In [1, p. 34 exercise 2.10] it is asked to prove that if \mathbf{A} is a complex n \times n positive definite matrix and \mathbf{B} is a full rank complex m \times n matrix with m \leq n, then \mathbf{BA}\mathbf{B}^H is also positive definite.
Solution:
Because the matrix \mathbf{A} is positive definite for any \mathbf{y}\neq 0:
Q_{1}=\mathbf{y}^H\mathbf{A}\mathbf{y} > 0. (1)

In order to prove that the matrix \mathbf{BA}\mathbf{B}^H is also positive definite we have to show that Q_{2}=\mathbf{x}^H\mathbf{BA}\mathbf{B}^H\mathbf{x} > 0 for any \mathbf{x} \neq \mathbf{0}. Thus it is sufficient to show that \mathbf{z}=\mathbf{B}^H\mathbf{x}\neq \mathbf{0} for any \mathbf{x} \neq \mathbf{0}, because then due to (1) the positive defineteness of the matrix \mathbf{B}\mathbf{A}\mathbf{B}^H follows. Let \mathbf{e}_i be the standard basis in \mathbb{C}^m. Then any vector in \mathbb{C}^m may be written as a linear combination of the standard base \mathbf{v}_k=\sum_{i=1}^{m}r_{ik} \mathbf{e}_i. Let T:\mathbb{C}^m \rightarrow \mathbb{C}^n be the linear transformation associated to the matrix \mathbf{B}^H. Because rank(\mathbf{B}^H)=m the dimension spanned by any transformed linear independent base will equal m (definition of the rank of a matrix [2, p. 216]), thus T(\mathbf{e}_i),i=1,...,m will be linear independent [2, propsition 4.3, p. 112] and the transform is injective which means that for \mathbf{v}_1,\mathbf{v}_2 \in \mathbb{C}^m with T(\mathbf{v}_1)=T(\mathbf{v}_2) \Rightarrow \mathbf{v}_1 =\mathbf{v}_2. The proof of this statement can be derived by the following reasoning:
T(\mathbf{v}_1)=T(\mathbf{v}_2)
T(\sum\limits_{i=1}^{m}r_{i1} \mathbf{e}_i)=T(\sum\limits_{i=1}^{m}r_{i2} \mathbf{e}_i)
\sum\limits_{i=1}^{m}r_{i1} T(\mathbf{e}_i)=\sum\limits_{i=1}^{m}r_{i2} T(\mathbf{e}_i)
\sum\limits_{i=1}^{m}(r_{i1}-r_{i2}) T(\mathbf{e}_i)=\mathbf{0} \Rightarrow
r_{i1}-r_{i2}=0 \Leftrightarrow (2)
\mathbf{v}_1=\mathbf{v}_2

Relation (2) is obtained due to the linear independence of the vectors T(\mathbf{e}_i). From this it follows that only the zero vector \mathbf{x}=\mathbf{0} maps to \mathbf{z}=T(\mathbf{x})=\mathbf{B}^H\mathbf{x}=\mathbf{0}=T(\mathbf{0}). Furthermore for any \mathbf{x} \neq \mathbf{0} \mathbf{z}=\mathbf{B}^H\mathbf{x} \neq \mathbf{0} which proofs that the matrix \mathbf{BA}\mathbf{B}^H is positive definite. QED.

[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.
[2] Lawrence J. Corwin and Robert H. Szczarba: “Calculus in Vector Spaces”, Marcel Dekker, Inc, 2nd edition, ISBN: 0824792793.