bebdmt Umar
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p(x) = det(xIn – A)
Where the determinant operation is ‘det’ and for the scalar element of the base ring, the variable is taken as x. As the entries of the matrix are (linear or constant) polynomials in x, the determinant is also an n-th order monic polynomial in x.
Also, read:
The Cayley–Hamilton theorem states that substituting the matrix A for x in polynomial, p(x) = det(xIn – A), results in the zero matrices, such as:
p(A) = 0
It states that a ‘n x n’ matrix A is demolished by its characteristic polynomial det(tI – A), which is monic polynomial of degree n. The powers of A, found by substitution from powers of x, are defined by recurrent matrix multiplication; the constant term of p(x) provides a multiple of the power A0, where power is described as the identity matrix.
The theorem allows An to be articulated as a linear combination of the lower matrix powers of A. If the ring is a field, the Cayley–Hamilton theorem is equal to the declaration that the smallest polynomial of a square matrix divided by its characteristic polynomial.
For 1 x 1 matrix A(a1,1) the characteristic polynomial is given by \(\begin{array}{l}p(\lambda )=\lambda – a\end{array} \)
So, p(A) = (a) – (a1,1) = 0 is obvious.
2.) 2 x 2 Matrices
Let us look this through an example
\(\begin{array}{l}A = \begin{pmatrix} 1 & 2\\ 3 & 4 \end{pmatrix}\end{array} \)
\(\begin{array}{l}p(\lambda )=det(\lambda I_{2}-A)= det\begin{pmatrix} \lambda -1 & -2\\ -3 & \lambda -4 \end{pmatrix} = (\lambda -1)(\lambda -4)-(-2)(-3)=\lambda ^{2}-5\lambda -2\end{array} \)
The Cayley-Hamilton claims that if, we define
\(\begin{array}{l}p(X) = X^{2}-5X-2I_{2}\end{array} \)
then,
We can verify this result by computation
For a generic 2 x 2 matrix,
\(\begin{array}{l}A=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\end{array} \)
The resultant polynomial is given by:
So the Cayley-Hamilton theorem states that
\(\begin{array}{l}p(A)=A^{2}-(a+d)A+(ad-bc)I_{2}=\begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}\end{array} \)
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