An $n$-square complex matrix $\mathbf{A}$ is normal iff it is orthogonally diagonalisable or unitarily diagonalisable, i.e., there exists a unitary matrix $\mathbf{U}$ such that

$\mathbf{A} = \mathbf{U}\text{diag}(\lambda_1,\ldots,\lambda_n)\mathbf{U}^\dagger,$

where

• $\lambda_i$ are the eigenvalues (i.e., spectrum) of $\mathbf{A}$;
• $\mathbf{U}$ consists of the orthonormal eigenvectors of $\mathbf{A}$ in its columns in the same order as $\lambda_i$.

In particular,

• $\mathbf{A}$ is positive semidefinite $\implies \lambda_i\geq0$.
• $\mathbf{A}$ is Hermitian $\implies \lambda_i$ are real.
• $\mathbf{A}$ is unitary $\implies |\lambda_i|=1$.

The Pauli-X matrix is a normal operator:

$\mathbf{X} = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}.$

Manually or using NumPy, we can determine the eigenpairs of $\mathbf{X}$ to be $(1, 1/\sqrt{2}[1 \; 1]^\top)$ and $(-1, 1/\sqrt{2}[-1 \; 1]^\top)$. Thus,

$\begin{split} \mathbf{X} &= 1\cdot\frac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}\cdot\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\end{bmatrix} - 1\cdot\frac{1}{\sqrt{2}}\begin{bmatrix}-1\\1\end{bmatrix}\cdot\frac{1}{\sqrt{2}}\begin{bmatrix}-1&1\end{bmatrix} \\ &= \frac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}\cdot\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\end{bmatrix} - \frac{1}{\sqrt{2}}\begin{bmatrix}1\\-1\end{bmatrix}\cdot\frac{1}{\sqrt{2}}\begin{bmatrix}1&-1\end{bmatrix} = |+\rangle\langle+| - |-\rangle\langle-|. \end{split}$