 3 & 3 \ 3 & 3 & 3 & 3 \ 3 & 3 & 3 & 3\end{pmatrix}$$, then we have that: $$\mathbf{X} = \begin{pmatrix}1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1\end{pmatrix}$$ In this case we have $n = 3$, $m = 4$, $\mathbf{A}$ is positive definite and $\mathbf{B}$ is symmetric? – Thomas Russell Jun 02 '15 at 12:14
  • @Shaktal $B$ isn't positive definite, it's semi-positive definite. – Git Gud Jun 02 '15 at 12:42
  • @GitGud But the question only states that $\mathbf{A}$ has to be positive definite, which the identity matrix is? – Thomas Russell Jun 02 '15 at 12:45
  • @Shaktal I actually changed the question after GitGud's initial comment, initially I had stated that both matrices are positive definite. I basically need only $A$ to be positive definite and $B$ symmetric. – george7 Jun 02 '15 at 12:56
  • If $A$ is positive definite, then $B$ must also be positive semi-definite, otherwise taking the determinant of both sides would lead to a contradiction. You may want to look at this question. – Victor Liu Jun 02 '15 at 17:02
  • @VictorLiu Thank you very much for your answer. I had already had a look at the question you referred to before submitting my question. In the question you mentioned, unfortunately no transpose appears and $X$ is a square matrix, which is not in our case. – george7 Jun 02 '15 at 17:40

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EDIT. Note that if $A$ is real symmetric $\geq 0$, then necessarily $B$ is real symmetric $\geq 0$ with $rank(B)\leq rank(A)$.

Thus we assume that $A,B$ are symmetric $\geq 0$ matrices that satisfy $rank(B)=r\leq rank(A)=s\leq n<m$. There are invertible matrices $P\in M_m,Q\in M_n$ s.t. $A=Qdiag(I_s,0_{n-s})Q^T,B=Pdiag(I_r,0_{m-r})P^T$and therefore $P^{-1}XQdiag(I_s,0)(P^{-1}XQ)^T=diag(I_r,0)$. Since we know $P,Q$, It suffices to find the $Y=P^{-1}XQ\in M_{m,n}$ s.t. $Ydiag(I_r,I_{s-r},0_{n-s})Y^T=diag(I_r,0_{s-r},0_{n-s},0_{m-n})$. With respect to the previous block matrices, let $Y=[Y_{i,j}],1\leq i\leq 4,1\leq j\leq 3$. By identification of the diagonal elements of RHS and LHS , $Y_{2,1}Y_{2,1}^T+Y_{2,2}Y_{2,2}^T=0$; that implies $Y_{2,1}=0,Y_{2,2}=0$. In the same way, $Y_{3,1}=0,Y_{3,2}=0$ and $Y_{4,1}=0,Y_{4,2}=0$. Finally $Y=\begin{pmatrix}Y_{1,1}&Y_{1,2}&Y_{1,3}\\0&0&Y_{2,3}\\0&0&Y_{3,3}\\0&0&Y_{4,3}\end{pmatrix}$ where $Y_{1,1}Y_{1,1}^T+Y_{1,2}Y_{1,2}^T=I_r$.