Eigenvalues and Eigenvectors of $AA^T$ vs $ADA^T$ Assume $D$ is a diagonal matrix. Are the Eigenvalues and Eigenvectors of $AA^T$ and $ADA^T$ related in anyway i.e. are the Eigenvectors same, Eigenvalues same, or is their any spectral relationship between these matrices? Also, the diagonal elements of the matrix $D$ are all greater than 1.
PS: I came across this when trying to prove something related to the operator norms of the two matrices. Any suggestions or references would be very helpful. Thanks.
 A: If $ A\in \mathbb{C}^{n \times n}$ 
We have that 
$$A^{T}A = (U \Sigma V^{T})^{T}(U \Sigma V^{T}) = (V^{T})^{T}\Sigma^{T} U^{T}(U\Sigma V^{T}) $$
now we know that $U^{T}U= I $
$$V \Sigma^{T} U^{T} \Sigma V^{T} = V \Sigma^{T} \Sigma V^{T} $$
and 
$ \Sigma^{T} \Sigma = \Sigma^{2} $
$$ A^{T}A = V \Sigma^{2} V^{T} = V \Lambda V^{T}$$
Where $ \Lambda $ is the matrix of eigenvalues $ VV^{T} =  V^{T} V = I$ 
Now 
$$ ADA^{T} = (U \Sigma V^{T})D(U \Sigma V^{T})^{T}  $$
$$ AD A^{T} = (U\Sigma V^T)D(V \Sigma U^{T}) $$
I can't come up any obvious connections other then $D=I$ then the eigenvalues are the same
I can only think of a relation perhaps 
$$A^{T}A = V \Lambda V^{T} $$
$$ \|V \Lambda V^{T} \|   \leq \| V \| \Lambda \| \| V^{T} \|$$
$$ \| A^{T}A\|  \leq \| \Lambda\| =\max_{1 \leq i \leq n} | | \lambda_{i}|$$
now suppose $D= cI$ where $c$ is a scalar 
$$ A^{T}D A = (V\Sigma^{T}U^{T} cI U\Sigma V^{T} $$
$$ \| A^{T}DA \| \leq \| V\| \Sigma^{T} \| \|U^{T} \| \|cI \| \|  U\| \| \Sigma \| \|V^{T} \|$$
$$ \| A^{T} D A  \| \leq  \|\Sigma \|  \| cI\| \| \Sigma \| $$
Now $ \Lambda = \Sigma^{2}$
$$\| A^{T} D A  \| \leq \max_{1 \leq i \leq n} | |c| \max_{1 \leq i \leq n} |\sigma_{i}|  = |c| \max_{1 \leq i \leq n} |\lambda_{i}| $$
The eigenvalues are scaled up in this case by the c. Or you can observe at the least the maximum eigenvalue will be?
