What does it mean to find a basis that "diagonalizes" a transformation?

Question

I'm having a hugely hard time wrapping my head around this statement. I am trying to figure it out on my own but I just don't get it. The terminology is weird to me and I can't really picture what it "looks like" in my head.

Could someone refer me to some literature that may clear this up?

Could someone give a "diagonalization-for-dummies" answer?

Answer 1

It means you find a set of independent vectors $(e_1, e_2, \cdots , e_n)$ such that $$ f(e_1) = \lambda_1 e_1,$$ $$ f(e_2) = \lambda_2 e_2,$$ $$ \cdots $$ $$ f(e_n) = \lambda_n e_n$$ Thus in this basis the matrix of the transformation is like this: $$ \begin{bmatrix} \lambda_1 & 0 & 0 & \cdots & 0\\ 0 & \lambda_2 & 0 & \cdots & 0 \\ \vdots & & \ddots & & \vdots \\ \vdots & & & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda_n\\ \end{bmatrix} $$

Answer 2

In general, if $v_1,...,v_n$ is a basis for a vector space, and $L$ is a linear map on the space, then $L(v_k)$ will be some complicated sum of all the basis elements. It is best when $L(v_k)$ is a multiple of itself, that is as good as it can be. We call that "diagonolizable" basis.