There is an exam problem that says as follows:
Let $A$ be a square matrix of order 3, with real entries and whose eigenspaces are $E_{\lambda_1} =$ {$(x, y, z)\in \mathbb{R}³$ : $x+y=0, z=0$}; $E_{\lambda_2} =$ {$(x, y, z)\in \mathbb{R}³$ : $x-y-2z=0$}. Determine: a) A base for $E_{\lambda_1}$. b) A base for $E_{\lambda_2}$. c) If the matrix is diagonalizable. d) If the matrix is orthogonally diagonalizable.
Now, I've already proved that the matrix is indeed diagonalizable through linear independence of the eigenvectors that I get finding the basis of both eigenspaces. I'm stuck on letter d) because I don't know the matrix $A$ therefore I cannot know if the matrix is orthogonally diagonalizable. Is there any way to solve letter d) with the information I already have (that is, not knowing about the symmetry of the matrix $A$)?
I could try to orthonormalize the vectors I already have, but it'll tell me nothing about if $A$ is or not orthogonally diagonizable, right?
We begin by assuming $A$ is orthogonally diagonalizable. This means that there are three mutually orthogonal linearly independent eigenvectors corresponding to $A$. Also, one of these eigenvectors has to be in the space $E_{\lambda_1} = <(1,-1,0)>$ (where we have used the pigeonhole principle). Let $(u,v,w)$ be an eigenvector of A such that it is orthogonal to $(1,-1,0)$ and lies in the space $E_{\lambda_2}$. Observing that any vector in the latter space can be expressed as $(u,v,w) = (2x,2y,x-y)$, we thus have:
$2x + (-2y) = 0$ $=> x = y$ $=> (u,v,w) = (u,u,0)$ $=> (u,v,w) \in E' = <(1,1,0)>$
thus implying $E'$ is the only possible subspace of $E_{\lambda_2}$ orthogonal to $E_{\lambda_1}$. However, it can be proved that the given eigenspaces are complements of each other (in $\mathbb{R}^3$, of course) and, hence, no other eigenspaces of $A$ can exist. (Try doing this yourself.) This leads us to a contradiction as we find two of our assumed eigenvectors to be linearly dependent, since both have to be vectors in $E'$. Clearly, our assumption is wrong and, thus, $A$ is not orthogonally diagonalizable.