During a test today I had this question:

Given $$M = \begin{pmatrix} A & C \\ 0 & B\\ \end{pmatrix}$$ where $$A$$ and $$B$$ are $$n \times n$$ diagonalizables matrices without eigenvalues in common, prove that $$M$$ is diagonalizable.

No information about $$C$$ was given. First, I tried

$$\det(M - \lambda I) = \det(A - \lambda I)\cdot \det(B - \lambda I)$$

So

$$p_M(\lambda)=p_A(\lambda)\cdot p_B(\lambda)$$

So the set of eigenvalues of $$M$$ is the union of the eigenvalues of $$A$$ and $$B$$ (given they don't have any in common). My next step was to do

$$M^k = \begin{pmatrix} A^k & C'\\ 0 & B^k\\ \end{pmatrix}$$

so if $$m_M(x)$$ is the minimal polynomial of $$M$$, we have that $$m_A(x)|m_M(x)$$ and $$m_B(x)|m_M(x)$$. But how can I conclude that hence the minimal polynomial of M will have just linear factors? I know that $$m_M(x) = m_A(x)\cdot m_B(x)\cdot Q(x)$$, but how can I show that $$Q(x) = 1$$? If $$m_M(x) = m_A(x)\cdot m_B(x)$$ it´s clear that M is diagonalizable, but I can´t see how to prove this.

In fact, in the end my approach was the same of the first answer, but after I still looking to a solution using the minimal polynomial.

• Maybe you only need $p_M$. Have you taken a look at the algebraic and geometric multiplicities? Commented Nov 17, 2022 at 0:33
• @RodrigodeAzevedo but M has order 2n and A and B just n. How can I proof that the geometric multiplicities will remain the same? Commented Nov 17, 2022 at 0:38
• Can you prove that the algebraic multiplicities remain the same? Commented Nov 17, 2022 at 0:46
• @RodrigodeAzevedo this follows from the fact that $p_A(x)$ and $p_B(x)$ don´t share roots Commented Nov 17, 2022 at 0:56

The eigenvalues are union of eigen values of $$A,B$$ and the matrix is diagonalizable. This can be seen as follows:
Let $$v$$ be an eigen vector of $$A$$ then $$[v;0]$$ is an eigen vector of $$M$$. Now if $$w$$ is an eigen vector of B then $$M[w';w] =[Aw'+Cw, Bw]= [Aw'+Cw, \lambda_B w]$$. we need $$[Aw'+Cw, \lambda_B w] = [\lambda_B w';\lambda_B w]$$. Now solve $$Aw'+Cw = \lambda_B w'$$=>$$(A-\lambda_B I)w' = -Cw$$. Since $$\lambda_B$$ is not an eigen value of $$A$$, $$A-\lambda_B I$$ is non-singular. So we can solve for $$w'$$. So we have produced all the linearly independent eigen vectors for $$M$$. This proves the diagonalizability.
You may continue as follows. Let $$m_A(x)=\prod_{i=1}^r(\lambda_i-x),\ m_B(x)=\prod_{j=1}^s(\mu_j-x)$$ and $$(x_1,x_2,\ldots,x_{r+s})=(\lambda_1,\ldots,\lambda_r,\mu_1,\ldots,\mu_s)$$. Define $$M_i=M-x_iI$$ and likewise for $$A_i$$ and $$B_i$$. One may prove by mathematical induction that for every $$k\le r+s$$, $$M_1M_2\cdots M_k=\pmatrix{A_1A_2\cdots A_k&\sum_{i=1}^k A_1A_2\cdots A_{i-1}CB_{i+1}B_{i+2}\cdots B_k\\ 0&B_1B_2\cdots B_k}.$$ However,
• when $$i\le r$$, we have $$B_{i+1}B_{i+2}\cdots B_{r+s}=(B_{i+1}\cdots B_r)(B_{r+1}\cdots B_{r+s})=(B_{i+1}\cdots B_r)m_B(B)=0$$;
• when $$i>r$$, we have $$A_1A_2\cdots A_{i-1}=(A_1A_2\cdots A_r)(A_{r+1}\cdots A_{i-1})=m_A(A)(A_{r+1}\cdots A_{i-1})=0$$.
Therefore $$(m_Am_B)(M)=M_1M_2\cdots M_{r+s}=0$$ and $$m_M|m_Am_B$$. Now use the assumption that $$A$$ and $$B$$ are two diagonalisable matrices that do not share any eigenvalue to conclude that $$m_Am_B$$ and in turn $$m_M$$ are products of distinct linear factors.