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.

• Have you tried simple 2x2 examples, before asking? – Cosmas Zachos Jul 20 '18 at 21:47
• Yes. I tried small matrices and it seems that the Eigenvalues of $ADA^T$ are greater than eigenvalues of $AA^T$. But I can't prove this analytically. – Ben Jul 20 '18 at 22:01
• Note that $ADA^T = (A\sqrt{D})(A\sqrt{D})^T$, so you're trying to compare singular values of $A$ and $A\sqrt{D}$, where $\sqrt{D}$ is an essentially arbitrary diagonal matrix consisting of values greater than $1$. I haven't parsed it properly, but maybe this answer might help? math.stackexchange.com/a/1793562/248286 – Theo Bendit Jul 20 '18 at 22:47

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?

• Thanks for the answer Geronimo. I added a small point to the question: diagonal elements of the matrix $D$ are greater than 1. Would this change anything? – Ben Jul 20 '18 at 22:00
• The diagonal elements of D are the eigenvalues of D. I imagine this would have small scaling factor on the other matrices and you may be able to show that through some norm comparisons. I think when I wrote this down I wrote $A^{T}A$ however the eigenvalues of $A^{T}A$ are equal to $AA^{T}$ ... – Shogun Jul 20 '18 at 22:04