Let $A$ be a square matrix of size $n$ such that all the entries of $A$ are $1$. Find the characteristic polynomial of $A$.
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2Related: https://math.stackexchange.com/q/153457/339790 – Rodrigo de Azevedo Jul 06 '20 at 12:14
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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 $$\chi_A(x)=x^{n-1}(x-n)=x^n-nx^{n-1}.$$
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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
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