Minimal and characteristic polynomial question

Question

$$A =\left(\begin{array}{rrrr} 0 & 1 & -1 & 1\\ -1 & 2 & -1 & 1\\ 0 & 0 & 1 & -1\\ 0 & 0 & 0 & 0 \end{array}\right).$$

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.

Answer 1

If you compute the determinant of $A-\lambda I$ outright you should get:

$$\begin{align*} \det\begin{bmatrix} -\lambda & 1 & -1& 1\\ -1 & 2-\lambda & -1 & 1\\ 0&0 & 1-\lambda & -1 \\ 0&0&0& -\lambda \end{bmatrix} &= -\lambda \det\begin{bmatrix}-\lambda & 1 & -1\\ -1 & 2-\lambda & -1 \\ 0 & 0 & 1-\lambda \end{bmatrix}\\ &= -\lambda(1-\lambda) \det\begin{bmatrix} -\lambda & 1\\ -1 & 2-\lambda\end{bmatrix} \end{align*}$$ by cofactor expansion along the bottom rows. The last term is:

$$ -\lambda(1-\lambda)[\lambda^2-2\lambda +1] = \lambda(\lambda -1)[(\lambda-1)^2] = \lambda(\lambda -1)^3. $$ From here, you know that the minimal polynomial is one of $\lambda (\lambda -1), \lambda (\lambda -1)^2, \lambda (\lambda -1)^3$ since it must share the same roots as the characteristic polynomial. Since there are only three you can plug in and check.

Answer 2

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}