Here is a question I heared from others:

Given four distinct positive real numbers $a_1,a_2,a_3,a_4$ and set $$a:=\sqrt{\sum_{1\leq i\leq 4}a_i^2}$$ $A=(x_{i,j})_{1\leq i\leq3,1\leq j\leq4}$ is a $3\times4$-matrix specified by $$ x_{i,j}=a_i\delta_{i,j}+a_j\delta_{4,j}-\frac{1}{a^2}(a_i^2+a_4^2)a_j $$ where $\delta_{i,j}$ is the Kronecker symbol or visually $$ A=\begin{pmatrix}a_1 &0&0&a_4\\ 0&a_2&0&a_4\\0&0&a_3&a_4\end{pmatrix}-\frac{1}{a^2} \begin{pmatrix} a_1(a_1^2+a_4^2) & a_2(a_1^2+a_4^2) & a_3(a_1^2+a_4^2) & a_4(a_1^2+a_4^2)\\ a_1(a_2^2+a_4^2) & a_2(a_2^2+a_4^2) & a_3(a_2^2+a_4^2) & a_4(a_2^2+a_4^2)\\ a_1(a_3^2+a_4^2) & a_2(a_3^2+a_4^2) & a_3(a_3^2+a_4^2) & a_4(a_3^2+a_4^2)\\ \end{pmatrix} $$

**The question is to show that the $3\times3$-matrix $B=AA^T$ admits three distinct eigenvalues**.($A^T$ is the transpose of $A$)

What I am curious about is how many methods can be utilized to show a matrix has different eigenvalues?

As for this question my idea is to calculate the characteristic polynomial $f$ of $B$ along with $f'$ which is a quadratic polynomial via Sagemath and show that neither of roots of $f'$ belongs to $f$. Or equivalently to calculate the resultant $R(f,f')$ of $f$ and $f'$ and show that $R(f,f')$ doesn't vanish for any distinct positive $a_i$'s.

But the difficulties are both ways involve hideous calculation which I don't think I can write down by hand. So I'm wondering if there is a tricky way to get to that point? (e.g. an algebraic-geometry method?)

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