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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.
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  • $\begingroup$ By "What is $\langle\ , \ \rangle_H$", do you mean "What $\langle\ , \ \rangle_H$ measures". Is it right? $\endgroup$
    – DiegoMath
    Commented Jun 4, 2021 at 17:33
  • $\begingroup$ More like what it really is. I'm not fond of the notations yet. $\endgroup$
    – muw
    Commented Jun 4, 2021 at 17:36
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    $\begingroup$ $\langle \,\,, \,\,\rangle_{H}$ has been defined by $\langle E(u), E(v)\rangle_{H}$ in the question. $\endgroup$
    – Anurag A
    Commented Jun 4, 2021 at 17:42
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    $\begingroup$ It essentially defines an inner product on the quotient space $V/H$ using only the components of vectors in $H^\perp$. $\endgroup$
    – user1551
    Commented Jun 4, 2021 at 17:55
  • $\begingroup$ So how do you treat it? It just looks so bizarre to me. $\endgroup$
    – muw
    Commented Jun 4, 2021 at 18:38

1 Answer 1

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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 $V.$

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    $\begingroup$ 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. $\endgroup$
    – Erick Wong
    Commented Jun 5, 2021 at 1:41
  • $\begingroup$ Please check my proof for number 1 if it's correct. $\endgroup$
    – muw
    Commented Jun 5, 2021 at 3:38
  • $\begingroup$ 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 $\rho\in \mathbb{R}, \rho E(v) = E(\rho v)$. $\endgroup$
    – muw
    Commented Jun 5, 2021 at 3:42
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    $\begingroup$ @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. $\endgroup$
    – md2perpe
    Commented Jun 5, 2021 at 6:17
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    $\begingroup$ @muw. No, you still need to know that things work with the basis vectors. $\endgroup$
    – md2perpe
    Commented Jun 5, 2021 at 11:22

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