# Minimal polynomial for T is a product of distinct linear factors

There proof given in my textbook is as follows:

Let T be a diagonalisable linear operator and let $c_1, ..., c_k$ be the distinct eigenvalues of T. Then it is easy to see that the minimal polynomial for T is the polynomial $$p = (x-c_1)...(x-c_k)$$

If $\alpha$ is an eigenvector, then one of the operators $T-c_1I,...,T-c_kI$ sends $\alpha$ into 0. Therefore $$(T-c_1 I)...(T-c_kI)\alpha = 0$$

for every eigenvector $\alpha$. There is a basis for the underlying space which consists of eigenvectors of T and hence $$p(T) = (T -c_1I)...(T-c_kI) = 0$$

I cannot understand this proof at all. I understand that one of the operators $T-c_1I,...,T-c_kI$ sends $\alpha$ into 0, but I can't see why we we multiply them together, it still works. And I also don't understand why there is a basis for the underlying space. Can someone kindly explain to me please. Thanks

The main point that is no stated explicitly in the argument, but is used nonetheless, is that appying any $T-cI$ to an eigenvector for$~\lambda$ results in another eigenvector for$~\lambda$, or in $0$ (the latter in case $c=\lambda$). Indeed the line spanned by such an eigenvector$~\alpha$ is invariant under $T$ (and of course under $I$), and all its nonzero elements are also eigenvectors for$~\lambda$. (One could even define eigenvectors as those vectors that span a $1$-dimensional $T$-stable subspace; this is indeed the initial definition I give in my LA course.) Concretely, applying the product of factors $T-c_iI$ to$~\alpha$ maybe transforms it into another eigenvector several times, but when it meets its own eigenvalue it gets killed, and stays dead (zero) from then on.

That there exists a basis of the whole space consisting of eigenvectors for$~T$ is the definition of $T$ being diagonalisable (as the matrix of$~T$ on any basis will be diagonal if and only if that basis consists entirely of eigenvectors for$~T$).

• Does that mean that the null space of $T-cI$ overlaps? I still don't get that part – user10024395 Oct 14 '14 at 12:35
• @user136266: overlaps what? – Marc van Leeuwen Oct 14 '14 at 13:09

You write

I understand that one of the operators $T-c_1I,...,T-c_kI$ sends $\alpha$ into $0$, but I can't see why we we multiply them together, it still works.

Note that the factors $T-c_iI$ all commutes with one another : for any $i,j$ we have $$(T-c_iI)(T-c_jI)=T^2-c_iT-c_jT+c_ic_jI=(T-c_jI)(T-c_iI).$$ In particular, if $\alpha$ is an eigenvector, then $(T-c_iI)\alpha=0$ for some $i$, and you can rewrite the product with the term $(T-c_iI)$ at the end, so that \begin{align}(T-c_1I)...(T-c_kI)\alpha & =(T-c_I)\dots(T-c_{i-1}I)(T-c_{i+1}I)\dots(T-c_kI)(T-c_iI)\alpha \\& = (T-c_I)\dots(T-c_{i-1}I)(T-c_{i+1}I)\dots(T-c_kI) 0 \\ & = 0\end{align} since every linear operator preserves $0$.

• I know this is an old question, but I found it while writing this for a duplicate, and I thought it might be useful for some people to have a different answer. – Arnaud D. Nov 27 '17 at 10:23