Proof about properties of eigenvalues

Question

Prove: If $a,b,c,d \in\mathbb{Z}$ such that $a+b=c+d$, then $A=\begin{pmatrix} a&b\\ c&d\\ \end{pmatrix}$ has integer eigenvalues. $$$$I need some help getting started. My idea is this: For there to be integer eigenvalues, $$(a+d)$$ and $$\sqrt{(a-d)^2+4bc}$$ must both be even, or must both be odd since division by $2$ will produce an integer this way. Note I am using the adaptation of the quadratic equation to the characteristic equation of a $2\times2$ matrix.

Also I see that the equation $a+b-(c+d)=0$ must be satisfied $\forall a,b,c,d \in\mathbb{Z}$.

Any other pointers would be appreciated!

Answer 1

Simply note that for $a+b=c+d=\lambda_1$

$$\begin{pmatrix} a&b\\ c&d\\ \end{pmatrix}\begin{pmatrix} 1\\ 1\\ \end{pmatrix} =\begin{pmatrix} a+b\\ c+d\\ \end{pmatrix} =\lambda_1\begin{pmatrix} 1\\ 1\\ \end{pmatrix} $$

therefore, by definition, $\lambda_1$ is an (integer) eigenvalue for $A$ with eigenvector $(1,1)$.

Then observe that

• $\lambda_1+\lambda_2=Tr(A)=a+d \implies \lambda_2=a+d-\lambda_1$