From the eigenvalue/eigenvector equation: \begin{equation} A \left|v\right> = \lambda \left|v\right> \end{equation} it is straightforward to show that if $\vert v\rangle$ is an eigenvector of $A$, then, any multiple $N\vert v\rangle$ of $\vert v\rangle$ is also an eigenvector since the (real or complex) number $N$ can pull through to the left on both sides of the equation.

Notice that for any vector $\vert v\rangle=\begin{pmatrix}a\\b\end{pmatrix}$, the operation: \begin{eqnarray} \langle v\vert v\rangle &=& \begin{pmatrix}a^* &b^*\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}\\ &=&\vert a\vert^2 + \vert b\vert^2 \end{eqnarray} always yields a positive, real number. Thus, we can use the square root of this operation to define the norm or length of the vector, $\vert \vert v\rangle\vert$. \begin{equation} \vert \vert v\rangle\vert=\left\{\langle v\vert v\rangle\right\}^{\frac{1}{2}} \end{equation}

It is always possible to choose the number $N$ above to find an eigenvector with length $1$. Such an eigenvector is called normalized.