Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this 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

2 Answers 2

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 $a\otimes b = ab'$.

share|improve this answer

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

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.