# Jordan decomposition of an endomorphism with minimal polynomial

Let $V$ be a vector space over a field $F$ of characteristic zero. Below is a proof of the fact that if $T\in\text{End}(V)$ and $T$ admits a minimal polynomial with distinct roots $\lambda_1,\ldots,\lambda_n$, then $$V=N_1\oplus\cdots\oplus N_n,$$ where $N_k=\ker (T-\lambda_k\,I)$.

Question: does the result still hold if we relax the condition on the characteristic? On the simplicity of the roots?

Here is the proof. Consider the polynomial $$q(t)=\sum_{k=1}^n\prod_{j\ne k}\frac{(t-\lambda_j)}{(\lambda_k-\lambda_j)}$$ This is well-defined because $\lambda_1,\ldots,\lambda_n$ are distinct. We have $q(\lambda_k)=1$ for all $k=1,\ldots,n$. As the degree of $q$ is $n-1$, we deduce that $q(t)=1$ for all $t$. Then $q(T)=I$. For any $x\in V$, we can write $x=q(T)x$, i.e. $$\tag{1} x=\sum_{k=1}^n\,\prod_{j\ne k}\frac{(T-\lambda_j\,I)\,x}{(\lambda_k-\lambda_j)} .$$ It is clear that $\prod_{j\ne k}\frac{(T-\lambda_j\,I)\,x}{(\lambda_k-\lambda_j)}\in N_k$, since $$(T-\lambda_k\,I)\,\prod_{j\ne k}\frac{(T-\lambda_j\,I)\,x}{(\lambda_k-\lambda_j)}=p(T)x=0.$$ Finally, if $y\in N_k\cap N_j$ for $k\ne j$, then $$0=Ty-\lambda_ky=Ty-\lambda_jy,$$ so $(\lambda_k-\lambda_j)y=0$. As $\lambda_k-\lambda_j\ne0$, we have $y=0$. So the decomposition $(1)$ is indeed a direct sum.

The fact that $p$ is the minimal polynomial guarantees that $N_k\ne\{0\}$ for all $k$: if, for example, $N_1=\{0\}$, then $(T-\lambda_2\,I)\cdots(T-\lambda_n\,I)x=0$ for all $x\in V$, and then $p$ would not be minimal.

• This proof works in any characteristic. Now, if $T\neq 0$, but $T^2=0$, then $ker(T)\neq V$. So we can not drop the simplicity. – Daniel Oct 6 '15 at 21:57
• Should that $\prod_{j \neq k}$ be there? – Ben Grossmann Oct 6 '15 at 22:41
• @Omnomnomnom: good point, edited. Thanks! – Martin Argerami Oct 6 '15 at 23:43
• @Daniel: thanks! – Martin Argerami Oct 6 '15 at 23:44

The proof works over any characteristic. In particular, you are only every dividing by $\lambda_j - \lambda_k$, and this is fine as long as $\lambda_j \neq \lambda_k$.
The simplicity assumption cannot be dropped. The matrix $$\pmatrix{0&1\\0&0}$$ is a consistent counterexample.