What does it mean, geometrically speaking to find the limit of a matrix raised to power? I know from 3B1B's videos that a matrix essentially refers to rotating and skewing graphs, with the matrix giving us the new coordinates of the unit vectors. So, if
$$A=
\begin{bmatrix}
a& 0  \\
0 & b \\
\end{bmatrix}
$$,
with $a>1$ and $b>1$
then we have that ${\lim \limits_{n \to \infty}(A^n)}^{-1}=O$.
Where 'O' is the null matrix
Therefore, the entire system collapses onto the origin in the limiting case.
However, I'm unable to visualize all intermediate points of this transformation. Is there any program that will allow me to do this and is there any way to do this manually?
In short, what would the equivalent of the graphical representation of such a transformation look like?
 A: I won’t share many programs for visualizing this, mostly because that depends on what exactly about the limit you want to visualize. Mathematica is useful for sketching diagrams. I’ve never used SageMath, but that is a free alternative.
Here’s a way to think about the situation manually.
A matrix $A: \mathbb{R}^2 \to \mathbb{R}^2$ is determined entirely by where it sends $e_1 = \binom{1}{0}$ and $e_2 = \binom{0}{1}$:
$$
  A\left(x e_1 + y e_2\right) = \begin{pmatrix}
    a & b\\
    c & d
  \end{pmatrix}\begin{pmatrix}
    x \\ y
  \end{pmatrix} = (ax + by) e_1 + (cx + dy)e_2.
$$
In the case of your diagonal matrix $ A = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$, we see that:
$$
  A(x e_1 + y e_2) = (ax)e_1 + (by) e_2.
$$
$A$ acts by multiplying the coordinates independently. Therefore $A^n (x e_1 + y e_2) = (a^n x) e_1 + (b^n y) e_2$. The inverse of $A^n$ is the matrix which sends $(a^n x) e_1 + (b^n y) e_2$ to the starting input: $x e_1 + y e_2$. We see that we need to divide the first coordinate by $a^n$ and the second by $b^n$:
$$
  \left(A^n\right)^{-1} = \begin{pmatrix} 
    a^{-n} & 0 \\
    0 & b^{-n}
  \end{pmatrix}
$$
In the comments you mention that $|a| > 1$ and $|b| > 1$, therefore $\lim_{n\to\infty} a^{-n} = \lim_{n\to\infty} b^{-n} = 0$. Thus, as we let $n \to \infty$, the entries of $(A^n)^{-1}$ go to zero. This determines what the limit matrix does to $e_1$ and $e_2$: since all the entries go to zero:
$$
  \lim_{n\to\infty} \left( A^n \right)^{-1} = 0.
$$
Alternatively, since we know what the entries of $(A^n)^{-1}$ are, we can see what the intermediate points are:
$$
  (A^n)^{-1} (x e_1 + y e_2) = \left(\frac{x}{a^n} e_1 + \frac{y}{b^n} e_2\right).
$$
Visually, $(A^n)^{-1}$ shrinks the horizontal axis by $a^n$ and the vertical axis by $b^n$. In the limit, both axes are shrunk by a factor of $\infty$. That is to say, the limit matrix sends every point to the origin.
