1.24 Verify the Cayley-Hamilton theorem that every matrix is zero of its characteristic polynomial

Problem, given the matrix

\[ \left ( {\begin {array} [c]{cc}1 & 2\\ 3 & 2 \end {array}} \right ) \]

Verify that matrix is a zero of its characteristic polynomial. The Characteristic polynomial of the matrix is found, then evaluated for the matrix. The result should be the zero matrix.

Mathematica

Remove["Global`*"] a = {{1,2},{3,2}}; n = Length[a]; p = CharacteristicPolynomial[a,x]	\(x^2-3 x-4\)
(-4 IdentityMatrix[n] - 3 a + MatrixPower[a,2])//MatrixForm	\[ \left ( {\begin {array}{cc} 0 & 0 \\ 0 & 0 \\ \end {array}} \right ) \]
Another way is as follows a = {{1,2},{3,2}}; p = CharacteristicPolynomial[a,x]; cl = CoefficientList[p,x]; Sum[MatrixPower[a,j-1] cl[[j]], {j,1,Length[cl]}]	\[ \left ( {\begin {array}{cc} 0 & 0 \\ 0 & 0 \\ \end {array}} \right ) \]

 

Matlab

MATLAB has a build-in function polyvalm() to do this more easily than in Mathematica. Although the method shown in Mathematica can easily be made into a Matlab function

clear; A=[1 2;3 2]; p=poly(A); poly2str(p,'x') polyvalm(p,A)

ans = x^2 - 3 x - 4 ans = 0 0 0 0