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

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.

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    $\begingroup$ Please don’t use images; use MathJax. Here’s a tutorial. Images can’t adapt to the display, they cannot be searched, and screenreaders cannot process them, making your post inaccessible to people who use them. $\endgroup$ May 22, 2021 at 20:58
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    $\begingroup$ $(\lambda + 1 + \sqrt{2})(\lambda + 1 - \sqrt{2})(1 - \lambda)(-\lambda)$ is simply incorrect. Use Laplace expansion. $\endgroup$
    – Meowdog
    May 22, 2021 at 20:59
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    $\begingroup$ The sum of the eigenvalues equals the trace of the matrix. The eigenvalues you calculated were $-1-\sqrt{2}$, $-1+\sqrt{2}$, $1$, and $0$. They add up to $-1$; but the trace of your matrix is $3$. So your calculation of the characteristic polynomial is definitely incorrect. $\endgroup$ May 22, 2021 at 21:02

2 Answers 2

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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}

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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.

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    $\begingroup$ @VLC: You don’t find the minimal polynomial with determinants. “Cofactor expansion along the bottom row” is just one way to compute determinants; whether you want to use it depends on the matrix. If you have difficulties computing determinants, perhaps that’s where you should start your questions, rather than at computing minimal and characteristic polynomials. $\endgroup$ May 22, 2021 at 21:08
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    $\begingroup$ @ArturoMagidin Ok so the point is that we can use cofactor expansion and see how the characteristic polynomial looks before it get's into it's final form, and the terms there are candidates for minimal polynomial $\endgroup$
    – VLC
    May 22, 2021 at 21:10
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    $\begingroup$ @VLC: I do not understand your comment; I think you are very confused. “Cofactor expansion” is a way to calculate any determinant. The determinant of $A-\lambda I$ is the characteristic polynomials, not “how the characteristic polynomial looks before it get’s[sic] into it’s[sic] final form.” The minimal polynomial divides the characteristic polynomial, yes, and it has other properties. But if you don’t know how to calculate determinants, I think that’s where your problems begin. $\endgroup$ May 22, 2021 at 21:12
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    $\begingroup$ By "smallest" monic polynomial I mean of smallest degree. So no, the minimal polynomial is unique. $\endgroup$ May 22, 2021 at 21:12
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    $\begingroup$ Maybe take a look at math.stackexchange.com/questions/2275809/… $\endgroup$ May 22, 2021 at 21:16

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