Let $\mathbb{K}$ be a field, $1\leq n\in \mathbb{N}$ and let $V$ be a $\mathbb{K}$-vector space with $\dim_{\mathbb{R}}V=n$.

Let $\phi :V\rightarrow V$ be a linear map.

I want to show that the following two statements are equivalent:

  • There is a basis $B$ of $V$ such that $M_B(\phi)$ is an upper triangular matrix.

  • There are subspaces $U_1, \ldots , U_n\leq_{\mathbb{K}}V$ such that $U_i\subset U_{i+1}$ and $U_i$ is $\phi$-invariant.

Could you give me a hint for that?


1 Answer 1


Hint: If $\mathcal B = \{v_1,\dots,v_n\}$ is a basis of $V$ and $M_{\mathcal B}(\phi)$ is upper-triangular, then the subspaces $U_i = \operatorname{span}\{v_1,\dots,v_i\}$ are invariant subspace of $\phi$.

  • $\begingroup$ How does this follow? I got stuck right now. $\endgroup$
    – Mary Star
    Jun 21, 2020 at 16:52
  • $\begingroup$ @MaryStar Another hint: show that if $M_{\mathcal B}$ is upper triangular, then we can write $$ \phi(v_i) = c_1 v_1 + \cdots + c_i v_i $$ for some coefficients $c_1,\dots,c_i$. Can you see why this implies that $U_i$ is $\phi$-invariant? Try to show this inductively: why is $U_1$ invariant? If $U_1$ is invariant, then how do we see that $U_2$ is invariant? And so forth. $\endgroup$ Jun 21, 2020 at 16:56
  • $\begingroup$ Ahh we have that $v_i\in U_i$ and that $\phi (v_i)\in U_i$ and this means that $U_i$ is $\phi$-invariant, right? $\endgroup$
    – Mary Star
    Jun 21, 2020 at 17:00
  • $\begingroup$ @MaryStar well it's clear that $\phi(v_i) \in U_i$, but it's important that we use some of the context to show that $U_i$ is invariant. In particular, we can show that $U_i$ is invariant by noting that $\phi(v_j) \in U_i$ for all $j \leq i$, since the vectors $v_1,\dots,v_i$ form a basis of $U_i$. $\endgroup$ Jun 21, 2020 at 17:01
  • $\begingroup$ Ah so to show that $U_i$ is $\phi$-invariant do we have to show that $\phi (v_j)\in U_i$ for all $j\leq i$ ? $\endgroup$
    – Mary Star
    Jun 21, 2020 at 18:11

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