Operators over real vector spaces are self adjoint right?

**durrrrrrrr** · July 17th 2011, 05:53 AM

From Linear Algebra Done Right:
"Positive Operators
An operator T ∈ L(V ) is called positive if T is self-adjoint and ⟨T v , v ⟩ ≥ 0
for all v ∈ V. Note that if V is a complex vector space, then the condition that T be self-adjoint can be dropped from this definition (by 7.3)."

If V is a real vector space then <Tv,v>=<v,Tv> so T is automatically self-adjoint. Doesn't this mean that the condition that T is self-adjoint can be dropped whatever the situation?

**durrrrrrrr** · July 17th 2011, 10:26 AM

No responses? I thought this was a really simple question

**Jose27** · July 17th 2011, 11:18 AM

Consider $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ with the usual inner product with $T(x_1,x_2)=(x_2,-x_1)$ then $\langle x,Tx \rangle =0$ for all $x\in \mathbb{R} ^2$ but $T$ is not self-adjoint because $\langle y,Tx \rangle = y_1x_2-x_1y_2 \neq y_2x_1-y_1x_2 = \langle Ty,x \rangle$ (the equality holds only when $y=\lambda x$ for some $\lambda$ )

**FernandoRevilla** · July 17th 2011, 12:30 PM

In general, if $T\in\textrm{End}(E)$ with $(E,<,>)$ finite dimensional euclidean space, $B=\{u_1,\ldots,u_n\}$ is basis of $E$ , $G$ is the Gram matrix with respect to $B$ and $A$ is the matrix of $T$ with respect to $B$ , is easy to prove that $T$ is self adjoint iff $GA=A^tG$ . Choosing for example $B$ orthonormal we have $G=I$ so, any non symmetric matrix represents a non self adjoint operator.