Prop: Suppose that $E_\lambda$ is an eigenspace of a linear operator $T$ with dimension $d$. Then $(t-\lambda)^d$ divides $\chi_T$.
So I already know that if $T$ is our linear operator, if we consider the restriction of $T$ onto a $T$-invariant subspace, then $\chi_{T|_{E_\lambda}}$ is going to divide the characteristic polynomial of $T$. My question is why is it that $$ \chi_{T|_{E_\lambda}} = (t-\lambda)^d. $$ I understand that as an Eigenspace $T$ acts on vectors in $V$ as multiplication by $\lambda$, but I don't see why this implies the form of the characteristic polynomial of the restriction. Is there a way that I can justify for an eigenbasis $\beta$ that $$ [T|_{E_\lambda}]_\beta^\beta = \begin{pmatrix} \lambda & 0 & ... & 0 \\ 0 & \lambda & ... & 0 \\ 0 & & \ddots & \\ 0 &... & & \lambda \end{pmatrix} $$ which in my mind would give $$ \det [t \cdot I - T|_{E_\lambda}]_\beta^\beta = \det \begin{pmatrix} t-\lambda & 0 & ... & 0 \\ 0 & t-\lambda & ... & 0 \\ 0 & & \ddots & \\ 0 &... & & t-\lambda \end{pmatrix} = (t-\lambda)^d. $$ If I left out any necessary background information let me know and I will fill it in. Thanks in advance for any help and clarifications!