Question

Verify cayley hamiltan theorem for A=[ 124212421]

Answer 1

$A=\begin{pmatrix} 1&2&4\\ 2&1&2\\ 4&2&1 \end{pmatrix}$

The characteristic polynomial of this matrix is

$P_A(\lambda)=\det(A-\lambda I)=\det\begin{pmatrix} 1-\lambda&2&4\\ 2&1-\lambda&2\\ 4&2&1-\lambda \end{pmatrix}=$

$(1-\lambda)^3+16+16-16(1-\lambda)-4(1-\lambda)-4(1-\lambda)=$

$(1-\lambda)^3-24(1-\lambda)+32$

By the Cayley - Hamilton theorem, it must be $P_A(A)=0$. Let's verify this equality.

$P_A(A)=(I-A)^3-24(I-A)+32I$

$I-A=\begin{pmatrix} 0&-2&-4\\ -2&0&-2\\ -4&-2&0 \end{pmatrix}$

$(I-A)^2=\begin{pmatrix} 20&8&4\\ 8&8&8\\ 4&8&20 \end{pmatrix}$

$(I-A)^3=\begin{pmatrix} -32 & -48 & -96\\ -48 & -32 & -48\\ -96 & -48 & -32 \end{pmatrix}$

$P_A(A)=\begin{pmatrix} -32 & -48 & -96\\ -48 & -32 & -48\\ -96 & -48 & -32 \end{pmatrix}- 24\begin{pmatrix} 0&-2&-4\\ -2&0&-2\\ -4&-2&0 \end{pmatrix}+$

$+\begin{pmatrix} 32&0&0\\ 0&32&0\\ 0&0&32 \end{pmatrix}=\begin{pmatrix} 0&0&0\\ 0&0&0\\ 0&0&0 \end{pmatrix}$

As we can see, the Cayley - Hamilton equality is true.