<s>[INST] How to find characteristic polynomial of this matrix? 
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.
 [/INST] Working across the top row....
$$\begin{align}\begin{vmatrix} -\lambda&0&1\\1&-3-\lambda&1\\0&1&-3-\lambda&\end{vmatrix} &= -\lambda\left[(-3 -\lambda)(-3 -\lambda) -1\right] -0[\,] + 1\left[1-0 \right]=0 \\
&= -\lambda\left[9 + 6\lambda +\lambda^2 -1 \right] + 1= 0 \\
&= -\lambda^3 -6\lambda^2 -8\lambda +1 = 0\\
&= \lambda^3 +6\lambda^2 +8\lambda -1 = 0
\end{align}$$
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