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Let $A$ be an $n\times n$ symmetric positive definite matrix and let $B$ be an $m\times n$ matrix with $\mathrm{rank}(B)= m$. Show that $BAB'$ is symmetric positive definite.

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closed as off-topic by 6005, Carl Mummert, msteve, Mnifldz, hardmath Aug 11 '15 at 20:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

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Please try to not make your post sound like you are giving us an assignment. – sxd Sep 30 '11 at 23:16
The correct spelling is poove. – Will Jagy Sep 30 '11 at 23:50

The result is symmetric since $(BAB')'= BAB'$

Case 1: $m<n$

$$\begin{bmatrix} .&. &. &. \end{bmatrix} \begin{bmatrix} . &. &. &.\\. &.&.&.\\. &.&. &.\\.&.&.&. \end{bmatrix} \begin{bmatrix} .\\.\\.\\. \end{bmatrix} $$ This matrix is always positive definite since you can always denote the outer factors as $B'x = y$ and since $A$ is positive definite $y'Ay$ > 0

Case 2: $m=n$

$$ \begin{bmatrix} . &. &. &.\\. &.&.&.\\. &.&. &.\\.&.&.&. \end{bmatrix} \begin{bmatrix} . &. &. &.\\. &.&.&.\\. &.&. &.\\.&.&.&. \end{bmatrix} \begin{bmatrix} . &. &. &.\\. &.&.&.\\. &.&. &.\\.&.&.&. \end{bmatrix}' $$ We obtain $BAB'= C$. If $B$ is invertible then this is called congruence transformation. From Sylvester's Law of Inertia, the number of positive, zero and negative eigenvalues of C are equal to of $A$. Therefore, if $A$ is positive definite, so is $C$. Since matrix $B$ is full rank, it is invertible. If $B$ was not invertible, then there exists a nonzero $x\in\mathbb{R}^n$ such that $B'x=0$ Thus, $BAB'$ becomes only positive semi definite.

Case 3: $m>n$

Then $B$ cannot have a rank of $m$ but suppose the question was modified and only B is full rank is given. $$ \begin{bmatrix} . &. &. &.\\. &.&.&.\\. &.&. &.\\.&.&.&.\\.&.&.&.\\.&.&.&. \end{bmatrix} \begin{bmatrix} . &. &. &.\\. &.&.&.\\. &.&. &.\\.&.&.&. \end{bmatrix} \begin{bmatrix} . &. &. &.&.&.\\. &.&.&.&.&.\\. &.&. &.&.&.\\.&.&.&.&.&. \end{bmatrix} $$ Similarly, this case leads only to a positive semidefinite product. The quickest way is to observe that The product is a matrix of $m\times m$ and the rank of this matrix is at max $n$ since we cannot arrive to a full rank matrix with a product of elements that have ranks less than the resulting matrix.You can think of the tensor(or outer) product of two vectors $a\otimes b = ab'$.

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Take $\xi$ different from zero and show that $\xi' B A B' \xi$ is always positive. If not, then $\mathrm{rank}(B) < m$.

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