0

I want to find the minimal polynomial of the following matrix: $A = (a_{ij})\in M_n(K) $ with $ a_{ij} = a \not = 0$ for every i and j

I found the matrix diagonalizable as dim ImA = 1 and the matrix is not nilpotent, which implies that it is diagonalizable because of this https://math.stackexchange.com/a/3479804/1286043

However, I have no idea how to get the minimal polynomial of the matrix. How can I find it?

PS.: Consider K a field with characteristic 0

Pedro Luiz com Z
  • 1
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Jun 01 '24 at 21:51
  • you want to get the minimal polynomial for which matrix? please clarify – AnCar Jun 01 '24 at 21:59
  • Put the problem in the post, not just the title. That depends on the characteristic of $K$. The kernel has dimension $n-1$, and $(1,\ldots,1)$ is an eigenvector with eigenvalue $na$. If the characteristic of $K$ divides $n$, then the matrix is not diagonalizable. Otherwise, assuming $a\neq 0$, it is. The minimal polynomial equals $x(x-na)$. – Arturo Magidin Jun 01 '24 at 22:01
  • @AnCar yes, I want to find the minimal polynomial, I edited the question, sorry for the confusion. – Pedro Luiz com Z Jun 01 '24 at 22:18
  • @ArturoMagidin, I get why the x-na is part of the minimal polynomial, but why the x is there? – Pedro Luiz com Z Jun 01 '24 at 22:20
  • Because $0$ is also an eigenvalue. – Arturo Magidin Jun 01 '24 at 22:20
  • @ArturoMagidin about including the hypothesis just fixed it – Pedro Luiz com Z Jun 01 '24 at 22:20

2 Answers2

2

I will leave the $n=1$ case to you. Assume $n\gt 1$.

If $a\neq 0$, the rank is $1$, so $0$ is an eigenvalue with geometric multiplicity $n-1$. The trace is $na$, so the remaining eigenvalue is $na$.

If $na\neq 0$ in $K$, then that means the matrix is diagonalizable with eigenvalues $0$ (multiplicity $n-1$) and $na$ (multiplicity $1$), so the minimal polynomial is $x(x-na)$.

If $na=0$ in $K$, which happens when $\text{char}K\mid n$, then the eigenvalue $0$ has algebraic multiplicity $n$ and geometric multiplicity $n-1$, so the matrix is not diagonalizable. The Jordan form has at least $n-1$ blocks, so the largest block has size at most $2$. So if the matrix is not diagonalizable, then the minimal polynomial must be $x^2$. But over $K$, this equals $x(x-na)$. So regardless of the characteristic of $K$, the minimal polynomial is $x(x-na)$.

Is $a=0$, the matrix is the zero matrix with minimal polynomial $x$.

Arturo Magidin
  • 417,286
0

If your matrix $A$ has the same value in each entry, call it $a$, then what are the possible eigenvalue-eigenvector pairs for it? If $v=(v_1,v_2,\dots,v_n)$, note that $Av=\lambda v$ implies $a\sum_{j=1}^n v_j = \lambda v_k$ for each $1\leq k\leq n$.

Now we have two cases.

Case 1: $\lambda\neq 0$. This quickly implies that $v_1=v_2=\dots=v_n$ and that $\lambda =na$. The 'only' eigenvector for this will be (up to scaling) $(1,1,\dots,1)$.

Case 2: $\lambda=0$. Here the only restriction we have is that $v_1+v_2+\dots+v_n=0$. The solutions form an $n-1$ dimensional subspace and we can take for eigenvectors, for example $(1,-1,0,0,\dots,0)$, $(1,0,-1,0,\dots,0)$ and so on and finally $(1,0,0,0,\dots,-1)$.

We see that we have $n$ linearly independent eigenvectors, so they form a basis, which means the matrix is diagonalizable. Since the only eigenvalues are $0$ and $na$, the characteristic poly is $x^{n-1}(x-na)$. But since the matrix is diagonalizable, the minimal poly can only have linear factors, so the minimal poly is $x(x-na)$.

AnCar
  • 1,864
  • It is not diagonalizable when $na=0$. – Arturo Magidin Jun 01 '24 at 22:35
  • 1
    Indeed, I did not consider the possibility of the characteristic of the field being finite. Your answer covers that nicely. I kept my answer up because it also constructs the explicit eigenvectors in the usual cases $\mathbb K=\mathbb C$ or $\mathbb K = \mathbb R$. – AnCar Jun 01 '24 at 22:59