$$A =\left(\begin{array}{rrrr}
0 & 1 & -1 & 1\\
-1 & 2 & -1 & 1\\
0 & 0 & 1 & -1\\
0 & 0 & 0 & 0
\end{array}\right).$$
We have this matrix, we want to find the characteristic and the minimal polynomial.
I have done the following:
First I used the formula $\det(A -\lambda I) = 0$, then I got this determinant into upper triangular form by getting rid of the $-1$ in the second row. Then I multiplied the diagonal and got :
$$(\lambda +1+ \sqrt{2})(\lambda + 1- \sqrt{2})(1-\lambda)(-\lambda) = 0 = p(\lambda)$$
Wolfram alpha gives the same result.
However the solution says that the characteristic and minimal polynomial should be
$$p_A(\lambda) = \lambda(\lambda-1)^3,\quad m_A(\lambda) = \lambda(\lambda -1)^2.$$
I do not understand how they got there or where is the mistake in my process.
Best Answer
You have$$A-\lambda\operatorname{Id}=\begin{bmatrix}-\lambda & 1 & -1 & 1 \\ -1 & 2-\lambda & -1 & 1 \\ 0 & 0 & 1-\lambda & -1 \\ 0 & 0 & 0 & -\lambda\end{bmatrix},$$whose determinant is clearly $0$ if $\lambda=0$. Otherwise\begin{align}\det(A-\lambda\operatorname{Id})&=-\lambda\begin{vmatrix}1&\frac1\lambda&-\frac1\lambda&\frac1\lambda\\ -1 & 2-\lambda & -1 & 1 \\ 0 & 0 & 1-\lambda & -1 \\ 0 & 0 & 0 & -\lambda\end{vmatrix}\\&=-\lambda\begin{vmatrix}1&-\frac1\lambda&\frac1\lambda&-\frac1\lambda\\0& 2-\lambda-\frac1\lambda & -1+\frac1\lambda & 1-\frac1\lambda \\ 0 & 0 & 1-\lambda & -1 \\ 0 & 0 & 0 & -\lambda\end{vmatrix}\\&=-\lambda\left(2-\lambda-\frac1\lambda\right)(1-\lambda)(-\lambda)\\&=\lambda ^4-3 \lambda ^3+3 \lambda ^2-\lambda\\&=\lambda(\lambda-1)^3.\end{align}