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.


Note that composition of injective maps is injective. Here a non zero vector is sent to a zero vector, which means that at least one of the maps has a non trivial kernel, and hence is not injective.

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?

  • $\begingroup$ I think I understand. I think my main problem coming into this is I'm not exactly sure what it means for a map to be of a basis. I know what a basis is, in terms of vectors, but the concept seems a little fuzzy to me when applied to a matrix. $\endgroup$ – Astrum Jan 19 '14 at 5:41
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    $\begingroup$ A matrix is always computed with respect to a basis. The first column of a matrix is the vector you get when you apply the linear operator to the first basis vector etc. I am sure you will get better at this concept with time. $\endgroup$ – voldemort Jan 19 '14 at 5:43
  • $\begingroup$ Oh, that's starting to make a lot more sense now! Thanks! So, the first column of the transformation is what you get when applied to the first basis vector, the second column is from the second basis vector and so on? $\endgroup$ – Astrum Jan 19 '14 at 5:49
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    $\begingroup$ yes that's it. :) $\endgroup$ – voldemort Jan 19 '14 at 5:50

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