Find a eigenvalues of matrix $A$ without using characteristic polynomial

Question

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?

Answer 1

Using the usual dodge of trying out a few simple linear combinations of the columns, one can quickly discover that by luck or by design, $(1,0,1)^T$ is an eigenvector with eigenvalue $3$. (I checked that combination first since the $2$ and $-2$ in the second row cancel.) Comparing this to $\operatorname{tr}A=11$, there are two possibilities: if $3$ is the double eigenvalue, then the other one must be 5; if the other eigenvalue is the double, then it must be $4$. Test these against $\det A$ to find the correct one.

Answer 2

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 $b=5$.