If the origin is asymptotically stable for $z' = Az$, prove that there exists $δ > 0$ (depending on A) such that if $||A − B|| < δ$ then the origin is asymptotically stable for $z' = Bz$.
My thoughts: Since $z' = Az$ is stable we know that the real parts of the eigenvalues of A are all less than 0. I know that I need a delta that makes sure that the real part of the eigenvalues of B are not greater than 0. I can't figure out how to write this though mathematically.
On
Basically you need to show that the mapping "$A$ maps to the eigenvalue of largest real part of $A$" is continuous (as a function from $\mathbb{C}^{n \times n}$ to $\mathbb{C}$) in your norm.
An intuitive way to see this result for "most" matrices is to use the implicit function theorem; this lets you define this eigenvalue implicitly through the characteristic equation. But this only works if the eigenvalue of second highest real part of $A$ has a strictly smaller real part than the highest one.
Elsewhere you need to be a bit more clever. One way is to change $B$ to the basis in which $A$ is diagonal and then apply Gerschgorin's theorem. But then this only works when $A$ is diagonalizable. So for instance neither of these approaches work for $A=\begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}$. But there is yet another way out of that, which I will leave for you to figure out.
The characteristic polynomial $P_A(\lambda) = \det(A - \lambda I)$ is a polynomial in the entries of the matrix $A$, and in particular is a continuous function of $\lambda$ and those entries. Let $\Gamma$ be a closed positively oriented contour in the complex plane such that all eigenvalues of $A$ are inside $\Gamma$. If $\|B-A\|$ is small enough, $|P_B(z) - P_A(z)| < |P_A(z)|$ for all $z$ on $\Gamma$. By Rouché's theorem, $P_B$ and $P_A$ have the same number of roots inside $\Gamma$ (counted by multiplicity), and thus all eigenvalues of $B$ are inside $\Gamma$.