# Consider a linear operator $L$ and some polynomial of it, $L'=p(L)$. Show that the minimal polynomial of $L'$ has smaller degree than that of $L$.

Consider a linear operator $L$ and some polynomial of it, $L'=p(L)$. Show that the minimal polynomial of $L'$ has degree less than or equal to the minimal polynomial of $L$.

First, start working over the algebraic closure of whatever field we were on originally.

Better proof: Put $L$ into its Jordan normal form. Then we can restrict our attention to each Jordan block as we have invariant subspaces under $L$ and thus also under $L'$. Thus the dimension of the subspaces of the corresponding blocks cannot grow. The degree of each linear factor in the minimal polynomial gives the size of the largest Jordan block for the corresponding eigenvalue, and thus it does not grow and so the degree

An attempt at a similar proof not using Jordan form: The eigenvectors of $L$ must be eigenvectors of $L'$. A little more is true, the eigendecomposition is essentially entirely preserved. I.e. if $v_1, v_2$ were eigenvectors of $L$ corresponding to the same eigenvalue $\lambda$, then any linear combo of them is an eigenvector of $L'$ with eigenvalue $p(\lambda)$. So vectors that were eigenvectors of $L$ will continue to be eigenvectors of $L'$. However, polynomials are not one to one, so eigenvalues may converge. Now let $\Pi (x-\lambda_i)^{n_i}$ be the minimal polynomial for $L$. I want to claim the minimal polynomial for $L'$ divides $\Pi (x-p(\lambda_i))^{n_i}$. All I need is to show that $\Pi (L'-p(\lambda_i))^{n_i}=0$. The issue I have is getting polynomials to play nice. Since they aren't linear, I'm struggling to get this to work out.

Any other proof ideas or hints would be lovely!

Let $m$ be the minimal degree of $L$. Consider the $m+1$ operators $(L')^0, (L')^1, \ldots (L')^m$. Reduce each one of these mod the minimum polynomial of $L$. Show that these $m+1$ operators are linearly dependent.