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

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The discussion centers on proving that for a linear map φ on a vector space V, if there exists a basis such that the matrix representation is upper triangular, then there exists a non-zero vector v in V and a scalar λ in the field K such that φ(v) = λv. The equivalence of having an upper triangular matrix and the existence of a chain of φ-invariant subspaces is established. It is noted that the subspace U1 must be non-trivial and one-dimensional, allowing the selection of a non-zero vector v. The argument emphasizes the need for clarity in these assumptions to complete the proof correctly.
mathmari
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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:
 
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mathmari said:
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.
 
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