Let \begin{equation*} A = \begin{bmatrix} 0.4 & -0.3\\ 0.4 & 1.2 \end{bmatrix} \end{equation*} a) I found the eigenvalues to be $1$ and $0.6$, with corresponding eigenvectors $(-1/2,1)$ (for $1$) and $(-3/2,1)$ (for $0.6$).
b) Is A diagonalizable? I said yes because there are two linearly independent eigenvectors.
c) Express $x_0=(-1,-2)$ as a linear combination of eigenvectors of A (I did this, you solve it and get $(-4,-2)$.
d) Use you answer to (b) to describe the long-term behavior or $A^nx_0$, that is predict the value of $A^nx_0$ when $n$ is large. (I did this, and since one eigenvalue is $<1$, and the other is $1,$ you end up with just the vector $(2,-4)$.
e) How would your answer to (d) change if we used a different initial vector $x_0$? This is where I'm stuck. I know that the component of $x_0$ in the direction of the eigenvector of $1$ remains constant and the component in the direction of the eigenvector of $0.6$ keeps getting smaller and eventually fades away entirely, but how do I answer that directly to the question? How do I frame the answer in other words?
Since there are distinct real eigenvalues $\lambda_1$ and $\lambda_2$, $A$ can be written as $\lambda_1P_1+\lambda_2P_2$, where $P_1$ and $P_2$ are projections onto the respective eigenspaces such that $P_1P_2=P_2P_1=0$. With this decomposition, we have $A^n=\lambda_1^nP_1+\lambda_2^nP_2$. Since $\lambda_1=1$ and $|\lambda_2|\lt1$, then $$\lim_{n\to\infty}A^n=P_1$$ that is, in the long term, $A^nx_0$ approaches the projection of $x_0$ onto the eigenvector of $1$. Basically, under repeated multiplication by $A$, the component of $x_0$ in the direction of the eigenvector of $1$ remains constant and the component in the direction of the eigenvector of $0.6$ keeps getting smaller and ventually fades away entirely.