Diagonalization

You now have two facts sitting next to each other, and putting them together is the point of Act IV.

From step 18: changing basis changes how a transformation looks. From step 19: every transformation has directions it does not rotate.

So stop fighting it. Use the eigenvectors as the basis.

Doing it

Our matrix again, with eigenvectors $(1,1)$ at eigenvalue 3 and $(1,-1)$ at eigenvalue 1:

$$A = \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix}$$

Build $P$ from those eigenvectors as columns, exactly the change-of-basis matrix from step 18:

$$P = \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix}$$

Now translate $A$ into that language with the round trip you already know:

$$P^{-1} A P = \begin{bmatrix} 3 & 0 \ 0 & 1 \end{bmatrix}$$

Every off-diagonal entry is gone. In the eigenbasis, the transformation is nothing but "multiply the first coordinate by 3, the second by 1". No mixing, no rotation, no shear.

The messy matrix was never messy. You were describing a pure stretch in a language badly suited to it.

Why anyone bothers

Try computing $A^{100}$ directly and you will multiply matrices until you die. In the eigenbasis it is trivial, since a diagonal matrix raised to a power just raises each entry:

$$D^{100} = \begin{bmatrix} 3^{100} & 0 \ 0 & 1 \end{bmatrix}$$

so $A^{100} = P D^{100} P^{-1}$. Repeated application is the whole game in Markov chains, population models, and every system that evolves step by step — and eigenvalues make it a single exponent.

The eigenvalues also tell you the long-run story at a glance. Any component along $(1,1)$ grows by 3 each step and dominates; anything along $(1,-1)$ stays put. After many steps almost everything points along the largest eigenvalue's direction.

One catch: this needs enough independent eigenvectors to form a basis, and not every matrix has them. But one family always does, and it is the family that matters most.