Give conditions on a,b,c, and d such that A has two, one, and no eigenvalues?

Question

I am given that matrix $$A= \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} $$

and I need to find conditions on a,b,c, and d such that A has

1. Two distinct real eigenvalues
2. One real eigenvalue
3. No real eigenvalue

I was trying to think this through and all I can figure out is that would it have two distinct eigenvalues if det(A-$\lambda$I)=0?

Answer 1

$$det(A-\lambda I)=0$$ $$(a-\lambda)(d-\lambda)-bc=0$$ $$ad-a\lambda-d\lambda +\lambda² -bc=0 $$ $$\lambda² -(a+d)\lambda+(ad-bc)=0 $$ Making analysis under discriminant and you will see...

$(a+d)²-4(ad-bc)>0$ it is $\Delta >0 <=>$ $2$ real eigenvalues

$(a+d)²-4(ad-bc)=0$ it is $\Delta =0 <=>$ $1$ real eigenvalues

$(a+d)²-4(ad-bc)<0$ it is $\Delta <0 <=>$ $0$ real eigenvalues

Well, all this is because $R$ is incomplete field, in case you have Complex matrices you lost the characteristic of order set, but you have that all polynomials have that same number of roots that their's degree, then you have one(multiplicity 2) or two eigenvalues.

Answer 2

So it seems you did compute the characteristic polynomial $X^2-(a+d)X+ad-bc$ and its discriminant, though it seems you used an incorrect formula for the latter. The correctly computed discriminant is $(a+d)^2-4(ad-bc)=a^2+d^2-2ad+bc$ which is also $(a-d)^2+4bc$. Supposing the entries $a,b,c,d$ are themselves real, you will be in cases 1, 2, 3 according as this discriminant is positive, zero, or negative. (In case of complex entries $a,b,c,d$, it is unlikely there are any real eigenvalues at all, but the precise conditions become rather complicated to state; notably "one real eigenvalue" could either be a double root, or be paired with another non-real complex root.)