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

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


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

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.

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

2 Answers 2


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.

  • $\begingroup$ Oh yes, I did this in this way! Nice! $\endgroup$ Commented Nov 17, 2022 at 3:48

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .