Orthogonal eigenvectors of the matrix Whether there is a variable $\alpha$ such that the eigenvectors of the matrix $A$ are orthogonal?
$A=\begin{bmatrix} 3&1&1\\0&2&1\\0&\alpha&4\end{bmatrix}$
 A: EDIT. I assumed (without reason) that the OP was looking for a basis of $\mathbb R ^3$ of eigenvectors. I think that the answer can be interesting itself, so I leave it here. If it's off-topic, feel free to delete this (@moderators).
There's no need to calculate the eigenvectors, since it can be shown directly that such a basis does not exist.
In fact, let $T:V\to V$ be an endomorfism of a metric vector space, endowed with a scalar product $s(\cdot,\cdot) $, and suppose that $T$ admits an orthogonal basis of eigenvectors, $\{e_i\}$, with eigenvalues $\{\lambda _i\}$.
Let's calculate $s(T(v),w)$ for $v,w\in V$.
$$s(T(v),w)=s(\sum v_i T(e_i), \sum w_j e_j)=s(\sum \lambda _iv_i e_i, \sum w_j e_j)\\=\sum _{i,j}\lambda _iv_iw_j \delta_{ij}=^{\text {why?}}\sum _{i,j}\lambda _j v_i w_j \delta _{ij}=...=s(v,T(w)).$$
So the endomorfism is symmetric.
Now, if $V=\mathbb R ^n$ with the canonical scalar product $s(x,y)=y^T x$, and $T=A\cdot$, for $A\in \text M _{n\times n}(\mathbb R)$, then what we have proved is that if $A$ has $n$ orthogonal eigenvectors, then $s(Ax,y)=s(x,Ay)$. But $$y^TAx=s(Ax,y)=s(x,Ay)=(Ay)^T x=y^T A^T x,$$
and this implies $A=A^T$.
Moral: an endomorfism which has a basis of orthogonal eigenvectors is symmetric. In the case of $\mathbb R ^n$, where the endomorfism is given by a matrix, the matrix itself is symmetric.
The converse of what I've proved is also true (and much more useful): see the Spectral theorem.
A: Find the expressions for the eigenvectors in terms of alpha (I've replaced all alpha's with a's for less typing):
$$v_1 = (1,0,0),$$ 
$$v_2 = \left(\frac{1-a+\sqrt{1+a}}{a\sqrt{1+a}}, \frac{-1-\sqrt{1+a}}{a}, 1\right),$$
$$
v_3 = \left(\frac{-1+a+\sqrt{1+a}}{a\sqrt{1+a}},\frac{-1+\sqrt{1+a}}{a},1\right).
$$
Now, using the orthogonality conditions for the pairs of vectors, you should find three equations in terms of $a$ from the inner product expressions, and if any of them conflict, then there exists no such $a$. If there is one such $a$, then it is possible.
A: WolframAlpha gives the eigenvalues of the matrix as $\lambda_1 = 3, \lambda_2 = 3 + \sqrt{\alpha+1}, \lambda_3 = 3 - \sqrt{\alpha + 1}$.  Thus, for $\alpha = -1$, the matrix will have only a single eigenvalue $\lambda = 3$.  Thus, all the eigenvectors lie in the same eigenspace.  
Can you show that there is an orthogonal set of vectors which spans this eigenspace?
