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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.

José Carlos Santos
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VLC
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    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. – Arturo Magidin May 22 '21 at 20:58
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    $(\lambda + 1 + \sqrt{2})(\lambda + 1 - \sqrt{2})(1 - \lambda)(-\lambda)$ is simply incorrect. Use Laplace expansion. – Hyperbolic PDE friend May 22 '21 at 20:59
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    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. – Arturo Magidin May 22 '21 at 21:02

2 Answers2

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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}

José Carlos Santos
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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.

Arturo Magidin
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quickquestion4
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  • Can I ask you where they got the minimal polynomial form from ? – VLC May 22 '21 at 21:00
  • Yes, I added an addendum. – quickquestion4 May 22 '21 at 21:02
  • Please let me know how my answer should be improved to avoid a downvote... – quickquestion4 May 22 '21 at 21:03
  • Is this general procedure to do for the minimal polynomial, by that I mean the cofactor expansion along the bottom rows. – VLC May 22 '21 at 21:06
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    @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. – Arturo Magidin May 22 '21 at 21:08
  • The expansion is one way of computing the determinant. The determinant of $A-\lambda I$ is by definition the characteristic polynomial, but not necessarily the minimal polynomial. The minimal polynomial is the smallest (monic) polynomial $\mu$ for which $\mu(A)=0$. It is a general fact $\mu|\chi$ and $\chi | \mu^k$ for some $k\in\mathbb{N}$ where $\chi$ is the characteristic polynomial. – quickquestion4 May 22 '21 at 21:10
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    @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 – VLC May 22 '21 at 21:10
  • @quickquestion4 So basically there can be many minimal polynomials right ? – VLC May 22 '21 at 21:11
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    @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. – Arturo Magidin May 22 '21 at 21:12
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    By "smallest" monic polynomial I mean of smallest degree. So no, the minimal polynomial is unique. – quickquestion4 May 22 '21 at 21:12
  • @ArturoMagidin I know how to compute the determinant, but it is my first time encountering minimal polynomials, and it looks like that until now I didn't really know what they are until now I am getting some intuition about them. – VLC May 22 '21 at 21:14
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    Maybe take a look at https://math.stackexchange.com/questions/2275809/calculate-minimal-polynomial-of-a-matrix/2611741 – quickquestion4 May 22 '21 at 21:16
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    @VLC Your questions about expanding along the bottom row suggest that you do not have a clear idea of how to compute determinants, so forgive me if I do not take you word for it that you know how to compute them. It seems to me that you actually have some issues doing so. The determinant of $A-\lambda I$ gives you the characteristic polynomial. The minimal polynomial must (i) divide the characteristic polynomial; (ii) be a multiple of each irreducible factor of the characteristic polynomial; (iii) annihilate the matrix; (iv) be of smallest possible degree doing all that; (v) be monic. (cont) – Arturo Magidin May 22 '21 at 21:20
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    @VLC If the characteristic polynomial is $\lambda(\lambda-1)^3$, then you have three possibilities for the minimal polynomial, in light of (i)-(v) above: it must be of the form $\lambda(\lambda-1)^i$ for $i=1$, or $i=2$, or $i=3$. There are ways of figuring out which one that involve theory and dimensions of eigenspaces and generalized eigenspaces; or Jordan canonical forms; or you can just check: see if $A(A-I)$ is the zero matrix; if so, minimal is $\lambda(\lambda-1)$; if not, multiple by $(A-I)$ and see; if zero matrix, minimal is $\lambda(\lambda-1)^2$; if not, min is last possibility. – Arturo Magidin May 22 '21 at 21:23
  • @ArturoMagidin Can I ask one last question, why is the polynomial $\lambda$ not a candidate for a minimal polynomial ? – VLC May 23 '21 at 11:47
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    @VLC: Item (ii) Every irreducible factor of the characteristic polynomial must divide the minimal polynomial; so the minimal polynomial must be a multiple of both $\lambda$ and $\lambda-1$. More basically: if $\mathbf{v}$ is an eigenvector associated to $1$, how could $A$ possibly annihilate it? – Arturo Magidin May 23 '21 at 16:45