Study on the matrix $A$ satisfying $(A-\alpha I)(A-\beta I)=O$

Question

Suppose that $A$ is an $n\times n$ square matrix satisfying $(A-\alpha I)(A-\beta I)=O$, where $\alpha$ and $\beta$ are distinct nonzero real numbers. I would like to study some of the characteristics of $A$, including the invertibility and diagonalizability.

Firstly expanding it we have $A^2-(\alpha+\beta)A+\alpha\beta I = O$. Then we observe that $A$ is invertible: Rewrite $A^2-(\alpha+\beta)A+\alpha\beta I = 0$ as $A(A-(\alpha+\beta)I)=-\alpha\beta I$, we have $$A^{-1}=\frac{\alpha+\beta}{\alpha\beta} A - \frac{1}{\alpha\beta} I.$$ But I am also interested in the invertibility of matrices that are closely related to $A$, such as $A-\alpha I$, $A-\beta I$ and $2A-(\alpha+\beta)I$ (which is $A-\alpha I + A-\beta I$).

Now we may study the eigenvalues of $A$. Let $v$ be an eigenvector of $A$ corresponding to the eigenvalue $\lambda$. Then from $A^2-(\alpha+\beta)A+\alpha\beta I = O$ we have \begin{align} (A^2-(\alpha+\beta)A+\alpha\beta I)v &= Ov \\ (\lambda^2-(\alpha+\beta)\lambda +\alpha\beta )v &= 0 \end{align} which gives $\lambda^2-(\alpha+\beta)\lambda +\alpha\beta=0$, i.e. $\lambda = \alpha, \beta$, since $v$ is nonzero. Then at least one of $A-\alpha I$ and $A-\beta I$ is noninvertible. However, $2A-(\alpha+\beta)I$ is invertible since $\frac{\alpha+\beta}{2}$ is not an eigenvalue of $A$.

The first question. I know that $\alpha$ and $\beta$ are the only eigenvalues of $A$. But is it possible to find their multiplicities? If the characteristics polynomial of $A$ is $p(t)$, then by Cayley-Hamilton we have $$p(A)=O=(A-\alpha I)^{m_1}(A-\beta I)^{m_2}$$ and so $p(t)=(t-\alpha I)^{m_1}(t-\beta I)^{m_2}$ ,where $m_1$ and $m_2$ are the multiplicities of $\alpha$ and $\beta$ respectively.

The second question. I guess $A$ is indeed diagonalizable. But I can't proceed to prove it.

Any help is highly appreciated.

Answer 1

You are wrong in claiming that $\alpha,\beta$ are eigenvalues of $A$. Since $A$ satisfies the equation $(A-\alpha I)(A-\beta I)=0$, the minimal polynomial of $A$ must divide $(x-\alpha)(x-\beta)$. So, at least one of $\alpha$ and $\beta$ is an eigenvalue of $A$, but it is not necessarily true that both of them are eigenvalues of $A$. (An easy counterexample: $A=\alpha I$.)

At any rate, since $\alpha$ and $\beta$ are distinct, the minimal polynomial of $A$, which divides $(x-\alpha)(x-\beta)$, must be a product of distinct linear factors. It follows that $A$ is diagonalisable.

As for invertibility, as we only know that at least one of $\alpha,\beta$ is an eigenvalue of $A$, we can guarantee that $A$ is invertible only when both $\alpha$ and $\beta$ are nonzero. If exactly one of them is zero, say, $\alpha\ne0=\beta$, then $A$ may be singular (e.g. when $A=0$) or non-singular (e.g. when $A=\alpha I$).

Now, for your first question, yes, the multiplicity of any eigenvalue of $A$ in the characteristic equation is by definition the algebraic multiplicity of $A$, but as we've seen in the above, that $(A-\alpha I)(A-\beta I)=0$ doesn't make both $\alpha$ and $\beta$ eigenvalues of $A$.

So, the answer to your question is "no", unless both $\alpha$ and $\beta$ are eigenvalues of $A$. Since $A$ is diagonalisable, this should be trivial if you look at the diagonalised matrix.

Answer 2

$A$ is diagonalizable iff its minimal polynomial splits into linear factors. Recall few more facts. Roots of minimal polynomial are the eigen values of $A$. Also minimal polynomial divides characteristic polynomial. So in this case the only eigen values of $A$ are $\alpha$ and $\beta$ both of which are non zero. So $A$ is invertible [Since, A is invertible iff 0 is not an eigen value]. Also note that minimal polynomial is a product of linear factors. Hence $A$ is diagonalizable as well. Note that $m_1+m_2=n$. Beyond this nothing more can be said about the multiplicities of $\alpha$ and $\beta$.