# What is a $\langle , \rangle_{H}$? And how do you prove that it is an inner product?

We were given this problem:

For the inner product space $$V$$ and its subspace $$H$$, let $$Q = \{E(v)\,|\,v\in V\}$$ be the vector space.

Define addition in $$Q$$ by: for $$v, w\in V, E(v)\oplus E(w) = E(v + w)$$.

Define scalar multiplication in Q by: for $$v \in V$$ and $$\rho\in \mathbb{R}, \rho E(v) = E(\rho v)$$.

For $$v\in V$$, define $$E(v) = \{v+h\,|\,h\in H\}$$.

Define for $$u,v\in V$$, and $$u = u_1 + u_2, v = v_1 + v_2$$ where $$u_1, v_1 \in H$$ and $$u_2, v_2\in H^{\bot}$$, \begin{align*} \langle E(u), E(v)\rangle_H = \langle u_2, v_2\rangle \end{align*}

Prove or disprove: $$\langle \, ,\,\rangle_H$$ is an inner product on $$Q$$.

What is $$\langle \,,\,\rangle_H$$? Anyway, here are my initial thoughts in solving the problem:

To prove $$\langle , \rangle_H$$ is an inner product on $$Q$$, we have to prove the following:

1. For all $$E(a), E(b), E(y)\in V$$, $$\langle E(a), E(b) + E(y)\rangle_H = \langle E(a), E(b)\rangle + \langle E(a), E(y)\rangle_H$$. \begin{align*} \langle E(a), E(b) + E(y)\rangle_H = E(a)_H \cdot E(b + y)_H &= \begin{pmatrix} a_1\\ a_2\\ \vdots\\ a_n \end{pmatrix}\cdot \begin{pmatrix} b_1 + y_1\\ b_2 + y_2\\ \vdots\\ b_n + y_n \end{pmatrix}\\ &= \sum_{i=1}^{n} a_i(b_i+y_i)\\ &= \sum_{i=1}^{n} a_ib_i + \sum_{i=1}^{n} a_iy_i\\ &= \begin{pmatrix} a_1\\ a_2\\ \vdots\\ a_n \end{pmatrix} \cdot \begin{pmatrix} b_1\\ b_2\\ \vdots\\ b_n \end{pmatrix} + \begin{pmatrix} a_1\\ a_2\\ \vdots\\ a_n \end{pmatrix} \cdot \begin{pmatrix} y_1\\ y_2\\ \vdots\\ y_n \end{pmatrix}\\ &= E(a)_H\cdot E(b)_H + E(a)_H \cdot E(y)_H\\ &= \langle E(a), E(b)\rangle + \langle E(a), E(y)\rangle_H \end{align*}
2. For all vectors $$a, b\in V$$, $$\langle a, b\rangle = \langle b, a\rangle$$. \begin{align*} \langle a, b\rangle = E(a)_H\cdot E(b)_H &= \begin{pmatrix} a_1\\ a_2\\ \vdots\\ a_n \end{pmatrix}\cdot\begin{pmatrix} b_1\\ b_2\\ \vdots\\ b_n \end{pmatrix}\\ &= \sum_{i=1}^{n} a_ib_i\\ &= \sum_{i=1}^{n} b_ia_i\\ &= \begin{pmatrix} b_1\\ b_2\\ \vdots\\ b_n \end{pmatrix}\cdot \begin{pmatrix} a_1\\ a_2\\ \vdots\\ a_n \end{pmatrix}\\ &= E(b)_H\cdot E(a)_H\\ &= \langle b, a\rangle \end{align*}
3. For all vectors $$a, b$$ and for all real numbers $$r \in \mathbb{R}, \langle ra,b\rangle = r\langle a,b\rangle$$. \begin{align*} \langle ra, b\rangle = E(ra)_H\cdot E(b)_H &= \begin{pmatrix} ra_1\\ ra_2\\ \vdots\\ ra_n \end{pmatrix}\cdot \begin{pmatrix} b_1\\ b_2\\ \vdots\\ b_n \end{pmatrix}\\ &= \sum_{i=1}^{n} (ra_i)b_i\\ &= r\sum_{i=1}^{n} a_ib_i\\ &= r\begin{pmatrix} a_1\\ a_2\\ \vdots\\ a_n \end{pmatrix}\cdot \begin{pmatrix} b_1\\ b_2\\ \vdots\\ b_n \end{pmatrix}\\ &= r(E(a)_H\cdot E(b)_H)\\ &= r\langle a,b\rangle \end{align*}
4. For all vectors $$a\in V, \langle a,a \rangle \geq 0$$ and $$\langle a,a \rangle = 0$$ iff $$a = 0$$. \begin{align*} \langle a, a\rangle = E(a)_H \cdot E(a)_H &= \begin{pmatrix} a_1\\ a_2\\ \vdots\\ a_n \end{pmatrix}\cdot \begin{pmatrix} a_1\\ a_2\\ \vdots\\ a_n \end{pmatrix}\\ &= \sum_{i=1}^{n} {a_i}^2 \geq 0 \end{align*} But then again, I have no idea what $$\langle\,,\,\rangle_H$$ is.
• By "What is $\langle\ , \ \rangle_H$", do you mean "What $\langle\ , \ \rangle_H$ measures". Is it right? Jun 4, 2021 at 17:33
• More like what it really is. I'm not fond of the notations yet.
– muw
Jun 4, 2021 at 17:36
• $\langle \,\,, \,\,\rangle_{H}$ has been defined by $\langle E(u), E(v)\rangle_{H}$ in the question. Jun 4, 2021 at 17:42
• It essentially defines an inner product on the quotient space $V/H$ using only the components of vectors in $H^\perp$. Jun 4, 2021 at 17:55
• So how do you treat it? It just looks so bizarre to me.
– muw
Jun 4, 2021 at 18:38

