## Section5.4Normalization of Eigenvectors

From the eigenvalue/eigenvector equation:

\begin{equation} A \left|v\right> = \lambda \left|v\right>\tag{5.4.1} \end{equation}

it is straightforward to show that if $\vert v\rangle$ is an eigenvector of $A\text{,}$ 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 $\vert v\rangle=\begin{pmatrix}a\\b \end{pmatrix}\text{,}$ the operation:

\begin{align} \langle v\vert v\rangle \amp = \begin{pmatrix}a^* \amp b^* \end{pmatrix} \begin{pmatrix}a\\b \end{pmatrix}\notag\\ \amp = \vert a\vert^2 + \vert b\vert^2\tag{5.4.2} \end{align}

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, $\vert \vert v\rangle\vert\text{.}$

\begin{equation} \vert \vert v\rangle\vert=\left\{\langle v\vert v\rangle\right\}^{\frac{1}{2}}\tag{5.4.3} \end{equation}

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