How is the characteristic equation det(A - λI) = 0 derived from the eigenvector definition Av = λv?

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

A 90-degree rotation moves every non-zero vector in the plane off its original line (span). Since eigenvectors must remain on their span, no real vector qualifies.

Conditions: Matrix represents a pure 90-degree rotation in 2D; Searching for real-valued eigenvectors

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