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?

The distributive laws apply to matrix (or matrix-vector) multiplication.
$$\eqalign{(A+B) C &= AC + BC\cr
           A(C+D) &= AC + AD\cr}$$
whenever $A$,$B$,$C$,$D$ have the right dimensions for these to make sense.
