Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $T$ be an arbitrary operator on a finite dimensional inner product space $(V,\langle\,,\,\rangle)$. Set $R=(1/2)(T^*+T)$, $S=(1/2)i(-T+T^*)$. Prove that if $T=R_1+iS_1$, where $R_1$, $S_1$ are self-adjoint, then $R_1=R$, $S_1=S$

($T^*$ is the adjoint of $T$)

share|cite|improve this question
$T^*=R_1^*-iS_1^*=R_1-iS_1$. Now solve for $R_1$ and $S_1$. – wj32 Nov 30 '12 at 10:42

Taking the dual $d:T\mapsto T^*$ is an $\mathbf R$-linear involution on the space $\def\End{\operatorname{End}}\End_\mathbf C V$ of all complex-linear operators on $V$ (considered here as a real vector space), that is $d$ is linear with respect to real scalars and it satisfies $d\circ d=\operatorname{id}_{\End V}$. Now linear involutions are always* diagonalisable with no other eigenvalues than $1$ and $-1$ (because the polynomial $X^2-1$ that annihilates involutions has those values as simple roots). The eigenvalues for $\lambda=1$ are the self-adjoint operators, the eigenvalues for $\lambda=-1$ are the anti-self-adjoint operators. Moreover multiplication by $i$ interchanges the (real) spaces of self-adjoint and anti-self-adjoint operators.

Now since $d$ is diagonalisable, $\End_\mathbf C V$ is the sum of these two eigenspaces (actually, the easily verified fact that for all $T$ one has $T=R+iS$ with $R,S$ as in the question already shows this), and a sum of eigenspaces is always a direct sum. This means that $R$ and $S$ in this decomposition, under the requirement of being self-adjoint, are always unique Q.E.D.

Added. I just wanted to give the general background, but since you asked about an explicit proof, it just boils down to showing that the sum of the subspaces of self-adjoint and anti-self-adjoint operators is direct. This means showing that if $0=R_0+iS_0$ with $R_0,S_0$ self-adjoint, then $R_0=S_0=0$. But with $-R_0=iS_0$ the first member is self-adjoint and the second is anti-self-adjoint, and they are equal; that can only happen for the null operator.

*except in characteristic $2$

share|cite|improve this answer
I think there are easier and more explicit proofs~ – i_a_n Nov 30 '12 at 9:11
At your service. – Marc van Leeuwen Nov 30 '12 at 9:37

Assuming $T=R_1+iS_1$, with $R_1$ and $S_1$ self-adjoint, for $u,v\in V$ we have $$ \langle S_1(u),v\rangle=\langle -iT(u)+iR_1(u),v\rangle =-i\langle T(u),v\rangle +i\langle R_1(u),v\rangle. $$

Also $$ \langle S_1(u),v\rangle=\langle u,S_1(v)\rangle=\langle u,-iT(v)+iR_1(v)\rangle=i\langle u,T(v)\rangle-i\langle u,R_1(v)\rangle=i\langle T^*(u),v\rangle-i\langle R_1(u),v\rangle. $$ Thus $$ 0=-i\langle T(u),v\rangle +i\langle R_1(u),v\rangle-i\langle T^*(u),v\rangle+i\langle R_1(u),v\rangle=\langle -iT(u)-iT^*(u)+2iR_1(u),v\rangle. $$ In particular, $$ 0=\langle i(2R_1-(T+T^*))(u),i(2R_1-(T+T^*))(u)\rangle, $$ so $$i(2R_1-(T+T^*))(u)=0.$$ Therefore $$ R_1=\frac{1}{2}(T+T^*). $$

Proving $S_1$ is similar.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.