TheoremProved

Cayley–Hamilton theorem

Statement

Every square matrix $A$ satisfies its own characteristic polynomial: if $p(\lambda)=\det(\lambda I - A)$ , then $p(A) = 0$ (the zero matrix).

Why is it true?

The characteristic polynomial $p(\lambda)$ is built exactly so that $p(\lambda)=0$ whenever $\lambda$ is an eigenvalue of $A$ . Plugging the matrix $A$ itself in for $\lambda$ looks like a huge leap, but it works because $A$ acts on each eigenvector the same way the scalar eigenvalue does, and (in the diagonalizable case) the eigenvectors span the whole space — so $p(A)$ kills every direction, meaning $p(A)$ is the zero matrix.

Proof sketch

For diagonalizable $A=PDP^{-1}$ with $D=\mathrm{diag}(\lambda_1,\dots,\lambda_n)$ , $p(A)=Pp(D)P^{-1}$ and $p(D)=\mathrm{diag}(p(\lambda_1),\dots,p(\lambda_n))=0$ since each $\lambda_i$ is a root of $p$ . The general (non-diagonalizable) case follows by a density/continuity argument, or directly via the adjugate identity $\mathrm{adj}(\lambda I-A)(\lambda I - A)=p(\lambda)I$ treated as a polynomial identity in matrices.

Stated by

Arthur Cayley

Topics that use this theorem

Eigenvalues and diagonalization

Related theorems

Spectral theorem (real symmetric matrices)Rank–nullity theorem

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Roger A. Horn, Charles R. Johnson (2012). Matrix Analysis · DOI:10.1017/CBO9781139020411