Which directions stay on the same line?

Source scope: §3.6. Worked numerical examples, experiments, and practice are teaching supplements unless identified as source results.

Direction can survive a transformation

For $A=\operatorname{diag}(2,-1)$, the horizontal vector $(1,0)$ becomes $(2,0)$; the vertical vector $(0,1)$ becomes $(0,-1)$. Both remain on their original lines. A negative scale reverses direction.

For a square real matrix $A$, a nonzero vector $x$ is an eigenvector with eigenvalue $\lambda$ if $Ax=\lambda x$. Zero is excluded because $A0=\lambda0$ for every $\lambda$. A zero eigenvalue is allowed: a nonzero direction can collapse to zero.

Not every vector is an eigenvector. Here $A(1,1)=(2,-1)$ is not a scalar multiple of $(1,1)$. A real matrix need not have real eigenvectors: a quarter-turn rotation has no nonzero real vector on an unchanged line.

Proposition 3: shift, invert, repeat

Source statement and proof. Suppose $Ax=\lambda x$ with $x\ne0$.

1. For any real $\gamma$, $(A+\gamma I)x=Ax+\gamma x=(\lambda+\gamma)x$. The same eigenvector has shifted eigenvalue $\lambda+\gamma$.
2. If $A$ is invertible, $x=A^{-1}Ax=\lambda A^{-1}x$. Invertibility forces $\lambda\ne0$, so division gives $A^{-1}x=\lambda^{-1}x$.
3. For every nonnegative integer $k$, $A^kx=\lambda^kx$. The case $k=0$ uses $A^0=I$; multiplying the $k$th relation by $A$ proves the next. If $A$ is invertible, applying the same argument to $A^{-1}$ proves the relation for negative integers too.

Source clarification: Proposition 3(iii) prints $k\in\mathbb Z$ without repeating the invertibility condition. Negative powers require it. For $k=0$, the scalar factor is one even when $\lambda=0$, by the identity-power convention.

Transfer

If $Ax=-2x$, what is $(A+3I)x$? What is $A^2x$?