The statement provided deals with eigenvectors, eigenvalues, and subspaces in the context of linear maps in vector spaces. Let's break it down step by step.

1. Context:

   • $V$ is a vector space, and $A: V \to V$ is a linear map.
   • $\lambda \in \mathbb{R}$, so $\lambda$ is a real number (presumably an eigenvalue of $A$).
   • $V_\lambda$ is defined as the subspace of $V$ generated by all eigenvectors of $A$ corresponding to the eigenvalue $\lambda$. Mathematically, we can define this subspace as: $V_\lambda = \{ v \in V : A(v) = \lambda v \text{ for some } v \neq 0 \}.$

2. Restatement of the claim:
   The statement claims that "every non-zero element of $V_\lambda$ is an eigenvector of $A$ having $\lambda$ as eigenvalue."

3. What is this subspace $V_\lambda$?
   The subspace $V_\lambda$ consists of all linear combinations of eigenvectors corresponding to $\lambda$. If $v_1, v_2, \dots, v_n$ are eigenvectors corresponding to the eigenvalue $\lambda$, then any element of $V_\lambda$ will be a linear combination of these vectors: $v = c_1 v_1 + c_2 v_2 + \cdots + c_n v_n,$ where $c_1, c_2, \dots, c_n \in \mathbb{R}$.

4. Proof or verification of the statement:
   We need to show that any non-zero element of $V_\lambda$ is an eigenvector corresponding to the eigenvalue $\lambda$.

   Let $v \in V_\lambda$ be a non-zero element. By definition, $v$ is a linear combination of eigenvectors corresponding to the eigenvalue $\lambda$. Thus, for some eigenvectors $v_1, v_2, \dots, v_n$ and scalars $c_1, c_2, \dots, c_n$, we have: $v = c_1 v_1 + c_2 v_2 + \cdots + c_n v_n.$ Since each $v_i$ is an eigenvector of $A$ corresponding to $\lambda$, we know: $A(v_i) = \lambda v_i \quad \text{for all } i.$ Now, applying $A$ to the linear combination $v$, we get:

   = c_1 A(v_1) + c_2 A(v_2) + \cdots + c_n A(v_n) = c_1 \lambda v_1 + c_2 \lambda v_2 + \cdots + c_n \lambda v_n = \lambda (c_1 v_1 + c_2 v_2 + \cdots + c_n v_n) = \lambda v.$$ Therefore, $$A(v) = \lambda v$$, which shows that $$v$$ is an eigenvector corresponding to the eigenvalue $$\lambda$$.

Thus, the statement is true: every non-zero element of $V_\lambda$ is an eigenvector of $A$ with $\lambda$ as the corresponding eigenvalue.

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Do you want further details or have any questions?

Here are some related questions to deepen your understanding:

1. How are generalized eigenvectors defined, and how do they relate to the subspace $V_\lambda$?
2. What happens if $A$ has complex eigenvalues? How does the subspace structure change in $V$?
3. Can you describe the geometric multiplicity and algebraic multiplicity of an eigenvalue? How do they relate to the dimension of $V_\lambda$?
4. What is the significance of the eigenspace $V_\lambda$ in the diagonalization of a matrix?
5. How does the Jordan canonical form involve the subspace $V_\lambda$?

Tip: In linear algebra, eigenvectors and eigenvalues are crucial in simplifying linear transformations, particularly through diagonalization and understanding invariant subspaces.