If $A^2 + b A + c I = 0$ why does $A$ have to be diagonalizable?

Question

If $A \in \mathbb C^{n \times n}$, $b, c$ are scalars such that $c \ne \frac 14 b^2$ and $$A^2 + b A + c I = 0$$ why does $A$ have to be diagonalizable?

I am getting to a point where I think $A$ has to be a $2 \times 2$ matrix, but I don't know how to justify it. How can I prove it?

Answer 1

Here's another answer, with a direct proof.

Consider the polynomial $x^2+bx+c$. We are given that $b^2-4c\ne 0$, so $x^2+bx+c$ has distinct roots, $\lambda$ and $\mu$.

Any $n\times n$ complex matrix $A$ that satisfies this polynomial will be diagonalizable.

Proof.

There are two cases, either both $\lambda$ and $\mu$ are eigenvalues or one of them is not an eigenvalue. (Technically these cases aren't needed, but I think it makes it a bit clearer.)

First suppose one of them is not an eigenvalue. Without loss of generality, assume $\lambda$ is not an eigenvalue. Then $A^2+bA+c=(A-\lambda)(A-\mu)=0$. Since $\lambda$ is not an eigenvalue, $A-\lambda$ is invertible. Thus $A-\mu=0$, or $A=\mu$. Since $\mu=\mu I$ is diagonal, $A$ is diagonal already.

Now suppose that both $\lambda$ and $\mu$ are eigenvalues. Let $E_\lambda = \ker (A-\lambda)$ and $E_\mu =\ker(A-\mu)$ be the eigenspaces associated to each of them. I claim $\Bbb{C}^n=E_\lambda \oplus E_\mu$, which will show that $A$ is diagonalizable (since its eigenvectors span $\Bbb{C}^n$, so there is a basis of eigenvectors).

First, $E_\lambda \cap E_\mu =0$, since if $v\in E_\lambda \cap E_\mu$, then $$Av=\mu v =\lambda v,$$ so $(\mu-\lambda)v=0$, and since $\mu\ne \lambda$, $\mu-\lambda \ne 0$. Thus $v=0$.

Next, if $v\in \Bbb{C}^n$, we must show that it can be written as a sum of some elements $v_\lambda \in E_\lambda$ and $v_\mu\in E_\mu$.

Let $w=(A-\lambda)v$. Then $w\in E_\mu$, since $(A-\mu)w = (A-\mu)(A-\lambda)v=0v=0$. Then let $v_\mu = \frac{1}{\mu-\lambda} w$, so that $(A-\lambda)v_\mu =(\mu-\lambda)v_\mu=w$.

Then let $v_\lambda = v-v_\mu$, so that $v=v_\lambda+ v_\mu$. We just need to check that $v_\lambda \in E_\lambda$. Then $$(A-\lambda)v_\lambda = (A-\lambda)(v-v_\mu)=(A-\lambda)v-(A-\lambda)v_\mu = w-w=0,$$ as desired.

Answer 2

The matrix $A$ gets annihilated by the polynomial $P=X^2+bX+c$. Since $c\neq\frac{b^2}{4}$, $P$ factors into $(X-z_1)(X-z_2)$ with $z_1,z_2$ two distinct, complex numbers. The minimal polynomial of $A$ must divide $P$, hence it splits into linear factors with multiplicity $1$ and $A$ is diagonalizable.

Answer 3

By hypothesis the polynomial $x^2+bx+c$ is in the annihilator of the matrix $A$, an this polynomial factors as $x-\lambda)(x-\mu)$ over $\mathbf C$, since its discriminant is non-zero.

Observe that $x-\lambda$ and $x-\mu$ are coprime polynomials. Therefore we can apply the kernels lemma: if the vector space here is denoted $V$ $$V=\ker 0=\ker(A^2+bA+cI)=\ker(A-\lambda I)\oplus\ker(A-\mu I)$$ so that on these subspaces, the restriction of the matrix $A$ is $\lambda I$ and $\mu I$ respectively, and taking a basis in each of these subspaces to obtain a basis of $V$, $A$ takes the form of the block-diagonal matrix $$\begin{pmatrix}\lambda I& 0\\ 0&\mu I\end{pmatrix}$$

Answer 4

The conditions given show that $A$ satisfies $p(A) = 0$ for $p(x) = x^2+bx+c$, where $p$ has two distinct roots, say $\lambda_1,\lambda_2$.

Hence the minimal polynomial must divide $p$.

Consider the Jordan form of $A$. Each Jordan block is of size $1$, hence $A$ is diagonalisable.