Given eigenvalues of $\vec l^2$ and $\vec s^2$, calculate the eigenvalue for $\vec j^2$ There was an exam question that read approximatly:

Let $\vec j = \vec l + \vec s$. Given eigenvalues of $\vec l^2$ and $\vec s^2$, calculate the eigenvalue for $\vec j^2$.

We came up with
$$\vec j^2 = (\vec l + \vec s)^2  = \vec l^2 + \vec s^2 + 2 \vec l \cdot\vec s$$
How can we get the eigenvalue for $\vec j^2$ only with this information? I think that we would need $m_l$ and $m_s$ or so.
 A: There seems to be no way to proceed unless more information is given.
In fact, from what you have above (correcting the typo pointed out by Bernhard and ticster):
$$
\vec{j}=\vec{l}+\vec{s} \quad \Rightarrow \quad \vec{j}^2=\big(\vec{l}+\vec{s}\big)^2 = \vec{l}^2+\vec{s}^2 + 2\, \vec{l}\cdot\vec{s}\,,
$$
meaning that knowledge of the eigenvalues of $\vec{l}^2$ and $\vec{s}^2$ is not sufficient to obtain an answer. This can be understood by recalling that addition of momenta can result in different values for $j$, ranging from $(l+s)$ to $|l-s|$ in integer steps; i.e. the possibilities are (See, for instance, David J. Griffiths' Introduction to Quantum Mechanics, 2nd edition, section 4.4):
$$
j = (l+s), (l+s-1), (l+s-2), \dots,|l-s|,
$$
where $j$, $l$, and $s$ correspond to the variable appearing in the eigenvalues of $\vec{j}^2$, $\vec{l}^2$ and $\vec{s}^2$, respectively - e.g.:
$$
\vec{j}^2 \,\vert j \rangle = j(j+1) \hbar^2\,|j \rangle.
$$
Intuitively, one might appeal to the relative orientation of the angular momentum vectors in space, although this makes no sense in quantum mechanics, as the components of the vector do not commute and thus cannot jointly characterize the system.
In the case you did have at your disposal the values of $m_l$ and $m_s$, i.e. a state $\vert l\, m_l\, s\, m_s\rangle$, then, apart from immediately knowing $m_j$ to be $m_j = m_l + m_s$, you would be able to figure out the value of $j$ through the equation you worked out, by expanding the inner product $\vec{l}\cdot\vec{s}$ (for simple examples, check out the aforementioned section of David J. Griffiths' Intro to QM, or Stephen Gasiorowicz's Quantum Physics, 3rd edition, section 10-4).
Edit: rob's answer nicely points out the special cases where $j = l+s = |l-s|$, which occur when $l$ or $s$ or both are zero.
A: In the special case where $\ell^2$ or $s^2$ has eigenvalue zero, then $j^2$ is fixed. Otherwise you must know the projections $m_\ell,m_s$ to find $j$.
