Consider the usual dot product
$$
(x_{1}, x_{2}, x_{3}, x_{4})
\cdot
(y_{1}, y_{2}, y_{3}, y_{4})
=
x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3} + x_{4} y_{4}.
$$
Then it is easy to see that $A$ is symmetric iff for all $x, y$
$$
x A \cdot y = x \cdot y A.
$$
(just take the $x = e_{i}, y = e_{j}$, where $e_{i}$ is the vector which is all zero except for a $1$ in the $i$-th position).
Now prove that if $x$ is an eigenvector with respect to $\lambda$, $y$ is an eigenvector with respect to $\mu$, and $\lambda \ne \mu$, then $x, y$ are orthogonal:
$$
\lambda (x \cdot y)
=
(\lambda x) \cdot y
=
x A \cdot y
=
x \cdot y A
=
x \cdot (\mu y)
=
\mu (x \cdot y)
$$
and $\lambda \ne \mu$ implies $x \cdot y = 0$.
Therefore in your case the missing eigenvector $v$ relative to the eigenvalue $2$ must be orthogonal to both $(1, 0, 0, −1)$ and $(0, 1, 1, 0)$. It is easy to check that this means
$$
v = (a, b, -b, a)
=
a (1, 0, 0, 1) + b (0, 1, -1, 0),
$$
for some $a, b$.
Assuming $A$ is $n\times n$, if you are given $n$ linear independent eigenvectors and the corresponding eigenvalues, then the $QDQ^{-1}$ formulation works.
If the matrix does not have this spanning set of eigenvectors, then you need the generalized eigenvectors, c.f. https://en.wikipedia.org/wiki/Generalized_eigenvector, to construct a matrix $P$ and the almost diagonal Jordan normal form $J$ such that $A=PJP^{-1}$.
Jordan normal forms, c.f. https://en.wikipedia.org/wiki/Jordan_normal_form, exist for all $n\times n$ matrices over the field $\mathbb{C}$, so the answer is "yes" if you are also given generalized eigenvectors and "no" if you are only given the eigenvectors of a non-diagonalizable matrix.
Best Answer
The eigenvectors $X^TX$ can be computed as follows:
(1) Assume $v$ is an eigenvector $XX^T$ to eigenvalue $\lambda\ne0$. Then it holds $XX^Tv=\lambda v$, and $$ X^TX(X^Tv) = \lambda (X^Tv). $$ Since $\lambda \ne0$, it follows $X^Tv\ne0$, hence $X^Tv$ is an eigenvector of $X^TX$ with eigenvalue $\lambda$.
(2) The eigenvectors to eigenvalue zero are the elements in the null space of $X$: $X^TX v=0$ implies $v^TX^TXv=\|Xv\|^2=0$, which is $Xv=0$.
That is, you obtain the eigenvectors of $X^TX$ from