Every positive definite matrix can be written as $B^TB$ for some invertible $B$ 
Let $A$ be a positive definite symmetric matrix. Show that there exists an invertible matrix $B$ such that $A=B^TB$. [Hint: Use the Spectral Theorem to write $A = QDQ^T$. Then show that D can be factored as $C^TC$ for some invertible matrix $C$.]

I can't seem to get to the correct answer. I'm not entirely sure what is meant by the last line of the hint too. Could anyone please help me out?
 A: Do you know Cholesky Decomposition? You can even chose $B$ as a triangular matrix.
If you don't like an abstract proof you just can compute it, for example 
$$\begin{pmatrix} l_{11} & 0 \\
l_{21} & l_{22} \\ \end{pmatrix}\cdot \begin{pmatrix} l_{11} & l_{21}\\
0 & l_{22} \\ \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12}\\ a_{12} & a_{22} \\ \end{pmatrix}$$
when you chose $l_{ii}>0$ you will always have $\frac{n(n+1)}{2}$ equations and unknowns, hence a unique solution exists.
To get the Cholesky decomposition from the eigenvalue decomposition we use an lu dcomposition, as $Q$ is invertible we know that $Q=L\cdot U$ for some normed lower triangular matrix $L$, and some upper triangular matrix $U$.
So 
$$A= Q^T  D Q = U^T L^T  D L U $$
As $U^T$ and $U$ are invertible we have
$$U^{T^{-1}} A U^{-1}=L^T  D L $$
We see that 
$$U^{T^{-1}} A U^{-1}$$
is symmetric and positiv definit iff $A$ is symmetric and positiv definit, and we can get any positive definite matrix, hence we get the cholesky decomposition.
A: $A$ is symmetric, so it can be written as $A=QDQ^T$ (Eigenvalue decomposition of a symmetric matrix), where $D$ is a diagonal matrix with real elements on main diagonal.
$A$ is PSD, so elements on main diagonal of $D$ are nonnegative, so they have square root. Let $D= D_1D_1^T$, where $D_1$ is also diagonal.
You can write $C^T= QD_1$. You will have $A=C^TC$.
A: Let 
$$
D=\mbox{diag}(\lambda_1, \ldots,\lambda_n) \quad\mbox{and}\quad U=\mbox{diag}(\sqrt{\lambda_1}, \ldots,\sqrt{\lambda_n}).
$$
So check it yourself $B=QU$. 
