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

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?

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.
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 $$b=5$$.