# symmetric positive definite matrix question [closed]

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.

-

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

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

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 6005, Carl Mummert, msteve, Mnifldz, hardmath
If this question can be reworded to fit the rules in the help center, please edit the question.

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 $a\otimes b = ab'$.

-

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

-