True or false (and give a proof): There exists
$T ∈ L(\mathbb R^3)$ such that $T$ is not self-adjoint (with respect to the usual inner product) and such that there exists a basis of $\mathbb R^3$
consisting of eigenvectors of T.
Attempt: Although, I possess a counter-example. But, I am unable to find a flaw in this argument:
If there exists a basis of $\mathbb R^3$ consisting of eigenvectors of T, then, the matrix of $T$ denoted by $M(T)$ is a diagonal matrix. Since, $M(T^*), ~(T^* \text{is the adjoint operator)}$ is the conjugate transpose of $M(T)$, thus $M(T^*)=M(T)$ is a diagonal matrix, hence, $T$ is self-adjoint.
( The above argument seeking equality between $M(T)~ \text {and}~M(T^*)$ is valid because there exists an isomorphism between $T$ and $M(T)$.
What could be the error in this argument?
Thanks a lot for your help!
Best Answer
When you diagonalise a matrix, you change the basis:
$$D=PMP^{-1}$$
$D$ is diagonal, hence $D^T=D$ indeed, but from this you cannot conclude that $M^D=M$, unless $P$ is orthogonal (where $P^{-1}=P^T$).
An operator being self adjoint (over a Euclidean vector space) doesn't mean that its matrix is symmetric with respect to any basis, only with respect to an orthogonal basis.
Another way to see it is that if you make a non-orthogonal change of basis, you twist the inner product upon which the notion of adjoint operator relies.