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Question: Prove than a square matrix $A$, with complex entries, is diagonalizable if and only if the minimal polynomial of $A$ has distinct roots.

In this answer Prove that T is diagonalizable if and only if the minimal polynomial of T has no repeated roots., the last answer references being able to do this using Jordan Canonical Form. I was wondering if someone could shed some light on this. I know how the minimal polynomial and characteristic polynomial of a matrix affects the Jordan Canonical Form, and so I know that if $m_A(t)$ factors as a product of distinct linear factors, then we have a Jordan Block of size $1$ for each eigenvalue... so $A$ must then be diagonalizable..?

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    $\begingroup$ If the minimal poly factors into distinct (linear) factors, then ALL Jordan blocks have size $1$, which means the Jordan form is diagonal. There may be multiple such blocks with the same eigenvalue, if the characteristic poly has repeated roots. $\endgroup$
    – Ned
    Commented Sep 9, 2021 at 20:53
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    $\begingroup$ If the Jordan blocks are all size $1$, then the JNF literally is a diagonal matrix. Thus, $A$ is similar to a diagonal matrix, and hence is diagonalisable. $\endgroup$ Commented Sep 9, 2021 at 21:01
  • $\begingroup$ @TheoBendit Ah, okay, so if the minimal polynomial has distinct roots, the the JCF is just a diagonal matrix where are the Jordan Blocks are size $1$. For the other direction, could I just say that if $A$ is diagonalizable then each Jordan block has size $1$, thus the degree of any factor of the minimal polynomial is $1$, hence the minimal polynomial must be a product of distinct linear factors and so all roots are distinct. $\endgroup$
    – User7238
    Commented Sep 9, 2021 at 21:29
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    $\begingroup$ @User7238 Yes, that's right, assuming you're happy with each of those individual steps (i.e. you don't have to explain any of them further). $\endgroup$ Commented Sep 9, 2021 at 21:38
  • $\begingroup$ @TheoBendit Revisiting this problem, in my comment, I say "if $A$ is diagonalizable then each Jordan block has size $1$". I felt like at the time this was obviously true, but something isn't quite sitting right with me about it. If you'd like, if you justified that in an answer (even if it is very short), I would be happy to accept that answer so the question is not left "unanswered". :) $\endgroup$
    – User7238
    Commented Jan 7, 2022 at 22:36

2 Answers 2

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As requested, I'll answer the question in the comments:

If $A$ is diagonalizable then each Jordan block has size $1$.

If $A$ is diagonalisable, then $A$ is similar to a diagonal matrix. Diagonal matrices are automatically in Jordan Canonical Form; they are block diagonal matrices with $1 \times 1$ Jordan cells. This means that one of the Jordan Canonical Forms (they are not, strictly speaking, unique) is a diagonal matrix.

Now, as I say, the JCF is not unique (typically). Of course, there is some rigidity here though: the JCF is unique up to a permutation of the Jordan Blocks along the diagonal. So, if a Jordan block appears in one Jordan Canonical Form of a matrix, then it will appear in every other JCF of the matrix, just in various different positions.

So, to sum up, if $A$, a complex $n \times n$ matrix, is diagonalisable, then $A$ is similar to a diagonal matrix $D$. $D$ is a matrix already in Jordan normal form, as each of the $1 \times 1$ submatrices along the diagonal are (trivially) Jordan Blocks. By the uniqueness result alluded to above, this means that every JCF of the matrix has the same $1 \times 1$ Jordan Blocks.

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It is overkill to refer to the Jordan form to answer this question. That a diagonalisable operator$~T$ has a minimal polynomials that (is split and) has simple roots is easy to prove: as any polynomial$~P$ acts on any eigenspace of$~T$ for$~\lambda$ by the scalar$~P[\lambda]$, the annihilating polynomials of$~T$ are precisely those that have every eigenvalue$~\lambda$ as root, so that the minimal polynomial is the product of factors $X-\lambda$ for such$~\lambda$, each factor taken just once.

The converse is a bit harder. One can prove somewhat more generally that if a split polynomial with simple roots $P=\prod_{i=1}^k(X-a_i)$ annihilates$~T$, then the space is a direct sum of the subspaces $V_i=\ker(T-a_iI)$ for $i=1,\ldots,k$, so that $T$ is diagonalisable with its set of eigenvalues contained in $\{a_1,\ldots,a_k\}$. (This will apply in particular when $P$ is the minimal polynomial of$~T$, in case it is split with simple roots.) I know two ways to proceed, one using basic linear algebra technique and one using a bit of polynomial arithmetic.

The first method uses the fact that eigenspaces for distinct eigenvalues form a direct sum, and that the dimension of the kernel of a composition of linear maps is at most the sum of the dimensions of their individual kernels. Note that the kernel of $P[T]=(T-a_1I)\circ\cdots\circ(T-a_kI)$ is the whole space (since the composition is assumed to be zero), so that $\dim(V)\leq\dim(V_1)+\cdots+\dim(V_k)$ by the second mentioned property, while $\dim(V_1)+\cdots+\dim(V_k)=\dim(V_1\oplus\cdots\oplus V_k)$ by the first mentioned property. This shows that the sum of the eigenspaces$~V_i$ fills all of $V$ (and the inequality is an equality), so $T$ is diagonalisable.

For the second method, one may first show that whenever a polynomial $P$ annihilating $T$ decomposes as a product of two relatively prime factors $P=QR$, the space decomposes $V=V_1\oplus V_2$ as a direct sum of $V_1=\ker(Q[T])$ and $V_2=\ker(R[T])$, and the corresponding projections of $V$ on $V_1$ and $V_2$ are given by polynomials in$~T$. To this end write a Bézout relation $AQ+BR=1$, which exists because $Q,R$ are relatively prime, put $\pi_1=(BR)[T]$ and $\pi_2=(AQ)[T]$, so that $\pi_1+\pi_2=I$. Then $\pi_1$ vanishes on $V_2$, and has its image contained in $V_1$ (since $QBR$ is a multiple of $QR=P$, and therefore an annihilating polynomial of $T$) and vice versa for $\pi_2$. Now $v=\pi_1(v)+\pi_2(v)$ for every $v\in V$ shows that $V=V_1+V_2$, while if $v_1\in V_1$ and $v_2\in V_2$ one gets $\pi_1(v_1+v_2)=(\pi_1+\pi_2)(v_1)=v_1$ and similarly $\pi_2(v_1+v_2)=v_2$, so that $V=V_1\oplus V_2$ and $\pi_i$ is the corresponding projection of $V$ onto $V_i$ for $i=1,2$. Applying this result with $Q=X-a_1$ the first factor of $P$ and $R=\prod_{i=2}^k(X-a_i)$ the product of the remaining factors, one gets a direct sum decomposition of$~V$ into the eigenspace $\ker(T-a_1I)$ of $T$ for $a_1$ and a subspace $\ker(R[T])$ restricted to which $R$ is an annihilating polynomial, and which by induction on $k$ can be further decomposed as a direct sum of eigenspaces.

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