Extra (optional) exercises – Short Day

These exercises are optional extra material for students who have finished the exercises of the day and have the time and interest for more.

Extra exercise 1: Cayley-Hamilton for diagonalisable matrices

For a given polynomial \(p(Z) =a_0Z^0+a_1Z+\cdots+a_dZ^d \in \mathbb{F}[Z]\) and a square matrix \(\mathbf A \in \mathbb{F}^{n \times n}\) we define

\[p(\mathbf A)=a_0 \mathbf{I}_n+a_1 \mathbf{A}+\cdots+a_d \mathbf{A}^d \in \mathbb{F}^{n \times n}.\]

Question a

Let \(p_1(Z)\) and \(p_2(Z)\) be two polynomials from \(\mathbb{F}[Z]\). Show that \(p_1(\mathbf A)\cdot p_2(\mathbf A)=p_2(\mathbf A)\cdot p_1(\mathbf A)\).

Hint

The key here is that \(a_i \mathbf A^i \cdot b_j \mathbf A^j=b_j \mathbf A^j \cdot a_i \mathbf A^i\) holds for all non-negative integers \(i,j\) and all constants \(a_i,b_j \in \mathbb{F}\). First, check this claim.

Question b

Now assume that \(\mathbf A\) can be diagonalised and that \((\mathbf v_1,\dots,\mathbf v_n)\) is an ordered basis for \(\mathbb F^n\) consisting of eigenvectors. The characteristic polynomial of \(\mathbf A\) is denoted by \(p_{\mathbf A}(Z)\) as usual. Show that

\[p_{\mathbf A}(\mathbf A) \cdot \mathbf v_i=\mathbf 0 \quad \text{for all $i$ from $1$ to $n$}.\]

Hint

Corollary 12.3.4 in the textbook describes precisely how the characteristic polynomial of \(\mathbf A\) will look.

Question c

Based on your answer to Question b, conclude that \(p_{\mathbf A}(\mathbf A)=\mathbf 0\). In other words, conclude that \(p_{\mathbf A}(\mathbf A)\) is the zero matrix.

Remark: In fact, \(p_{\mathbf A}(\mathbf A)=\mathbf 0\) holds for every square matrix. This result is known as the Cayley-Hamilton theorem.