# $M_B(\phi)$ is an upper triangular matrix iff $U_i\subset U_{i+1}$ and $U_i$ is $\phi$-invariant

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?

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 $$\phi$$.
• @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. Jun 21, 2020 at 16:56
• 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? Jun 21, 2020 at 17:00
• @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 $U_i$. Jun 21, 2020 at 17:01
• Ah so to show that $U_i$ is $\phi$-invariant do we have to show that $\phi (v_j)\in U_i$ for all $j\leq i$ ? Jun 21, 2020 at 18:11