Given the matrix:
$$
A =
\begin{bmatrix}
2 & 0 \\
1 & 4
\end{bmatrix}.
$$
The eigenvalues are:
$$λ_1 = 2,$$
$$λ_2 = 4.$$
To find the eigenvectors:
$$v_1 = \operatorname{nullity}(A – λ_1I),$$
$$v_2 = \operatorname{nullity}(A – λ_2I).$$
Which yielded the following eigenvectors for $v_1$:
$$
v_1 =
\begin{bmatrix}
-\frac{2}{3}\\
1
\end{bmatrix}.
$$
However, I have seen this alternative eigenvector for $v_1$:
$$
v_1 =
\begin{bmatrix}
-2\\
3
\end{bmatrix}.
$$
Question: Are both eigenvectors for $v_1$ correct? In other words, can you scale an eigenvector by any real number and it still be an eigenvector of that matrix?
This would make sense, as I believe we are only really interested in the direction (eigenvectors) and the factor by which the matrix scaled (eigenvalues). Thus, I would argue that the magnitude of the values within an eigenvector are meaningless. For example, the following eigenvector for $v_1$ is just as correct:
$$
v_1 =
\begin{bmatrix}
-44444\\
66666
\end{bmatrix}
$$
Which perhaps begs the question:
Why aren't these called eigenvalues and eigendirections?
Best Answer
A formal proof of this statement could be the following: Let $A$ be a square matrix with eigenvector $v$ and eigenvalue $\lambda$ (i.e. $Av = \lambda v$). We would like to show that $cv = x$ is also an eigenvector, where $c \neq 0$ is a constant. This follows from
$$Ax = A(cv) = c (Av) = c (\lambda v) = \lambda (cv) = \lambda x.$$
Note that the case $c = 0$ has to be excluded since $x = 0$ is by definition never an eigenvector.