Linearly Independence Of Eigenvectors

Theorem: Let $T$ be a linear operator on a vector space $V$, and let $\lambda_1,\lambda_2,\lambda_3..\lambda_k$**be distinct eigenvalues of T. If $v_1,v_2,...v_k$ are eigenvectors of $T$ such that $\lambda_i$ corresponds to $v_i$, then $\{v_i\mid i = 0,1,2,3..k\}$ is linearly independent.

Proof : by mathematical induction, Suppose k =1 then $\{v_1\}$ is LI.

Suppose the theorem holds for the case of k-1 distinct eigenvalues.

and also Suppose tha $a_1v_1+a_2v_2+...+a_kv_k=0$ then Applying $T-\lambda_kI$ then we obtain

$a_1(\lambda_1-\lambda_k)v_1+a_2(\lambda_2-\lambda_k)v_1....a_{k-1}(\lambda_{k-1}-\lambda_k)v_{k-1}$.. (*)

(...)

Question. For the (*), I can' understand what "Applying $T-\lambda_kI$" means.

Any help? After the advice, will complete the proof for archiving.

It's like you have a function $f:X\rightarrow Y$, say $x,y\in X$ and $x=y$ then apply $f$ both sides to get $f(x)=f(y)$
$T-\lambda_k I$ is a function from $V\rightarrow V$
It's given $a_1v_1+a_2{v_2}\dots+a_k{v_k}=0$