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A friend of mine is studying physics in first semester and for his next assignment, he has to prove the following theorem:

Let $V$ be a finite dimensional vector space over an algebraically closed field $K$. Further, let $f: V \to V$ be an endomorphism. Then there exists a basis $B$ of $V$, such that $\mbox{Mat}_{B,B}(f)$ is an upper triangular matrix.

Now this theorem really stumbles me, because I know two proofs of it but they are way beyond first semester. They have only introduced elementary matrix/basis manipulation, basis change theorems and they know theorems about the existence of eigenvectors and eigenvalues ($K$ is algebraically closed so there has to exist an eigenvector). Is there a way to prove this theorem just with the mentioned work tools?

Thanks for your help!

Rodrigo de Azevedo
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SometimesBlind
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  • Do they know about quotient spaces? You can prove this by induction on $\dim V$ using just the fact that $f$ has an eigenvector and quotient spaces. – Eric Wofsey Jan 14 '16 at 23:10
  • No, they haven't defined quotient spaces yet, even though his professor said that there should be some kind of induction over dim V. – SometimesBlind Jan 14 '16 at 23:11
  • This question was on my quals, with the explicit instructions to not use Jordan Form (which would give it immediately). I remember you have to construct the basis one vector at a time. Once you figure out how to get the first vector, the rest follows easily. Edit: I think Eric's comment does it. – Future Jan 14 '16 at 23:16
  • Looks like he has to do it the old-fashioned annoying way, doesn't he? – SometimesBlind Jan 15 '16 at 00:13

1 Answers1

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Let $A$ be the matrix of $f$ with respect to a basis. You need to find an invertible matrix $S$ and an upper triangular matrix $T$ such that $A=STS^{-1}$.

Since the base field is algebraically closed, we can find an eigenvalue $\lambda$ and an eigenvector $v$. Complete $v$ to a basis of $V$, say $\{v=v_1,v_2,\dots,v_n\}$, and let $S_0=[v_1\ v_2\ \dots\ v_n]$. Then $$ S_0^{-1}AS_0= \begin{bmatrix} \lambda & \mathbf{x}^T \\ \mathbf{0} & A_1 \end{bmatrix} $$ for some vector $\mathbf{x}\in K^{n-1}$ and some $(n-1)\times(n-1)$ matrix $A_1$. By induction hypothesis, there is an invertible $(n-1)\times(n-1)$ matrix $S_1$ such that $$ T_1=S_1^{-1}A_1S_1 $$ is upper triangular. Consider $$ \hat{S}_1= \begin{bmatrix} 1 & \mathbf{0}^T \\ \mathbf{0} & S_1 \end{bmatrix} $$ Then $$ \hat{S}_1^{-1}= \begin{bmatrix} 1 & \mathbf{0}^T \\ \mathbf{0} & S_1^{-1} \end{bmatrix} $$ and \begin{align} \hat{S}_1^{-1}S_0^{-1}AS_0\hat{S}_1&= \begin{bmatrix} 1 & \mathbf{0}^T \\ \mathbf{0} & S_1^{-1} \end{bmatrix} \begin{bmatrix} \lambda & \mathbf{x}^T \\ \mathbf{0} & A_1 \end{bmatrix} \begin{bmatrix} 1 & \mathbf{0}^T \\ \mathbf{0} & S_1 \end{bmatrix}\\ &= \begin{bmatrix} 1 & \mathbf{0}^T \\ \mathbf{0} & S_1^{-1} \end{bmatrix} \begin{bmatrix} \lambda & \mathbf{x}^TS_1\\ \mathbf{0} & A_1S_1 \end{bmatrix} \\ &= \begin{bmatrix} \lambda & \mathbf{x}^TS_1\\ \mathbf{0} & S_1^{-1}A_1S_1 \end{bmatrix} \\ &= \begin{bmatrix} \lambda & \mathbf{x}^TS_1\\ \mathbf{0} & T_1 \end{bmatrix} \end{align} is upper triangular.

egreg
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