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I have having trouble with the following question:

Let $V$ be a finite-dimensional complex inner product space, and let $T$ be a linear operator on $V$. Prove that if $\langle T\alpha, \alpha\rangle$ is real for every $\alpha$ in $V$ then $T$ is self-adjoint.

I have been focusing on trying to show that if $\langle T\alpha, \alpha\rangle = \langle T^*\alpha, \alpha\rangle$ for all $\alpha$ then $T = T^*$, but have been unable to do so. I also feel like I'm missing something really obvious. How should I proceed.

providence
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3 Answers3

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Here are two hints. You want to use the following two things:

  1. $\langle T \alpha, \alpha \rangle = \langle \alpha, T^{\ast} \alpha \rangle$ (by the definition of the adjoint).
  2. $\langle T \alpha, \alpha \rangle = \overline{ \langle \alpha, T \alpha \rangle}$.

The tricky part is then proving the following claim:

Let $A$ be an operator on a complex inner product space. Then $A = 0$ if and only if $\langle Ax, x \rangle = 0$ for all $x$.

To prove this, you need to prove a variant of the polarization identity, which should be

$$ \langle Ax, y \rangle = \frac{1}{4} \left( \langle A(x + y), x + y \rangle - \langle A(x - y), x -y \rangle + i \langle A(x + iy), x + iy \rangle - i \langle A(x - iy), x - iy \rangle \right) $$

Apply this claim to $A = T - T^{\ast}$.

Edit: Alternatively, you can use the fact that $T - T^{\ast}$ is skew-skymmetric; that is, its adjoint is equal to its negative. Then it is a normal operator, and see if you can apply the spectral theorem.

Eric Wofsey
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Christopher A. Wong
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  • @ChristopherAWong I believe I've made it this far. The issue is proving that $\langle a, Ta\rangle = \langle a, T^a\rangle$ (or equivalently in this case that $\langle Ta, a\rangle = \langle T^a, a\rangle$) for all $a$ suffices to show that $T = T^*$. – providence Mar 02 '13 at 06:10
  • Well, $\langle T \alpha, \alpha \rangle $ is real, so it equals $\langle \alpha, T \alpha \rangle$. Then $\langle \alpha, T \alpha \rangle$ does in fact equal $\langle \alpha, T^{\ast} \alpha \rangle$. – Christopher A. Wong Mar 02 '13 at 06:12
  • Yes, I understand this much. However, how does that prove that the linear transformation $T$ is equal to the linear transformation $T^$. For all I know, it is possible that $\langle a, Ta\rangle = \langle a, T^a\rangle$ and $Ta \ne T^a$ for some $a$. For, to prove that $Ta = T^a$ I believe that I must show $\langle a, Tb\rangle = \langle a, T^*b\rangle$ for all $a, b \in V$. – providence Mar 02 '13 at 06:15
  • @providence $\langle a,Ta\rangle=\langle a, T^a\rangle,; \forall a \in \mathbb{C}$, necessarily means that $ Ta=T^a$. Proof: $\langle a,Ta\rangle=\langle a, T^a\rangle \to \langle T^a,a\rangle=\langle Ta,a\rangle \to \langle T^a,a\rangle-\langle Ta,a\rangle=0 \to \langle (T^-T)a,a\rangle=0,; \forall a \in \mathbb{C}$. This is the same as saying $T^-T=0 \to T^=T$. Then $T^*a=Ta$. – Clerni Dec 26 '20 at 14:03
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First note that $$ \langle T(a), a \rangle = \overline{\langle a, T(a) \rangle} = \langle a, T(a) \rangle = \langle T^*(a), a \rangle $$ for any $a \in V$. We get the second equality since the conjugate of a real number is itself and the other equalities are obtained from definitions. We can use the derived equality, $\langle T(a), a \rangle = \langle T^*(a), a \rangle$, for $a = x + y$ and $a = x + iy$ for any $x, y \in V$. Using properties of linear transformations and inner products, we get $$ \langle T(x + y), x + y \rangle = \langle T^*(x + y), x + y \rangle \\ \langle T(x) + T(y), x + y \rangle = \langle T^*(x) + T^*(y), x + y \rangle \\ \langle T(x), x \rangle + \langle T(x), y \rangle + \langle T(y), x \rangle + \langle T(y), y \rangle = \langle T^*(x), x \rangle + \langle T^*(x), y \rangle + \langle T^*(y), x \rangle + \langle T^*(y), y \rangle. $$ But $\langle T(x), x \rangle = \langle T^*(x), x \rangle$ and $\langle T(y), y \rangle = \langle T^*(y), y \rangle$, so those terms cancel to give $$ \langle T(x), y \rangle + \langle T(y), x \rangle = \langle T^*(x), y \rangle + \langle T^*(y), x \rangle.\tag{1} $$ Similarly, for $a = x + iy$, you just need the extra step of taking out the factor $i$ and you will get $$ -i\langle T(x), y \rangle + i\langle T(y), x \rangle = -i\langle T^*(x), y \rangle + i\langle T^*(y), x \rangle. $$ Dividing out by $-i$ gives $$ \langle T(x), y \rangle - \langle T(y), x \rangle = \langle T^*(x), y \rangle - \langle T^*(y), x \rangle. \tag{2} $$ Add equations $(1)$ and $(2)$ and divide by $2$ to get $$ \langle T(x), y \rangle = \langle T^*(x), y \rangle \\ \langle (T - T^*)(x), y \rangle = 0 $$ for any $x, y \in V$. So we can take $y = (T - T^*)(x)$ to get $\langle (T - T^*)(x), (T - T^*)(x) \rangle = 0$ which implies $(T - T^*)(x) = 0$ for any $x \in V$. Thus, $T - T^* = O$, ie. $T = T^*$ which means $T$ is self adjoint.

I don't think this is true for real inner product space since we can't use $x + iy$ so we can't prove this.

Pratyush Sarkar
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It is a proposition in Conway's book (a course in functional analysis, page 33 prop 2.12). The full proof is given there. Also it is false for real case. This example is given on the same page of Conway's book (Let A be a 2x2 matrix with first row (0,1) and second row (-1,0), so (Ah,h)=0 for all h in the Cartesian plane but A and A* are different). The counter example is actually the counter-clockwise 90 degrees rotation about the origin.

wiki
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