1.
If a matrix A is symmetric then the eigenvectors from different eigenspaces are orthogonal.
Is the reverse true?
If yes, can you provide proof.
If no, can you provide an example of a non-symmetric matrix whose eigenvectors from different
eigenspaces are arthogonal?
Solution
2 down vote A definition of "symmetric" which should prove useful here is: if < , > is a
canonically chosen inner product, we say T is symmetric if and only if for all vectors $u,v in
V$ we have < u, Tv> = < Tu, v>. Assume that the dimension of V is equal to the sum of the
dimensions of the eigenspaces $E_1,...,E_k$ associated to the eigenvalues
$lambda_1,...,lambda_k$. Then we can write any vectors $u,v in V$ as $sum_{i=1}^k u_i$
and $sum_{i=1}^k v_i$ respectively. < u, Tv> can then be rewritten <$sum u_i$, T $sum
v_i$> = < $sum u_i$, $sum T v_i$> = < $sum u_i$, $sum lambda_i v_i$>. Now we use
orthogonality. If the eigenspaces are orthogonal to eachother, then this expression becomes
$sum$ < $u_i$, $lambda_i v_i$> = $sum lambda_i$< $u_i$, $v_i$> = $sum$ <
$lambda_i u_i$, $v_i$> = < $sum lambda_i u_i$, $sum v_i$> = < Tu, v>. The reverse
implication isn't hard, simply consider the equality < u, Tv> = < Tu, v> when u, v are two
eigenvectors corresponding to two different eigenvalues.