Stability of eigenvalues/singular values on altering the matrix From Strang's Introduction to Linear Algebra (p375), there is a paragraph on the instability of eigenvalues in relation to the stability of singular values when $A$ is altered slightly. Moreover, it says, that the instability of eigenvalues is only if $AA^T$ is far from $A^TA$, but if those are very similar, the eigenvalues are also stable. The book goes on to present a $4\times4$ example with one value altered by $1/60000$ which changes the eigenvalues by $1/10$ but singular value change is only $1/60000$.
To put what I understood from this claim and the example in definite terms, the eigenvalues of $A$ are seriously unstable if $A^TA$ is significantly different $AA^T$, i.e. on altering $a_{ij}$ in $A$ by $\Delta a_{ij}$, the $\lambda_A$ change by values much greater ($\Delta \lambda_A >> \Delta a_{ij}$) but singular values change in the same order ($\Delta \sigma_A \approx\Delta a_{ij}$).
The overarching question I have is, I do not understand, intuitively or mathematically, how this came to be
My attempts:
I tried to characterize $\Delta \lambda_A$ by looking at how the coefficient of the polynomial $det(A-\lambda_A I) = 0$ change when $a_{ij}$ becomes $a_{ij} + \Delta a_{ij}$. But if I were to compare this to the singular value case, I would be looking at the change in coefficients of the polynomial $det(A^TA - \sigma_A^2I) = 0$, but in $A^TA$, the $\Delta a_{ij}$ term would be actually present at more than one entries and it would seem that it would actually lead to a greater change in the coefficients! I also am not sure how would a change in coefficients be related to the change in the zeros of the polynomial.
Furthermore, why would the relation between $AA^T$ and $A^TA$ affect $\Delta \lambda_A$ rather than $\Delta \sigma_A$, as one would expect since the singular value is quite strongly associated with the eigenvectors of $AA^T$ and $A^TA$, but the eigenvalues of $A$ appears to not have a direct relation with them.
Any more clarity on the intuition of the three bold statements would be appreciated.
 A: Let $A\in M_n(\mathbb{R})$ and $\lambda$ be a SIMPLE real eigenvalue of $A$. Let $Ax=\lambda x$ where $||x||^2=1$. Then $x^Tx=1,x^T(\Delta x)=0$.
$(\Delta A)x+A(\Delta x)=(\Delta \lambda)x+\lambda (\Delta x)$ implies that 
(*) $x^T(\Delta A)x+x^TA(\Delta x)=\Delta \lambda$ and 
(*bis) $x^T(\Delta A)x=\Delta \lambda$ when $AA^T=A^TA$. 
Let $AA^T=S$ (a symmetric matrix) and $Sy=\sigma y$, where $||y||^2=1$. Then 
$(\Delta S)y+S(\Delta y)=(\Delta \sigma)y+\sigma (\Delta y)$ implies that 
(**) $y^T(\Delta S)y=\Delta \sigma$.
Then $\Delta \sigma$ has same order as $\Delta S=(\Delta A)A^T+A(\Delta A)^T$, that is, same order as $\Delta A$.
On the other hand, $\Delta \lambda$ has same order as $\max(order(\Delta A),order(\Delta x))$ and as $order(\Delta A)$ when $AA^T=A^TA$.
Finally a matrix $A$ with instable eigenvalue $\lambda$ is obtained when $\Delta x$ is very large. The Strang's example is
$A=\begin{pmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0\end{pmatrix}$ and $\Delta A=\begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&0&0\\1/60000&0&0&0\end{pmatrix}$.
Note that $0$ is not a simple eigenvalue; yet instability is further caused by the magnitude of $\Delta x$ (the unique eigenvector of $A$ explodes in $4$ eigenvectors that are far from the first one -when we change $a_{4,1}=0$ into $1/60000$-).
EDIT. To get the simplicity of the eigenvalues (as I suppose it), just choose $a_{4,1} = 10^{-10}$ (then there are $4$ distinct eigenvalues) and change it to $a_{4,1}=1/60000$.
Then check that the eigenvector associated with the $> 0$ eigenvalue "moves" quickly. It's your business.
