Calculating the minimal polynomial of this matrix

Question

Given $\begin{pmatrix} 1 & 1 & \cdots & 1 \\ 2 & 2 & \cdots & 2 \\ \cdot & \cdot & \cdots & \cdot \\ n & n & \cdots & n \end{pmatrix}$, calculate the minimal polynomial.

I know that the characteristic polyomial is: $$P_A(x) = x^{n-1} \cdot (x - \frac{n(n+1)}{2}) $$

But how do I exactly know the minimal polynomial? I am sensing that it is by using the Cayley-Hamilton theory!

That means that it is $m_a(x) = x (x - \frac{n(n+1)}{2})$. But how can we calculate it if we don't know it is diagonalizable? — TheNotMe, Jun 03 '13 at 15:20
You can easily calculate the eigenvectors of $0$, whose dimension is $n-1$. — Ma Ming, Jun 03 '13 at 15:23
Eigenvectors $(1,-1,\dots)^T,(0,1,-1,\dots)^T,\dots,(0,0,\dots,1,-1)^T$ all have eigenvalue $0$. — Thomas Andrews, Jun 03 '13 at 15:32
The rank of $A$ is clearly $1$. So by rank-nullity: $\dim \ker A=n-1$. — Julien, Jun 03 '13 at 15:47
And what does that exactly say about the minimal polynomial? (sorry for too many questions, I'm merely trying to understand) — TheNotMe, Jun 03 '13 at 15:57
@TheNotMe: I can see how you became confused. All of their answers regard proving $A$ is diagonalizable. Note that the dimension of $\ker(\lambda I - A)$ is in general the rank of the eigenspace of $\lambda$. Here, since $\lambda = 0$, that amounts to looking at $\ker A$. — Doug, Nov 25 '14 at 22:14

score 5 · Accepted Answer · edited Jun 03 '13 at 18:28

5

If $$ A=\begin{pmatrix} 1 & 1 & ... & 1 \\ 2 & 2 & ... & 2 \\ . & . & ... & . \\ n & n & ...& n \end{pmatrix}, $$ then it is not hard to check that $$ A^2=\frac{n(n+1)}2\,A. $$ So $p(A)=0$, where $p(x)=x\left(x-\frac{n(n+1)}2\right)$. This polynomial is divided by the minimal polynomial $m_a(x)$. As the minimal polynomial has degree at least two (if $n\geq2$, because otherwise $A$ would be zero or a scalar multiple of the identity), we get $$ m_a(x)=x\left(x-\frac{n(n+1)}2\right). $$

edited Jun 03 '13 at 18:28

robjohn

345,667

answered Jun 03 '13 at 15:30

Martin Argerami

205,756

Didn't understand the connection between $A^2$ and between knowing that $p(x) = x(x-...)$. – TheNotMe Jun 03 '13 at 15:38
The minimal polynomial has the property that it divides any polynomial that annihilates the matrix. In more technical terms, the ideal $I={p:\ p(A)=0}$ is generated by the minimal polynomial. – Martin Argerami Jun 03 '13 at 16:12
I know. But how did you determine that it is $p(x) = x * ..$ that annihilates the matrix and not $p(x) = x^{k} *...$? – TheNotMe Jun 03 '13 at 16:16
Nevermind, I understood. Excuse my stupidity after 8 hours of lectures! – TheNotMe Jun 03 '13 at 16:26
@MartinArgerami: (+1) However, if $n=1$, the minimal polynomial is $x-1$. – robjohn Jun 03 '13 at 16:44
Good point, I added a small mention. Gotta love $1\times1$ matrices ;) – Martin Argerami Jun 03 '13 at 18:19

score 2 · Answer 2 · edited Apr 13 '17 at 12:20

This matrix is $$ A=\begin{bmatrix}1\\2\\3\\\vdots\\n\end{bmatrix}\overbrace{\begin{bmatrix}1&1&1&\cdots&1\end{bmatrix}}^{1\times n}\tag{1} $$ According to this answer, $$ \begin{align} \det(I_n\lambda-A) &=\lambda^{n-1}\det\left(I_1\lambda-\begin{bmatrix}\frac{n(n+1)}{2}\end{bmatrix}\right)\\ &=\lambda^{n-1}\left(\lambda-\frac{n(n+1)}{2}\right)\tag{2} \end{align} $$ However, as Martin Argerami notes, and is easily shown using $(1)$, $$ A^2=\frac{n(n+1)}{2}A\tag{3} $$ So the minimal polynomial of A would be $\lambda\left(\lambda-\frac{n(n+1)}{2}\right)$ if $n\gt1$. If $n=1$ then the minimal polynomial would be $\lambda-1$.

score 1 · Answer 3 · answered Jun 03 '13 at 18:41

Here is a more mechanical, more general (but less slick) approach:

Think about the linear map $\alpha$ that the matrix $A$ represents. First we find the matrix of $\alpha$ in Jordan Normal Form. It is then well known (and relatively easy to see by drawing out the matrix) that the if we write $p_\alpha(x)=\prod_{i=1}^k(x-\lambda_i)^{n_i}$ then $n_i$ is the size of the largest $\lambda_i$ block. We can also see that $\dim\big(\ker(\alpha-\lambda_i)\big)$ is the number of $\lambda_i$ blocks.

In this example, we have $0$ as an eigenvalue, since the determinant is zero as the columns are not linearly independent. If $e_i$ is the usual basis of $\mathbb{R}^n$ then $\dim\big(\ker(\alpha-0)\big) = \dim\big(\langle e_1-e_2,\dots,e_1-e_n\rangle\big) = n-1$ and $\dim\big(\ker(\alpha-\frac{n(n-1)}{2}\big)=1$. Thus all $0$ blocks are size $1$ and there is only one $\frac{n(n-1)}{2}$ block, also of size 1. Thus each root of the minimal polynomial appears only once.

In fact, we can in this case do something a little quicker (but in the same spirit). It is always true that the sum of the eigenspaces of a linear map is direct, here the eigenspaces have dimension $n-1$ and $1$ as shown above, so their sum is direct and has dimension $n$ so must be $\mathbb{R}^n$. Thus the matrix is diagonal, so all Jordan Blocks are size $1$ so each root of the minimal polynomial appears only once.

Calculating the minimal polynomial of this matrix

3 Answers3