Question: prove that if a,b,c and d are integers such that a+b=c+d, then the matrix A=[ a b, c d]
has integer eigenvalues.
Attempt at solution: So to find the eigenvalues of a 2×2 matrix we have to use det( λͅI – A) = 0
So my matrix would then be [ λͅ-a -b, -c λͅ-d]
and then to find the determinant the following calculation must be done ( λͅ-a)( λͅ-d)-(bc)
If this is then made equal to 0, then we can solve for the eigenvalues.
first I must multiply and expand the equation
so: λͅ^2 – d λͅ – a λͅ +ad -bc = 0
From here I have a couple of problems
1) I don't understand how to factor this and show that its root (the eigenvalues) will in fact be integers (non-fractional values)
2) how does the relation a+b= c+d come into this
I feel like I'm missing the key to this question, I would be grateful for any help
and btw this is elementary linear algebra
Additional Info: I have seen a version of this exact question where it says that I have to prove that the eigenvalues are integers namely λͅ1= a+b and λͅ2= a-c when I solve the eigenvalues out. although this isn't a part of my question above, it might hint to how I could find my solution. I'm not exactly sure :/
Best Answer
The characteristic equation is $\lambda^2 -(a+d) \lambda + ad-cb = 0$, hence we know that the solutions are $\lambda = {1 \over 2} (a+d \pm \sqrt{(a+d)^2-4 (ad-cb)})$.
We know that $\sqrt{(a+d)^2-4 (ad-cb)}$ is either imaginary, an integer or irrational.
One eigenvalue is straightforward to guess using the vector $(1,1)^T$. This is an integer, hence we know that both eigenvalues are real (since they occur in conjugate pairs otherwise). Since it is an integer, we know that $\sqrt{(a+d)^2-4 (ad-cb)}$ is an integer.
Furthermore, the formula above shows that the difference between the two pairs is $\sqrt{(a+d)^2-4 (ad-cb)}$, which is an integer. Hence the other eigenvalue is an integer.