## Section4.4Normalization of Eigenvectors

From the eigenvalue/eigenvector equation:

$$A \left|v\right> = \lambda \left|v\right>\tag{4.4.1}$$

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

$$\vert \vert v\rangle\vert=\left\{\langle v\vert v\rangle\right\}^{\frac{1}{2}}\tag{4.4.3}$$

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