Let $A$ be a $3\times3$ matrix and $u, v, w$ be linearly independent vectors in $\mathbb{R}^3$ such that:
$Au = 2u, Av = 2v, Aw = 0$.
Which of the statements are NOT necessarily true?
Option 1: $w$ is an eigenvector of $A$.
Option 2: $\text{span}({u,v})$ is the eigenspace of $A$ associated with eigenvalue 2.
Option 3: $u+w$ is an eigenvector of $A$ associated with eigenvalue 2.
Option 4: $2u$ is an eigenvector of $A$ associated with eigenvalue 2.
Option 5: All the statements are necessarily true.
I am not sure about option 2 and 3, but I think option 1 and 4 are true. Please help. Thank you.
Option 2 is certainly the most difficult. What do you know about the total sum of the geometric multiplicities of eigenvalues?
We know that $\text{span}(u, v)$ is at least a subset of the eigenspace of 2, so the geometric multiplicity of 2 is at least 2.
Assume that $\text{span}(u, v)$ is not all of the eigenspace of 2, then the geometric multiplicity of 2 is at least 3. But we know that $w$ is an eigenvector for 0, showing that the geometric multiplicity is at least 1. Is it possible for the geometric multiplicity of 2 to be at least 3 and that of 0 to be at least 1?
Option 3 is asking you whether $u+w$ is a nonzero vector that satisfies $A(u+w)=2(u+w)$. Is it?