symmetric positive definite matrix question 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 $BAB&#39;$ is symmetric positive definite.
The result is symmetric since $(BAB&#39;)&#39;= BAB&#39;$

Case  1: $m&lt;n$

$$\begin{bmatrix}
.&amp;. &amp;. &amp;.
\end{bmatrix}
\begin{bmatrix}
. &amp;. &amp;. &amp;.\\. &amp;.&amp;.&amp;.\\. &amp;.&amp;. &amp;.\\.&amp;.&amp;.&amp;.
\end{bmatrix}
\begin{bmatrix}
.\\.\\.\\.
\end{bmatrix}
$$
This matrix is always positive definite since you can always denote the outer factors as $B&#39;x = y$ and since $A$ is positive definite $y&#39;Ay$ > 0

Case 2: $m=n$

$$
\begin{bmatrix}
. &amp;. &amp;. &amp;.\\. &amp;.&amp;.&amp;.\\. &amp;.&amp;. &amp;.\\.&amp;.&amp;.&amp;.
\end{bmatrix}
\begin{bmatrix}
. &amp;. &amp;. &amp;.\\. &amp;.&amp;.&amp;.\\. &amp;.&amp;. &amp;.\\.&amp;.&amp;.&amp;.
\end{bmatrix}
\begin{bmatrix}
. &amp;. &amp;. &amp;.\\. &amp;.&amp;.&amp;.\\. &amp;.&amp;. &amp;.\\.&amp;.&amp;.&amp;.
\end{bmatrix}&#39;
$$
We obtain $BAB&#39;= C$. 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 $B&#39;x=0$ Thus, $BAB&#39;$ becomes only positive semi definite. 

Case 3: $m&gt;n$

Then $B$ cannot have a rank of $m$ but suppose the question was modified and only B is full rank is given. 
$$
\begin{bmatrix}
. &amp;. &amp;. &amp;.\\. &amp;.&amp;.&amp;.\\. &amp;.&amp;. &amp;.\\.&amp;.&amp;.&amp;.\\.&amp;.&amp;.&amp;.\\.&amp;.&amp;.&amp;.
\end{bmatrix}
\begin{bmatrix}
. &amp;. &amp;. &amp;.\\. &amp;.&amp;.&amp;.\\. &amp;.&amp;. &amp;.\\.&amp;.&amp;.&amp;.
\end{bmatrix}
\begin{bmatrix}
. &amp;. &amp;. &amp;.&amp;.&amp;.\\. &amp;.&amp;.&amp;.&amp;.&amp;.\\. &amp;.&amp;. &amp;.&amp;.&amp;.\\.&amp;.&amp;.&amp;.&amp;.&amp;.
\end{bmatrix}
$$
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 $a\otimes b = ab&#39;$.
Take $\xi$ different from zero and show that $\xi&#39; B A B&#39; \xi$ is always positive. If not, then $\mathrm{rank}(B) &lt; m$.