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Let $B,C$ square matrices above a field,and $D$ Rectangle matrix in the correct size above the same field. $A=\begin{pmatrix} B &D \\ 0& C \end{pmatrix}$

I need to prove that if $A$ is Diagonalizable so Do $C$ and $B$,

And if $B$ and $c$ are Diagonalizable matrices with no common eigenvalue, so $A$ is Diagonalizable.

So I want to use the claim that says that if a matrix is Diagonalizable so it's Minimal polynomial splits to different factors with degree 1.

The second claim is easy because We know that $M_{A}(x)=M_{B}(x)M_{C}(x)$ and it is easily can be shown, but is this claim works the other way? Can I use it also the other way? And if not, what should I do to prove it?

(Please write simple and clear as you can cause I lose it with explanations of Linear algebra in English, differently from calculus, that for some reason it's more easy and fluent to me)

Thank you very much!

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I think your third line should be "then so are $B$ and $C$" (I mean, if $A$ is diagonalizable, then of course $A$ is diagonalizable)... Also, it seems your shift key is suffering from intermittent misfires; might want to take a look at it. –  Arturo Magidin Aug 10 '11 at 18:42

2 Answers 2

up vote 3 down vote accepted

Theorem. $A$ is diagonalizable over $\mathbf{F}$ if and only if the minimal polynomial of $A$ splits over $\mathbf{F}$ and is square free.

Proof. Let $\mu(t) = (\phi_1(t))^{n_1}\cdots(\phi_k(t))^{n_k}$ be the minimal polynomial of $A$, where $\phi_i$ are pairwise distinct monic irreducible polynomials, $n_j\geq 1$; and let $\chi(t) = (\phi_1(t))^{m_1}\cdots(\phi_k(t))^{m_k}$ be the characteristic polynomial, so $n_j\leq m_j$ for each $j$.

(It is a theorem that every irreducible factor of the characteristic polynomial must divide the minimal polynomial, and the Cayley-Hamilton Theorem shows that the minimal polynomial must divide the characteristic polynomial).

If $A$ is diagonalizable, then the characteristic polynomial splits, hence the minimal polynomial splits; and it is easy to verify that $(A-\lambda_1I)\cdots(A-\lambda_kI) = \mathbf{0}$, so the minimal polynomial is square free.

Conversely, if the minimal polynomial splits, then so does the characteristic polynomial; thus, $A$ has a Jordan canonical form. The largest block associated to the eigenvalue $\lambda$ equals the largest power of $t-\lambda$ that divides the minimal polynomial. The matrix is diagonalizable if and only if all the blocks have size $1$, so if the minimal polynomial is square free, then every block is of size $1$, so $A$ is diagonalizable. QED

It is not hard to verify that the minimal polynomial of $C$ must divide the minimal polynomial of $A$; for $B$, the result follows simply because we are looking at the restriction of $A$ to an $A$-invariant subspace. It follows that if $A$ is diagonalizable, then $B$ and $C$ have minimal polynomials that divide a squarefree polynomial that splits, so the minimal polynomials of $B$ and $C$ are squarefree and split, so $B$ and $C$ are diagonalizable. Conversely, if you happen to know that the minimal polynomial of $A$ equals the product of the minimal polynomials of $B$ and $C$, then if $B$ and $C$ are diagonalizable and have distinct eigenvalues, then the minimal polynomials of $B$ and $C$ split and are squarefree, and in addition they are relatively prime. So the minimal polynomial of $A$ splits, and since it is the product of two relatively prime squarefree polynomials, it is itself squarefree; thus, $A$ is diagonalizable.

The condition that $B$ and $C$ have distinct eigenvalues cannot be omitted. For example, $$A = \left(\begin{array}{c|cc} 1 & 1 & 0\\ \hline 0 & 1 & 0\\ 0 & 0 & 2 \end{array}\right),\quad B=\left(1\right),\quad C = \left(\begin{array}{cc}1 & 0\\0 & 2\end{array}\right),\quad D = \left( 1\quad 0\right),$$ then $B$ and $C$ are each diagonalizable, but $A$ is not (it is already in Jordan canonical form, and is not diagonal).

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It was very clear and understandable answer. Thank you very much. –  user6163 Aug 10 '11 at 19:19

I would like to clarify a small part of Arturo's excellent answer.

Let $p$ be any polynomial. Because $A^n=\pmatrix{ B^n & * \\ 0 & C^n}$ (where $*$ stands for some unknown entry), we have that $p(A)=\pmatrix{ p(B) & * \\ 0 & p(C)}$. Now, let $M_A$ be the minimal polynomial of $A$. Our computation shows that $M_A(B)=0$ and $M_A(C)=0$. Therefore, the minimal polynomial of $B$ and the minimal polynomial of $C$ ($M_B$ and $M_C$) both divide $M_A$. However, we do NOT have that $M_A=M_B M_C$ unless the eigenvalues of $B$ are distinct from the eigenvalues of $C$. For example, take $A=I_{2n}$ and $B=C=I_n$. We have that $M_A=M_B=M_C=(x-1)$.

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I write: "... it follows that if $A$ is diagonalizable, then $B$ and $C$ have minimal polynomials that divide a squarefree polynomial that splits, so the minimal polynomials of $B$ and $C$ are squarefree and split, so $B$ and $C$ are diagonalizable." In what sense am I missing the answer to "If $A$ is diagonalizable, then so are $B$ and $C$"? –  Arturo Magidin Aug 10 '11 at 19:53
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@Arturo Apologies. I was not reading carefully, and missed the second half of the sentence. The assumption that $M_A=M_B M_C$ is not necessarily true without specifying the eigenvalues are different, and the first half of your sentence was taking it as a given, so I assumed (incorrectly) that you were using it in an answer to the second question, not the first. –  Aaron Aug 10 '11 at 20:04
    
In that, you have a point; Nir claimed he knows that $\mu_A=\mu_B\mu_C$, but as you point out this claim is unnecessary for this part. –  Arturo Magidin Aug 10 '11 at 20:07

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