Let $A$ be the $n \times n$ matrix over a field of characteristic 0, all of whose entries are 1. What are the eigenvalues of $A$, counted with their multiplicities?
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1possible duplicate of find the characteristic polynomial of a square matrix of size n with all entries of 1 – colormegone Aug 30 '15 at 23:31
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What is $Ax$, where $x =(x_1,\ldots,x_n)^\top$? It is an $n\times 1$ column vector in which every entry is $x_1+\cdots+x_n$. Thus $x$ is in the kernel, and hence $x$ is an eigenvector with eigenvalue $0$, if that sum is $0$. So what is the dimension of the space of $n$-tuples in which the sum of the components is $0$? For the mapping $(x_1,\ldots,x_n)\mapsto x_1+\cdots+x_n$, the dimension of the image is $1$ and the dimension of the domain is $n$; hence the dimension of the kernel is $n-1$. So the geometric multiplicity of $0$ as an eigenvalue of $A$, i.e. the dimension of the eigenspace, is $n-1$. That leaves room for a $1$-dimensional eigenspace with some other eigenvalue. And $(1,\ldots,1)^\top$ is mapped to $(n,\ldots,n)$, so that gives you the other one.
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"hence the dimension of the kernel is $n-1$". 1. We can see this as the independent solutions for the single equation (due to "the dimension of the image is 1" and "if that sum is 0") $x_1+\cdots+x_n=0$ has $n-1$ solutions, i.e. $n-1$ free variables. 2. It can be also directly got from Rank–nullity theorem. – An5Drama Feb 01 '24 at 08:51
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Let $A$ be the $n$-by-$n$ all ones matrix. Since each row is a repeated copy of the first row, determinant of $A$ is $0$. This implies that $0$ is an eigenvalue of $A$. Also, observe that nullity of $A$ is $n-1$. This implies that the geometric multiplicity of $0$ is $n-1$. The remaning eigenvalue has geometric multiplicity $1$. Note that if $x=[1 \cdots 1]^T$, then $Ax=n x$. Thus, $n$ is an eigenvalue of $A$.
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