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Hi: I'm reading another document on Hilbert spaces and progressing some but I've come to a theorem in this document, many of the steps of which I don't follow.

I will state the theorem below and then explain which steps are not clear to me.

Theorem 5.1. Let $H$ be a hilbert space and $\phi: H \rightarrow \mathbb{C}$, be a continuous functional. Then,

$\dim N(\phi)^\perp = 1$.

where $N(\phi) = \{h \in H| \phi(h) = 0\}$.

Proof: It is easy to see that $N(\phi)$ is a closed subspace of $H$.

Since, $\phi \neq 0, N(\phi) \neq H$.

Then, $N(\phi) \neq \{0\}$. ( recall that $H = N(\phi) \oplus N(\phi)^\perp$).

Take $x_1 \neq 0, x_2 \neq 0$ in $N(\phi)^{\perp}$.

Then, there exists a complex number $a \neq 0$ such that $\phi(x_1) + a \phi(x_{2}) = 0$.

Then, $\phi(x_1 + a x_{2} ) = 0$ which implies that $x_1 + a x_{2} \in N(\phi) \oplus N(\phi).^{\perp} = \{0\}$.

Thus, $\dim N(\phi)^\perp = 1$.

END OF THEOREM.

There are many steps in above that I don't follow.

1) Then, $N(\phi) \neq \{0\}$. I don't see where that comes from and does that indicate the null set or zero ?

2) $\phi(x_1) + a \phi(x_{2}) = 0$. Is this because the linear combo of 2 objects in a hilbert space is also in that space ?

3) $\phi(x_1 + a x_{2} ) = 0$. This is clearly crucial to the proof. I don't get it.

4) implies that $x_1 + a x_{2} \in N(\phi) \oplus N(\phi).^{\perp} = \{0\}$. Why does that imply it and again, is that the null set or zero ?

5) $\dim N(\phi)^\perp = 1$. Don't follow this either ?

Thanks a lot.

hal4math
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mark leeds
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2 Answers2

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(1) It seems from context that there is a typo and they meant to write $N(\phi)^\perp \ne \{0\}$. That is, $N(\phi)^\perp$ is not equal to the set containing only the zero vector. Since every linear subspace contains the zero vector, this means $N(\phi)^\perp$ must contain some elements other than the zero vector, which justifies the following line.

(2) Note that $\phi(x_1)$ and $\phi(x_2)$ are just (nonzero) complex numbers. Using simple algebra, you can very easily find what $a$ should be in order to make $\phi(x_1) + a \phi(x_2) = 0$.

(3) Recall that $\phi$ is a linear functional.

(4) I think they meant to write $x_1 + a x_2 \in N(\phi) \cap N(\phi)^\perp$. We have just shown that $\phi(x_1 + a x_2) = 0$, which is the very definition of $x_1 + a x_2 \in N(\phi)$. And we started out assuming that $x_1, x_2 \in N(\phi)^\perp$, and $N(\phi)^\perp$ is a linear subspace... Finally, recall that a set and its orthogonal complement always have only the zero vector in common. (Any vector common to both must be orthogonal to itself.)

(5) We have just shown that $x_1 + a x_2 = 0$ for arbitrary $x_1, x_2$, which is to say that any two vectors in $N(\phi)^\perp$ are linearly dependent. If the dimension of $N(\phi)^\perp$ was at least 2, we would be able to find two linearly independent vectors, and we've just shown that can't be done.

Nate Eldredge
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  • I went through your answers carefully and they were all great. But I think that I'm missing the forest from the trees (or whatever that saying is ) for the following reason. In part 5) of above, since x1 and x2 are complex, then we can find a linear combination that is equal to zero. Then if that's the case, we then know that the dimension is 1 because any two vectors in $N(\phi)^{\perp}$ are linearly dependent. But, if that's the argument, then what was the purpose of all the other theorem steps ? Thanks a lot. – mark leeds Sep 11 '19 at 03:03
  • @markleeds: Keep in mind that $x_1, x_2$ are vectors, not numbers. In general, given any two vectors, you may or may not be able to find a linear combination yielding zero. The whole proof up to this point was to show that in this particular case, we actually can do so. – Nate Eldredge Sep 11 '19 at 04:44
  • Thanks. I'll go over the whole thing again with that in mind. It's much appreciated. – mark leeds Sep 11 '19 at 05:52
  • I'm sorry to be such a pest but I have one more bother. You explained that $x_1$ and $x_2$ are vectors and that $\phi(x_1)$ and $\phi(x_2)$ are complex scalars. The latter is because $\phi$ is a function from $H$ to $C$. But how does one know that $x_1$ ans $x_2$ are vectors. Is that assumed ? Because sometimes I'll see cases where the hilbert space is the vector space of summable sequences over the complex numbers. Also, why can't one just assume that some linear combination can be found to make the two vectors add to a zero vector. Thanks. – mark leeds Sep 13 '19 at 02:44
  • "Vector" here just means "an element of the Hilbert space". If you're thinking of the space of square-summable sequences, then a "vector" is a sequence. Here I'm saying vector to emphasize that these are not just complex numbers (scalars). – Nate Eldredge Sep 13 '19 at 02:48
  • @markleeds: To the second question, you can even consider $H = \mathbb{C}^2$ and vectors like $x_1 = (1,0)$ and $x_2 = (0,1)$. You are not going to find any complex number $a$ such that $x_1 + a x_2 = (0,0)$. – Nate Eldredge Sep 13 '19 at 02:49
  • Thanks so much for your patience and help. I will review carefully. – mark leeds Sep 13 '19 at 14:18
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The other answer covers it, but here is another approach, too long for a comment:

$H=N(\phi)\oplus N(\phi)^\perp$ (because $H$ is a Hilbert space) and $H/N(\phi)\cong \mathbb C$ (By first isomorphism theorem), so the dimension of the quotient is $1$.

There is an $h_1\in H$ such that $h_1+N(\phi)$ generates $H/N(\phi)$. Now, $h_1=h+k$ for some $ h\in N(\phi),\ k\in N(\phi)^{\perp},$ so $h_1+N(\phi)=k+N(\phi).$

Therefore, if $x\in N(\phi)^{\perp},$ then $[x]=a(k+N(\phi))$ for some $a\in \mathbb C$, from which it follows that $x-ak\in N(\phi)$, but since each of these terms is contained in $N(\phi)^{\perp},$ we must have $x-ak=0$, and so $x=ak$. The result follows.

Matematleta
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  • Thanks much to both Nate and Matematleta. I will print out and read both answers carefully but I will check Nate's because I know that I will have a better chance of understanding it because there are more details. Thanks again. – mark leeds Sep 10 '19 at 05:48