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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

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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$~T$).

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  • $\begingroup$ Does that mean that the null space of $T-cI$ overlaps? I still don't get that part $\endgroup$ – user10024395 Oct 14 '14 at 12:35
  • $\begingroup$ @user136266: overlaps what? $\endgroup$ – Marc van Leeuwen Oct 14 '14 at 13:09
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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 $0$.

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  • $\begingroup$ 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. $\endgroup$ – Arnaud D. Nov 27 '17 at 10:23

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