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.

Can anything more general be said about quadratic equation in matrices?
 A: 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.
This is the most you can say though. Any matrix like I have described will be satisfy your polynomial, because all its Jordan blocks do.
