Suppose $T$ is an operator on a complex inner product space $V$ such that $\langle Tv, v \rangle = 0$ for all $v \in V$. Then $T = 0$.

The result above is false for real inner product spaces - a counterexample being the rotation operator.

Are real inner product spaces not a subset of complex inner product spaces?

It seems contradictory to me that a result can only be true for a more general case.

  • $\begingroup$ Did you see the proof in the complex case (complex scalars...) and the counter example in the real case? What isn't clear, then? $\endgroup$ – DonAntonio Jun 11 '18 at 20:40
  • $\begingroup$ My question isn't about the specific theorem itself, I only gave it as an example of something that holds for the complex case but not for the real case. Since real numbers are a subset of complex numbers, I am confused that something can be true for the complex case but not the real case. $\endgroup$ – tmakino Jun 11 '18 at 20:43

Actually, it's essentially the other way around: intuitively, being a complex inner product space is a more restrictive condition than being a real inner product space.

Let's start by ignoring the inner product structure, since that already reveals the key point:

If I have a vector space $V$ over $\mathbb{C}$, then $V$ can also be viewed as a vector space over $\mathbb{R}$ - since $\mathbb{R}\subseteq\mathbb{C}$, I can just restrict the scalar multiplication in the $\mathbb{C}$-sense to real scalars. The converse, however, is false: there's no obvious way to take a vector space $W$ over $\mathbb{R}$ and "extend the scalars" to turn it into a $\mathbb{C}$-vector space.

For a concrete example, $\mathbb{R}$ is certainly an $\mathbb{R}$-vector space in an obvious way; is there a natural way to view $\mathbb{R}$ as a $\mathbb{C}$-vector space?

The point is that in a $k$-vector space, I have to know how to multiply vectors by every element of $k$. Making $k$ bigger (e.g. going from $\mathbb{R}$ to $\mathbb{C}$) means that I need to know more things.

Similarly, any complex inner product space "is" a real inner product space (or can be viewed as one by "restricting scalars"). However, the converse is false: there is no obvious way to turn a real inner product space into a complex inner product space.

Essentially the same phenomenon is on display here and here.

  • $\begingroup$ I understand that you are unable to scalar multiply $w \in W$ by $\lambda \in \mathbb{C}$, which means being a vector space in $\mathbb{C}$ is a more restrictive condition. However, I don't understand how you are able to view $V$ as a real vector space. Addition and scalar multiplication will work, but how do you represent the vector's entries $a + bi$ in $\mathbb{R}$? $\endgroup$ – tmakino Jun 11 '18 at 21:13
  • $\begingroup$ @tmakino You don't have to "represent the vector's entries $a+bi$ in $\mathbb{R}$" - all a vector space is is some set together with a notion of addition and of scalar multiplication. $\mathbb{C}$ is a $\mathbb{R}$-vector space in the obvious way (scalar multiplication given by $c(a+bi)=ca+cbi$). $\endgroup$ – Noah Schweber Jun 11 '18 at 21:17

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