How to find characteristic polynomial of a matrix $A$ whose elements are of the form $a_{ij} = a_ia_j$ for all $1\leq i,j \leq n$ For $n > 1$,
$$A = \begin{pmatrix} {a_1}^2 & a_1a_2 & ... a_1a_n \\
 a_2a_1 & {a_2}^2 & ... a_1a_n \\.... \\
 a_na_1 & a_na_2 & ... {a_n}^2 \end{pmatrix}$$
The characteristic polynomial of matrix $A$ is $x^{n-1}(x-tr(A))$
My question is that how to find characteristic polynomial of matrix $A$
My Attempt:
I noticed that elements of $A$ are of the form $a_{ij} = a_ia_j$ for all $1\leq i,j \leq n$ So $$A = \begin{pmatrix} {a_1}^2 & a_1a_2 & ... a_1a_n \\
 a_2a_1 & {a_2}^2 & ... a_1a_n \\.... \\
 a_na_1 & a_na_2 & ... {a_n}^2 \end{pmatrix} = \begin{pmatrix} a_1 & a_1 & ... a_1 \\
 a_2 & a_2 & ... a_2 \\.... \\
 a_n & a_n & ... a_n\end{pmatrix}×\begin{pmatrix} a_1 & a_1 & ... a_1 \\
 a_2 & a_2 & ... a_2 \\.... \\
 a_n & a_n & ... a_n\end{pmatrix}$$
 A: Observe, that for $\mathbf a:=(a_1,a_2,\ldots,a_n)^T$, we have $A=\mathbf a\mathbf a^T$. Hence for $\mathbf v\in \mathbb R^n$ with $\mathbf a^T \mathbf v=\langle \mathbf a,\mathbf v\rangle=0$, we have $A\mathbf v=\mathbf 0$, i.e. $0$ is a eigen value with multiplicity $n-1$, because the space $\{\mathbf v \in \mathbb R^n\mid \langle \mathbf a,\mathbf v\rangle=0\}$ is an $n-1$-dimensional hyperplane.
Further $A\mathbf a=\langle \mathbf a,\mathbf a\rangle \cdot \mathbf a=\|\mathbf a\|^2\mathbf a=\operatorname{tr}(A)\mathbf a$, which means that $\operatorname{span}(\mathbf a)$ is an one dimensional eigenspace to the eigenvalue $\operatorname{tr}(A)$.
I am not sure, if your decomposition of the matrix is right.
A: If you look carefully the matrix $A$ is  a rank $1$ matrix with trace $a_1^2+a_2^2+\cdots +a_n^2$ . So the eigen values are $0, 0, \cdots, (n-1) $ times and $tr(A)$ . So obviously the characteristic polynomial is $x^{n-1}(x-tr(A))$.
A: $$A = \begin{pmatrix} {a_1}^2 & a_1a_2 & ... a_1a_n \\
 a_2a_1 & {a_2}^2 & ... a_1a_n \\.... \\
 a_na_1 & a_na_2 & ... {a_n}^2 \end{pmatrix} = \begin{pmatrix} a_1 & a_1 & ... a_1 \\
 a_2 & a_2 & ... a_2 \\.... \\
 a_n & a_n & ... a_n\end{pmatrix}×\begin{pmatrix} a_1 & a_1 & ... a_1 \\
 a_2 & a_2 & ... a_2 \\.... \\
 a_n & a_n & ... a_n\end{pmatrix}$$
Let $A = B×C$, where $$B = C = \begin{pmatrix} a_1 & a_1 & ... a_1 \\
 a_2 & a_2 & ... a_2 \\.... \\
 a_n & a_n & ... a_n\end{pmatrix}$$
We know that $Rank(BC) \leq \min (Rank (B),Rank(C))$
But here we see that $Rank(B) = Rank(C) =1$. Hence $Rank (A) = 1$. So $n-1$ times eigen values are $0$ and one eigen value is trace($A$). Therefore $Ch_A(x) = x^{n-1}(x-trace(A))$
