eigenvector equivalence for subspace of matrix?

There is an eigenvalue $\lambda_1$ in matrix A, which is $3x3$, with corresponding eigenvector $x_a = 1/\sqrt{2}(1,0,1)$. The same eigenvalue $\lambda_1$ can be found in the subspace of matrix A, which is $2x2$, with corresponding eigenvector $x_b = 1/\sqrt{2}(1,1)$.

Question: Could a zero be added to eigenvector $x_b$ such that it equals $x_a$?

Matrix A = \begin{pmatrix} 1 & 0 & \lambda \\ 0 & 0 & 0 \\ \lambda & 0 & 1 \\ \end{pmatrix} and the subspace of Matrix A = \begin{pmatrix} 1 & \lambda \\ \lambda & 1 \ \\ \end{pmatrix}

Depends on how you define equal, obvious $x_a \ne x_b$ in the exact sense since they DEFINITELY aren't the same thing (one has 3 dimensions the other has 2)!
That doesn't mean they are not exhibiting some similar behavior. If you restate "equal" as "equal under a particular model" then the question becomes interesting again. It might be the case that there is a function $g$ out there such that $g(x_b) = g(x_a)$ where $g$ encodes some information about the matrix you are working with and the matching eigenvalues.