Conditions for diagonalizability of a linear transformation T

Question

Given an n-dimensional vector space V, a linear transformation $T\colon V\rightarrow V$, with a list of distinct eigenvalues $\{\lambda_i\}_{i=1}^k$. Given a fixed eigenvalue $\lambda$, such that the eigenspace of $\lambda\ E_{\lambda}$ has dimension n-k+1, prove that T is diagonalizable.

I'm trying to prove that there exists a basis $\beta\subseteq V$ such that $\forall w_i\in \beta,\ \sum a_iT(w_i)=\lambda \sum a_iw_i$ for a collection of n scalars $a_i$ (I understand that is the definition of diagonizability).

So far I have the following:

Let $dim(E_{\lambda}):= p$. So $n=p+k-1$. So, given a basis $\gamma\subseteq E_{\lambda}\implies\forall u\in E_{\lambda}, \gamma_i\in\gamma,\sum_{i=1}^p c_iT(\gamma_i)=\lambda\sum_{i=1}^pc_i\gamma_i$.

Now, given a vector and basis $v\in V, \beta\subseteq V$ respectively, $\forall w_i\in\beta,\ T(v)=\sum_{i=1}^n a_iT(w_i)=\sum_{i=1}^p a_iT(w_i)+\sum_{i=p+1}^n a_iT(w_i)$

So far, my guess is, given a coordinate change, we can send $T(\gamma)$ to $T(\beta)$ such that we can make the following substitution:

$$T(v)=\lambda\sum_{i=1}^p a_iw_i+\lambda\sum_{i=p+1}^n a_iw_i$$

Thus finishing the proof, but I don't know why should the size of the base of the eigenspace be p for it to work (if it does work), or why should the base of V be a function of the amount of eigenvalues and the dimension of the eigenspace of one of those eigenvalues.

Thanks a lot

Answer 1

Assume your $k$ eigenvalues are numbered in such a way that the first one is this particular $\lambda,$ and denote by $p_1=n-k+1,p_2,\dots,p_k$ the dimensions of the corresponding eigenspaces. Since $p_2,\dots,p_k\ge1,$ $$\sum_{i=1}^kp_i\ge n-k+1+(k-1)\cdot1=n,$$ which is a (necessary and) sufficient condition of diagonalizability.

Answer 2

Adding to Anne Bauval's answer.

Since we concluded that $$\dim(E(\lambda_1,T)+\dots+E(\lambda_k, T))\ge n$$

We know that $\dim(E(\lambda_1,T)+\dots+E(\lambda_k, T))=n$ as $E(\lambda_1,T)+\dots+E(\lambda_k,T)$ is a subspace of $V$ and thus can't have dimension greater than $n$. Which implies that $E(\lambda_1,T)+\dots+E(\lambda_k, T)=V$

A basis of $E(\lambda_1,T)+\dots+E(\lambda_k, T)$(which is also a basis of $V$) consists of $n$ linearly independent eigenvectors of $T$. The matrix of $T$ with respect to this basis is a diagonal matrix and thus $T$ is diagonalizable.