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.
symmetric positive definite matrix question
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3The correct spelling is poove. – 2011-09-30
2 Answers
The result is symmetric since (BAB')'= BAB'
Case 1: $m
$\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 $\mathrm{rank}(B) < m$.