# Matrix of Linear Map/Operator - Triangular

While proving that any operator on a complex vector space has at least one eigenvalue, the very end of the proof states:

$0=c(T-\lambda _1I)\cdots(T-\lambda _m I)v$ means that the map $(T-\lambda _i I)$ is not injective for at least one $i$. Why does this make sense?

My book has done a rather poor job in explaining why a linear map relative to a basis is always upper triangular, and why the eigenvalues will be diagonal.

It also gives no concrete examples! I know what a triangular matrix looks like, but not how/why it is this way.

The following is a proposition from my text:

Suppose $T: V \rightarrow V$ and $(v_1, \ldots, v_k)$ is a basis of $V$. The the following are equivalent:

a)The matrix of $T$ with respect to $(v_1,\ldots,v_n)$ is upper triangular.

b) $Tv_k \in \operatorname{span} (v_1,\ldots,v_k)$for each $k=1,\ldots,n$

c) $\operatorname{span} (v_1,\ldots,v_k)$ is invariant under $T$ for each $k=1,\ldots,n$

I'm rather confused, if anybody could explain this to me, I'd really appreciate it. I'm not sure if my book is not very clear, or I'm just having problems with the material in general.

The second part follows from the definition of a "matrix representation". Suppose you write $T$ as a matrix with respect to the basis $v_1,..v_n$. If $Tv_k$ lies in the span of $v_1,..v_k$ then it implies that in the matrix representation, with respect to the above basis, the matrix is upper triangular, as the coeffecients of $Tv_k$ form the columns of the matrix $T$. By linearity of $T$ if $Tv_k$ lies in span of $v_1,..v_k$ then $c_1v_1+...c_kv_k$ again lies in span of $v_1,..v_k$, and so span $v_1,..v_k$ is invariant under $T$. Does this make more sense now?