a) I finished a problem that sort of highlighted the fact that if a real symmetric matrix $A_2$ = A + I, where A is also real and symmetric, then $A_2$ has the same eigenvectors as A, but its spectrum is shifted by 1. So, if the spectrum of A were {1,2,3}, the spectrum of $A_2$ would be {2,3,4}.
Are the above statements also true for non-symmetric matrices (real or complex)?
b) Also, a quick question about one of the statements of the Spectral Theorem: it says the vector space V is the direct sum of the eigenspaces (for a normal operator T in complex space or a self-adjoint operator T in real space). Does this mean every vector in V is an eigenvector of T?
Thanks in advance,
Question a: The answer to this one is yes. You can see this by applying the definition of an eigenvalue. In particular, if $Ax = \lambda x$ for some vector $x$, then $(A + I)x = (1 + \lambda)x$. Or equivalently, if $\det(A - \lambda I) = 0$, then $\det([A + I]- (\lambda + 1)I)$ must also be zero.
Question b: That's not quite what it means. What it does mean is that every $v$ can be expressed in the form $v = v_1 + \cdots + v_k$, where each $v_i$ is an eigenvector of $T$. Moreover, the spectral theorem allows us to select $v_i$ so that $\langle v_i,v_j \rangle = 0$ when $i \neq j$.