Can you factor out vectors? My prof introduced eigenvalues to us today:

Let $A$ be an $n \times n$ matrix. If there a scalar $\lambda$ and an $n-1$ non-zero column vector $u$, then
$$Au = \lambda u$$
then $\lambda$ is called an eigenvalue and $u$ is called an eigenvector.
$$Au - \lambda u = 0$$
$$\implies (A - \lambda I)u = 0$$
$$\implies \det(A - \lambda I) = 0$$

How did he get from $Au - \lambda u = 0$ to $\implies (A - \lambda I)u = 0$? It looks like he factored out the vector, but I thought you could only factor out constants? If you can factor out vectors can you explain why?

"Factoring" in this case is just applying the distributive property in reverse, and since the distributive property holds for matrix multiplication, this is indeed possible.
The one quirk that still might be an impediment for you is that the scalar multiplication $\lambda u$ is identical with the matrix product $(\lambda I) u$. Once this is known, what you wrote is just an application of the distributive law.
