Find the Rational form of $$ A=\begin{pmatrix} 1&2&0&4\\ \:\:\:4&1&2&0\\ \:\:\:0&4&1&2\\ \:\:\:2&0&4&1\end{pmatrix} $$
I don't wanna the solution, instead I would like to know a quickest way to calculate $\det (\lambda I-A)$. $$ \begin{vmatrix} x-1 &-2 &0 &-4 \\ -4&x-1 &-2 &0 \\ 0&-4 &x-1 & -2 \\ -2&0 &-4 &x-1 \end{vmatrix} $$
here I could see the polynomial but the procedure is quite long.
$\endgroup$2 Answers
$\begingroup$Add columns $\,2,3,4\,$ to the first one (this is a circulant matrix), and get:
$$\det(t I-A)=\begin{vmatrix}t-7&-2&0&-4\\ t-7&t-1&-2&0\\ t-7&-4&t-1&-2\\ t-7&0&-4&t-1\end{vmatrix}\stackrel{R_i-R_1}=\begin{vmatrix}t-7&-2&0&-4\\ 0&t+1&-2&4\\ 0&-2&t-1&2\\ 0&2&-4&t+3\end{vmatrix}=$$$${}$$
$$=(t-7)\begin{vmatrix} t+1&-2&4\\ -2&t-1&2\\ 2&-4&t+3\end{vmatrix}=(t-7)\left[(t^2-1)(t+3)+24+8t+8-4t-12-8t+8\right]=$$
$$=(t-7)\left[(t^2-1)(t+3)-4t+28\right]=(t-7)\left[t^3+3t^2-5t+25\right]=$$
$$=(t-7)(t+5)(t^2-2t+5)=(t+t)(t+5)(t-(1-2i))(t-(1+2i))$$
$\endgroup$ 2 $\begingroup$The quick way is to realise you have a circulant matrix, so you can immediately write down the determinant\begin{align*} &\det\begin{pmatrix} x-1 &-2 &0 &-4 \\ -4&x-1 &-2 &0 \\ 0&-4 &x-1 & -2 \\ -2&0 &-4 &x-1 \end{pmatrix}\\ &=(x-1-2-4)(x-1-2i+4i)(x-1+2+4)(x-1+2i-4i) \end{align*}since the eigenspaces are $\langle(1,i^k,i^{2k},i^{3k})\rangle=\langle(1,i^k,(-1)^k,(-i)^k)\rangle$, for $k=0,1,2,3$.
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