To begin with, $$E(v)$$ is a subset of $$V$$. Let us call such a set an $$E$$-set.

Next, $$\langle \cdot, \cdot \rangle_H$$ is a function that takes two $$E$$-sets and returns a number; given $$E(u)$$ and $$E(v)$$ it returns $$\langle u_2, v_2 \rangle,$$ where $$u_2,v_2$$ are the $$H^\bot$$ components of $$u,v$$ respectively.

Then, if you can show that $$\langle \cdot, \cdot \rangle_H$$ satisfies the criteria for being an inner product, then you have an inner product on the set of $$E$$-sets. You can then say things like "this $$E$$-set is orthogonal to this other $$E$$-set."

To prove $$\langle , \rangle_H$$ is an inner product on $$Q$$, we have to prove the following:

1. For all vectors $$a, b,$$ and $$y\in V$$, $$\langle a, (b + y)\rangle = \langle a,b\rangle + \langle a + y\rangle$$.

Not for all $$a,b,y\in V$$ but for all $$E$$-sets $$a,b,y,$$ or $$\langle E(u), E(v)\oplus E(w) \rangle_H = \langle E(u), E(v) \rangle + \langle E(u), E(w) \rangle_H$$ for $$u,v,w\in V.$$

Proof
$$\langle E(u), E(v) \oplus E(w) \rangle_H = \langle E(u), E(v+w) \rangle_H = \langle u_2, (v+w)_2 \rangle = \langle u_2, v_2+w_2 \rangle \\ = \langle u_2, v_2 \rangle + \langle u_2, w_2 \rangle = \langle E(u), E(v) \rangle_H + \langle E(u), E(w) \rangle_H.$$ Note that you need to show first that $$(v+w)_2=v_2+w_2.$$ Everything else follows from definition of $$\langle \cdot, \cdot \rangle_H$$ and from $$\langle \cdot, \cdot \rangle$$ being an inner product on $$V.$$

• Given the OP’s inexperience, I think it would also help to add a brief explanation of why the inner product is even a function at all (for instance, it would not be a function if the RHS was $\langle u_1, v_1 \rangle$, and the OP may not be in a position to appreciate this subtlety). It’s also not obvious whether they are familiar with addition and multiplication in $Q$ either, though that is really up to the OP to clarify. Jun 5, 2021 at 1:41
• Please check my proof for number 1 if it's correct.
– muw
Jun 5, 2021 at 3:38
• I noticed that I missed one tiny detail in the problem. I forgot to include the definition of addition and scalar multiplication in $Q$: Define addition in $Q$ by: for $v, w\in V, E(v)\oplus E(w) = E(v + w)$. Define scalar multiplication in Q by: for $v \in V$ and $\rho\in \mathbb{R}, \rho E(v) = E(\rho v)$.
– muw
Jun 5, 2021 at 3:42
• @muw. You rewrite things as column matrices. This means that you have decomposed $E(a)$ etc. in some basis. But doing this requires that $\langle \cdot, \cdot \rangle_H$ is linear in each argument, which is what you want to show. Jun 5, 2021 at 6:17
• @muw. No, you still need to know that things work with the basis vectors. Jun 5, 2021 at 11:22