## Group Theory and Linear Algebra

Last updated: 21 September 2014

## Lecture 14: The Cayley-Hamilton Theorem

Let $A$ be a matrix in Jordan normal form. Let ${m}_{A}\left(t\right)$ be the minimal polynomial of $A\text{.}$ Let ${c}_{A}\left(t\right)$ be the characteristic polynomial of $A\text{.}$ ${m}_{A}\left(t\right)$ is minimal degree such that $m\left(A\right)=0\text{.}$ ${c}_{A}\left(t\right)=\text{det}\left(t-A\right)\text{.}$

If $J= ( λ 1λ 1λ 1λ )$ is a Jordan block of type ${\left(t-\lambda \right)}^{4}$ then $J-λ = ( 0 10 10 10 ) , (J-λ)2 = ( 0000 0000 1000 0100 ) , (J-λ)3 = ( 0000 0000 0000 1000 ) and (J-λ)4 = 0.$ If $mA(t) = (t-λ1)m1 (t-λ2)m2⋯ (t-λk)mkand cA(t) = (t-λ1)c1 (t-λ2)c2⋯ (t-λk)ck$ then ${m}_{j}$ is the maximal size of a Jordan block with eigenvalue ${\lambda }_{j}$ in $A,$ and
${c}_{j}$ is the sum of the sizes of the Jordan blocks of eigenvalue ${\lambda }_{j}$ in $A\text{.}$
So, finding the minimal polynomial and characteristic polynomial of a matrix in Jordan normal form is easy.

$A= ( 2 12 2 2 12 12 3 13 3 13 13 13 ) .$ Then $m(t)=(t-2)3 (t-3)4and c(t)=(t-2)6 (t-3)6$ since the sizes of Jordan blocks of eigenvalue 2 are 2, 1, 3 (with maximum 3) and the sizes of Jordan blocks of eigenvalue $3$ are 2, 4 (with maximum 4).

Let $Q$ be an $n×n$ matrix. The Jordan normal form theorem says that: There exists $P$ such that $PQP-1=A$ is in Jordan normal form.

Let us compare ${m}_{Q}\left(t\right)$ and ${m}_{A}\left(t\right)$ and compare ${c}_{Q}\left(t\right)$ and ${c}_{A}\left(t\right)\text{.}$ $cQ(t) = det(t-Q), cA(t) = det(t-A) = det(t-PQP-1) = det(tPP-1-PQP-1) = det(P(t-Q)P-1) = det(P)det(t-Q) det(P-1) = det(t-Q) = cQ(t).$ If $\ell \left(t\right)$ is a polynomial then $ℓ(t) = ℓ0+ℓ1t+ℓ2 t2+⋯+ℓdtd,and ℓ(Q) = ℓ0+ℓ1Q+ ℓ2Q2+⋯+ ℓdQd,and ℓ(A) = ℓ0+ℓ1A+ ℓ2A2+⋯+ ℓdAd = ℓ0+ℓ1PQP-1 +ℓ2(PQP-1)2 +⋯+ℓd(PQP-1)d.$ Since $(PQP-1)2 = PQP-1PQP-1= PQ2P-1, (PQP-1)2 = PQP-1PQP-1PQP-1= PQ3P-1,…$ then $ℓ(A) = ℓ0+ ℓ1PQP-1+ ℓ2PQ2P-1+⋯+ ℓdPQdP-1 = P ( ℓ0+ℓ1Q+ ℓ2Q2+⋯+ ℓdQd ) P-1 = Pℓ(Q)P-1$ and $ℓ(Q)=P-1ℓ (A)P.$ So if $\ell \left(Q\right)=0$ then $\ell \left(A\right)=0$ and if $\ell \left(A\right)=0$ then $\ell \left(Q\right)=0\text{.}$

${m}_{Q}\left(t\right)$ is the smallest degree monic polynomial such that ${m}_{Q}\left(Q\right)=0\text{.}$

${m}_{A}\left(t\right)$ is the smallest degree monic polynomial such that ${m}_{A}\left(A\right)=0\text{.}$

The point: If $PQ{P}^{-1}=A,$ then ${m}_{Q}\left(t\right)={m}_{A}\left(t\right)\text{.}$
If $PQ{P}^{-1}=A,$ then ${c}_{Q}\left(t\right)={c}_{A}\left(t\right)\text{.}$

### The Cayley-Hamilton Theorem

Let $Q$ be an $n×n$ matrix and let $cQ(t)=det(t-Q)$ be the characteristic polynomial of $Q\text{.}$ Then $cQ(Q)=0.$

## Notes and References

These are a typed copy of Lecture 14 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 24, 2011.