Cayley-Hamilton Theorem

Topics: Algebra - Diagonalisation of a Linear Transformation

---

This theorem assures us that $T:V\to V$ a linear transformation (with $V$ a vector space on $\mathbb{K}$ with finite dimension) nullifies its own characteristic polynomial.

(theorem)

Let $T:V\to V$ be a linear transformation, and $f(t)$ be its characteristic polynomial.

Then, $f(T)=T_0$

(corollary)

Let $A\in M_{n\times n}(\mathbb{K})$ be the associated matrix of $T$, and $f(t)$ be its characteristic polynomial.

Then, $f(A)=0$.

Theorem applications §

Finding the inverse of $A$ §

We’ll find the inverse of $A=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$.

1. Find the characteristic polynomial §

$$f(\lambda )=\det \left(\begin{array}{cc}1-\lambda & 2\\ 2& 1-\lambda \end{array}\right)={\lambda }^2-2\lambda -3$$

We then substitute $\lambda$ for $A$, and multiply the constant term by the identity matrix:

$$A^2-2A-3I=0$$

(it’s equal to $0$ since the theorem tells us that the characteristic polynomial is nullified).

Continuing:

$$A^2-2A=3I$$

$$A(A-2I)=3I$$

We then multiply this by $A^{-1}$:

$$A^{-1}A(A-2I)=3A^{-1}I=3A^{-1}$$

$$A-2I=3A^{-1}$$

$$\frac{A-2I}{3}=A^{-1}$$

Operating with the actual matrix, we get:

$$\frac{1}{3}\left(\begin{array}{cc}-1& 2\\ 2& -1\end{array}\right)=\left(\begin{array}{cc}-\frac{1}{3}& \frac{2}{3}\\ \frac{2}{3}& -\frac{1}{3}\end{array}\right)$$

Voilà. We got that $A^{-1}=\left(\begin{array}{cc}-\frac{1}{3}& \frac{2}{3}\\ \frac{2}{3}& -\frac{1}{3}\end{array}\right)$.