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

Question

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..?

Answer 1

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.

Answer 2

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.