2

Let $A$ be a square matrix of size $n$ such that all the entries of $A$ are $1$. Find the characteristic polynomial of $A$.

2 Answers2

4

Key facts:

  • $A^2=nA$
  • $\mathrm{tr}(A)=n$

Thanks to the first property and the fact that $A\ne O$ and $A\ne nI$, the minimal polynomial of $A$ is $\mu_A(x)=x^2-nx=x(x-n)$ hence the eigenvalues of $A$ are $0$ and $n$.

Thanks to the second property, the eigenvalue $n$ is simple and the eigenvalue $0$ has multiplicity $n-1$. Thus, the characteristic polynomial of $A$ is $$\chi_A(x)=x^{n-1}(x-n)=x^n-nx^{n-1}.$$

Did
  • 279,727
0

Since $A$ has rank $1$, we have $\chi_A(x)=x^n-tr(A)x^{n-1}=x^n-nx^{n-1}$ (see this for instance).

Sahiba Arora
  • 10,847
lhf
  • 216,483