Is there a quadratic equation for matrices?

If you have an $$n \times n$$ matrix $$A$$ that satisfies

$$aA^2 + bA + cI_n = \mathbf{0}_n$$

does it let you say anything about $$A$$?

I'm thinking:

• If $$a= 0$$ then then $$A$$ must be a diagonal matrix, in particular it is a multiple of the identity matrix: $$A = (-c/b)I_n$$
• If $$b=0$$ then there can be infinitely many solutions for $$A$$, one of them is a multiple of the identity matrix: $$A = \pm\left(\sqrt{-c/a}\,\right)I_n$$ but I think these are not the only solutions in that case.

• there could be more than two square roots of a matrix Oct 22 '20 at 17:56

Yes. If a matrix satisfies a polynomial, it implies that all its eigenvalues satisfy this same polynomial. (For example in your situation, if $$v$$ is an eigenvector of $$A$$ with eigenvalue $$\lambda$$, then $$aA^2v + bAv + cI_nv = (a\lambda^2+b\lambda+c)v=0$$, and since $$v$$ is nonzero this implies that $$a\lambda^2+b\lambda+c=0$$.
In particular, you know that $$A$$ has at most two eigenvalues, and in particular they could be either of the roots of the polynomial $$aX^2+bX+c$$. (If it happens that this polynomial has only one root, then you know $$A$$ has only one eigenvalue.)
Using the Jordan form, it tells you that $$A$$'s Jordan form has only at most these two values on the diagonal; thus $$A$$ is similar to a Jordan form matrix with at most those two values on the diagonal.
Actually, you can know a little more. If the quadratic polynomial $$aX^2+bX+c$$ has two distinct roots, then the matrix must be diagonalizable. Indeed, a Jordan block of size $$m\times m$$ with eigenvalue $$\alpha$$ has minimal polynomial $$(X-\alpha)^m$$, and your matrix can't satisfy a polynomial one of its Jordan blocks doesn't satisfy, so if your polynomial has the form $$a(X-\alpha)(X-\beta)$$ with $$\alpha\neq \beta$$, then it cannot have a jordan block of size $$m>1$$. So all the Jordan blocks are size 1. To summarize, if your polynomial has 2 distinct roots, then $$A$$ is similar to a diagonal matrix with at most those 2 values on the diagonal.
If your polynomial has a double root, i.e., it's of the form $$a(X-\alpha)^2$$, then its Jordan form can have blocks of size up to 2. So in this case, it's similar to a Jordan matrix with only one value on the diagonal and blocks of size 1 and/or 2.