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I found a proof on Math Stackexchange that a matrix can be upper triangularized, but I was confused by the proof they gave. I copy pasted it:

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I know of a theorem from Axler's Linear Algebra Done Right which says that if $T$ is a linear operator on a complex finite dimensional vector space $V$, then there exists a basis $B$ for $V$ such that the matrix of $T$ with respect to the basis $B$ is upper triangular.

The proof of this theorem is by induction on the dimension of $V$. For dim $V = 1$ the result clearly holds, so suppose that the result holds for vector spaces of dimension less than $V$. Let $\lambda$ be an eigenvalue of $T$, which we know exists for $V$ is a complex vector space.

Consider $U = $ Im $(T - \lambda I)$. It is not hard to show that Im $(T - \lambda I)$ is an invariant subspace under $T$ of dimension less than $V$.

So by the induction hypothesis, $T|_U$ is an upper triangular matrix. So let $u_1 \ldots u_n$ be a basis for $U$. Extending this to a basis $u_1 \ldots u_n, v_1 \ldots v_m$ of $V$, the proof is complete by noting that for each $k$ such that $1 \leq k \leq m$, $T(v_k) \in $ span $\{u_1 \ldots u_n, v_1 \ldots v_k\}$.

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Why that last step? Is this really a correct proof?

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  • $\begingroup$ Are you asking why is $T(v_k)$ in the span? Isn't that a basis of the space? $\endgroup$ – N. S. May 18 '16 at 23:04
  • $\begingroup$ Yes, but why do $v_{k+1}, ... , v_m$ not show up? $\endgroup$ – D_S May 19 '16 at 0:18
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Yes, this is a correct proof but it omits a few details that are included in the proof in the book. In the book, this is Theorem 5. 13 and it is preceded by Proposition 5. 12:

Suppose $T \in \mathcal L(V)$ and $(v_1, \dots, v_n)$ is a basis of $V$. Then the following are equivalent:

(a) the matrix of $T$ with respect to $(v_1, \dots, v_n)$ is upper triangular

(b) $Tv_k\in \text{span} (v_1, \dots, v_n)$ for each $k = 1, \dots, n$

(c) $\text{span} (v_1, \dots, v_n)$ is invariant under $T$ for each $k =1, \dots, n$

This proposition is used in the proof of theorem 5. 13:

First show that $T u_i \in \text{span} (u_1, \dots, u_n)$ then show that $T v_i \in \text{span} (u_1, \dots, u_n, v_1, \dots, v_m)$.

Together these mean that $Tx \in \text{span} (u_1, \dots, u_n, v_1, \dots, v_m)$ for all $x \in (u_1, \dots, u_n, v_1, \dots, v_m)$ hence $T$ is upper triangular.

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