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Show that φ(v)=λv for a vector v and a coefficient λ

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,003
Hey!! 😊

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.

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.

Let $\phi$ satisfy the above properties. Then show that there is $0\neq v\in V$ and a $\lambda\in \mathbb{K}$ such that $\phi (v)=\lambda v$.

For that I have done the following:

We consider the subspace $U_1$. Since $U_1$ is $\phi$-invariant, it follows for $v\in U_1\subset V$ that $\phi (v)=\lambda v$, with $\lambda\in \mathbb{K}$.

Is that correct and complete? :unsure:
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
832
Hey!! 😊

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.

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.

Let $\phi$ satisfy the above properties. Then show that there is $0\neq v\in V$ and a $\lambda\in \mathbb{K}$ such that $\phi (v)=\lambda v$.

For that I have done the following:

We consider the subspace $U_1$. Since $U_1$ is $\phi$-invariant, it follows for $v\in U_1\subset V$ that $\phi (v)=\lambda v$, with $\lambda\in \mathbb{K}$.

Is that correct and complete? :unsure:
The two statements that you have mentioned are equivalent provided you assume that $U_1$ is not the trivial subsspace and the containments $U_i\subseteq U_{i+1}$ are strict.

With this, you, in your argument, need to mention $U_1$ is necessarily one dimensional (do you see why) and that $v$ can be chosen to be nonzero.