Existence of upper triangular matrix proof I am reading Linear ALgebra Done Right and Thm $5.27$ says: 

Suppose $\textsf V$ is a finite-dimensional complex vector space.
  Then a square matrix $\textsf T$ has an upper-triangular matrix with respect to some basis of $\textsf V$.

The proof given is based on induction and I am not getting much of a feel for it. Is there another proof of this theorem, I cant find any? Hopefully more intuitive. 
My understanding so far is that suppose $\textsf T$,of $\dim (\textsf{V})$, has null space as $\{0\}$. Then I can select any linearly independent set of vectors and make that my basis. So I choose vectors that satisfy the upper triangular matrix template such as $(x,0,0,0,0,.....),(y,z,0,0,0....),\dots$
Any vectors of this form are linearly independent and thus I select this as my basis.(This would imply upper-triangular matrices are not unique).
But what about the case where null space has dimension greater than zero. How is upper-triangular matrix possible here?
 A: 

Bonus Theorem. If $k$ is a field then the following are equivalent: (i) every square matrix over $k$ has an eigenvalue (in $k$) (ii) every square matrix is similar to an upper-triangular matrix (iii) the characteristic polynomial of every square matrix splits over $k$.


Note: As I suspected when I  wrote this, (i) is also equivalent to saying $k$ is algebraically closed. See here for that.
Anyway, (ii) implies (iii) is trivial, as is (iii) implies (i). For (i) implies (ii): Assume (i).
First note that if $B=(b_1,\dots,b_n)$ is a basis then 


Lemma 1. The matrix for $T$ wrt the basis  $B$ is upper triangular if and only if $Tb_k$ is a linear combination of $b_1,\dots,b_k$ for every $k$.


So it's enough to prove this:


Lemma 2. There exists a non--zero  $v\in V$ such that if  $E$ is the span of $v$ then there is a subspace $W$ with $V=E\oplus W$ and $TW\subset W$.


Proof that Lemma 2 implies the result: By induction on the dimension there is a basis $b_1,\dots,b_{n-1}$ for $W$ such that  $Tb_k$ is a linear   combination of $b_1,\dots,b_k$ for $1\le k\le n-1$. Let $b_n=v$.
Edit: This is seeming easier. At first I was wondering how to get $v$ as in Lemma 2. But no, we just concentrate on $W$ and then $v$ comes along for free. It's clear that Lemma 2 follows from Lemma 3:


Lemma 3. There is a subspace $W\subset V$ of codimension $1$ such that $TW\subset W$.


(Given $W$ as in Lemma 3, let $v$ be any non-zero element of $V\setminus W$ and Lemma 2 follows.)
And Lemma 3 is finally something we can prove:
Let $\Lambda$ be an eigenvector of the adjoint  $T^*:V^*\to V^*$, and let $W$ be the kernel of $\Lambda$. Then $W$ certainly has codimension $1$, and if $x\in W$ then $$\Lambda(Tx)=(T^*\Lambda)x=\lambda\Lambda x=0;$$hence $TW\subset W$.
A: I don't quite follow your argument assuming that the nullspace is trivial. But if you can do that case you're done:
Since $T$ has only finitely many eigenvalues there exists $\lambda$ such that $T-\lambda I$ is non-singular. So there is a basis making $T-\lambda I$ upper triangular; now $T$ is upper triangular in the same basis.
But alas it can't be entirely trivial in the non-singular case. Because the result implies that every $T$ has an eigenvalue. Since the argument above works in any infinite field, the proof in the non-singular case has to have some step that doesn't work if the scalar field is $\Bbb Q$ or $\Bbb R$.
