Why does a 90-degree rotation matrix in 2D have no real eigenvectors?

A matrix is diagonalizable if and only if it possesses a full set of linearly independent eigenvectors (an eigenbasis). The primary benefit is computational efficiency: calculating high powers of the matrix (e.g., $A^{100}$) becomes trivial because you simply raise the diagonal entries (eigenvalues) to that power, rather than performing repeated matrix multiplication.

Conditions: Matrix has $n$ linearly independent eigenvectors in $n$-dimensional space; Change of basis matrix $P$ formed by eigenvectors is invertible

Eigenvectors are special because they do not rotate off their original line (span) during a transformation; they only stretch, shrink, or flip. The zero vector is excluded from this definition because it maps to itself under any linear transformation ($A\vec{0}=\vec{0}$), providing no directional information about the transformation's effect.

Conditions: Vector $\vec{v}$ must be non-zero ($\vec{v} \neq \vec{0}$); Transformation is linear

Starting from $Av = \lambda v$, we rewrite the right side as $(\lambda I)v$ and move all terms to one side to get $(A - \lambda I)v = 0$. Since we seek non-zero solutions for $v$, the matrix $(A - \lambda I)$ must squash space into a lower dimension (have a non-trivial null space).

Conditions: $v$ is a non-zero eigenvector; $A$ is a square matrix; $I$ is the identity matrix