What is the upper bound on the error of the solution of a noise perturbed linear system $Mx=b$? Let $x$ be solution to the following linear system:
$$ Mx = b$$
and let $ \tilde{x}$ be the solution to the above linear system with some additive noise:
$$ M \tilde{x}= \tilde{b}$$
where $\tilde{b} = b + e$
let $\sigma_{min}(M)$ denote the smallest singular value of matrix $M$.
I was trying to verify rigorously why:
$$ \| x - \tilde{x} \| \leq \frac{ \| b - \tilde{b} \|}{\sigma_{min}(M)}$$ 
was true. However, I was unsure how to do it correctly.
This is what I have tried:
Notice that (if we assume a bound on the condition number, hence invertibility of M), then M is invertible and we get:
$$ \tilde{x}= M^{-1}\tilde{b} = M^{-1}(b+e) = x+M^{-1}e$$
Now $\| x - \tilde{x} \|$ is:
$$\| x - \tilde{x} \| = \| M^{-1}e \|$$
To relate the singular values of M to the equation above, lets bring in its SVD, $M^{-1} = (U \Sigma V^T)^{-1} = V \Sigma^{-1} U^T$:
$$\| x - \tilde{x} \| = \| M^{-1}e \| = \| V \Sigma^{-1} U^T e \| = \| \sum^{r}_{j=1} \frac{1}{\sigma_j} v_j u_j^T e\|$$
By triangle inequality:
$$ \| \sum^{r}_{j=1} \frac{1}{\sigma_j} v_j u_j^T e\| \leq \sum^{r}_{j=1} \| \frac{1}{\sigma_j} v_j u_j^T e\| $$
After this point a got a little stuck, however, I had a few ideas that looked promising but wasn't sure how to use them. It feels that if the singular vectors are orthonormal, then taking the norm yields 1 which together with the triangle inequality again could yield an upper bound of the sum of the reciprocal of the singular values? Seems like a true statement, however, might this upper bound to lose and its just the wrong direction for the proof?

To continue from the stuck-point, since $v_j$'s are orthonormal, the Pythagorean theorem gives
$$
\left\|\sum_{j=1}^r\frac{1}{\sigma_j}v_ju_j^Te\right\|_2^2=
\left\|\sum_{j=1}^r\frac{u_j^Te}{\sigma_j}v_j\right\|_2^2
=\sum_{j=1}^r\frac{|u_j^Te|^2}{\sigma_j}\leq\frac{1}{\sigma_{\min}(M)}\sum_{j=1}^r|u_j^Te|^2.
$$
The last touch is due to 
$$
\sum_{j=1}^r|u_j^Te|^2=\|U^Te\|_2^2\leq\|U\|_2^2\|e\|_2^2\leq\|e\|_2^2.
$$
