Can diagonalization of a positive definite matrix be used to compute its determinant and inverse? I have a positive definite matrix $A= E^T DE$. $A\in \mathcal{R}^{n \times n} $. 
$E$ is a orthonormal matrix $E \in \mathcal{R}^{m \times n}$ consisting of n orthonormal columns where $m >n $. And, $E^T E=I_{n \times n}$ (this follows directly from the orthonormality). 
Also, $D \in \mathcal{R}^{m \times m}$ is a diagonal matrix. 
Notice that this diagonalization is somewhat similar to the singular value decomposition(svd)but not quite as svd requires $A=udv^t$ where $u$ has orthonormal columns not rows as is the case here.  
Given, $E$ and $D$, is it possible to compute the inverse and determinant of A? 
 A: I don't see any simple method for computing the determinant. What I can confirm is that there is no function for $\det A$ purely in terms of $D$; the entries in $E$ must also be considered in any formula. To prove this, all we need is a diagonal matrix, and two orthogonal $m \times n$ matrices $E_1, E_2$ such that $\det A_1 \neq \det A_2$, where $A_i = E_i^T D E_i$.
Let
\begin{align}
D &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \\
E_1 &= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} \\
E_2 &= \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 1 \end{pmatrix}.
\end{align}
Then,
\begin{align}
A_1 &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} \\
&= \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \\
A_2 &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 1 \end{pmatrix} \\
&= \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix},
\end{align}
which have respective determinants $2$ and $3$.
