# Calculating the minimal polynomial of this matrix

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!

• Hint: this matrix is diagonalizable. Jun 3 '13 at 15:19
• 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? Jun 3 '13 at 15:20
• Yes. Just determine the dimension of $\ker A$. Jun 3 '13 at 15:23
• You can easily calculate the eigenvectors of $0$, whose dimension is $n-1$. Jun 3 '13 at 15:23
• julien can you please show me how to do that? Jun 3 '13 at 15:26

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 $$m_a(x)=x\left(x-\frac{n(n+1)}2\right).$$

• Didn't understand the connection between $A^2$ and between knowing that $p(x) = x(x-...)$. Jun 3 '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. Jun 3 '13 at 16:12
• I know. But how did you determine that it is $p(x) = x * ..$ that annihilates the matrix and not $p(x) = x^{k} *...$? Jun 3 '13 at 16:16
• Nevermind, I understood. Excuse my stupidity after 8 hours of lectures! Jun 3 '13 at 16:26
• @MartinArgerami: (+1) However, if $n=1$, the minimal polynomial is $x-1$.
– robjohn
Jun 3 '13 at 16:44

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 $\lambda-1$.

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.