[Math] Find a eigenvalues of matrix $A$ without using characteristic polynomial

eigenvalues-eigenvectorslinear algebramatrices

Let $A$ be the following matrix:

$A=\begin{bmatrix}
4&1&-1\\
2&5& -2\\
1&1&2\\
\end{bmatrix}$

Find the eigenvalues of $A$ if you know that algebraic multiplicity of one eigenvalue is $2$. But you must not use characteristic polynomial.

I have no idea how to solve this, because if I use trace and determinant I still get polynomial with third degree so is still a characteristic polynomial. If I add $A^T$ on $A$ I get a symmetric matrix which is positive definite, so the eigenvalues are positive, so maybe I can use spectral theorem because $A+A^T$ is symmetric but I still need eigenvectors, so nothing from that. Do you know something?

Best Answer

The characteristic polynomial of $A$ (which I will not compute) is a monic poynomial with integer coefficients. Since it has a double root, it is reducible in $\mathbb{Q}[x]$ and therefore its roots are rational numbers. But every rational root of a monic poynomial with integer coefficients is an integer. So, the eigenvalues are integers $a$ (with algebraic multiplicity $2$) and $b$ such that:

  • $a^2b=\det A=45$;
  • $2a+b=\operatorname{tr}A=11$.

Therefore, $a=3$ and $b=5$.

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