Let, $A=\begin{bmatrix} 4&0&1&0\\1&1&1&0\\0&1&1&0 \\0&0&0&4 \end{bmatrix}$. Knowing that $4$ is one of its eigenvalues, find the characteristic polynomial of $A$.
Well if $4$ is an eigenvalues of $A$, one should have $|A-4I_{4}|=0$ .
And so,
$\begin{vmatrix} 0&0&1&0\\1&-3&1&0\\0&1&-3&0 \\0&0&0&0 \end{vmatrix}=0$
It's clear that the previous equation is true (the determinant of $(A-4I_{4})=0$). Now that the factor $(\lambda-4)$ was pull out, one gets a new matrix by removing the null row and null column.
$A'=\begin{bmatrix} 0&0&1\\1&-3&1\\0&1&-3&\end{bmatrix}$
The characteristic polynomial of $A'$ will be a $3^{th}$ degree polynomial, which product with $(\lambda-4)$ equals to a $4^{th}$ degree polynomial.
Now, in order of finding the characteristic polynomial of $A'$ one must to solve the characteristic equation:
$\begin{vmatrix} -\lambda&0&1\\1&-3-\lambda&1\\0&1&-3-\lambda&\end{vmatrix}=0$
My doubt is on finding this determinant. I already tryed Laplace's transformations in order to make null row or a column, but I couldn't do it.
Can you give me a clue? Thanks.
Answer
...and how to find the charateristic polynomial of the original matrix, $A$
$$\begin{align}\mathrm{det}(A - \lambda \mathrm{I}) &= 0 \tag{1}\\
\begin{vmatrix}(4-\lambda)&0&1&0 \\
1&(1-\lambda)&1&0\\
0&1&(1-\lambda)&0\\
0&0&0&(4-\lambda) \end{vmatrix} &= 0 \tag{2}\\
(4-\lambda)\begin{vmatrix}(4-\lambda)&0&1\\
1&(1-\lambda)&1\\
0&1&(1-\lambda) \tag{3}\\
\end{vmatrix} &= 0 \\
(4-\lambda) \left[(4-\lambda) \left[(1-\lambda)^2 -1 \right] + 1\right] &= 0\tag{4} \\
(4-\lambda) \left[(4-\lambda)(\lambda^2 -2\lambda) + 1\right] &= 0\tag{5} \\
(4-\lambda) \left[(4\lambda^2 -8\lambda -\lambda^3+2\lambda^2) + 1\right] &= 0 \tag{6}\\
(4-\lambda)(-\lambda^3 + 6\lambda^2 - 8\lambda +1 ) &= 0 \tag{7}\\
(\lambda -4)(\lambda^3 - 6\lambda^2 + 8\lambda - 1)&= 0\tag{8}\\
\end{align}$$
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