# Thread: Question:Inner product axioms and orthogonal projection/distance

1. ## Question:Inner product axioms and orthogonal projection/distance

There's a theorem in my Linear Algebra book which states:

Let $\mathbf{u}$ and $\mathbf{v}$ be two vectors in an inner product space $V$, such that $\mathbf{v} \neq 0$. Then
$d(\mathbf{u}, \text{proj}_v{\mathbf{u}}) < d(\mathbf{u}, c\mathbf{v}), \quad c \neq \frac{\langle \mathbf{u},\mathbf{v}\rangle}{\langle \mathbf{v},\mathbf{v} \rangle}.$

Here $d$ means the distance between two vectors.

The proof is given like this:

Let $b = \langle \mathbf{u},\mathbf{v} \rangle / \langle \mathbf{v},\mathbf{v} \rangle$. Then you write:

$\parallel \mathbf{u} - c\mathbf{v} \parallel^2 = \parallel (\mathbf{u} - b\mathbf{v} ) + (b - c)\mathbf{v} \parallel^2,$

where $(\mathbf{u} - b\mathbf{v} )$ and $(b - c)\mathbf{v}$ are orthogonal. You can verify this by using the inner product
axioms to show that

$\langle (\mathbf{u} - b\mathbf{v} ),(b - c)\mathbf{v}\rangle = 0$

....rest of the proof.

And Inner product axioms are:
Let $\mathbf{u}, \mathbf{v},$ and $\mathbf{w}$ be vectors in a vector space $V$, let $c$ be any scalar. An inner product on $V$
is a function that associates a real number $\langle \mathbf{u},\mathbf{v} \rangle$ with each pair of vectors $\mathbf{u}$ and $\mathbf{v}$
and satisfies the following axioms.

1) $\langle \mathbf{u},\mathbf{v} \rangle = \langle \mathbf{v},\mathbf{u} \rangle$
2) $\langle \mathbf{u},\mathbf{v} + \mathbf{w} \rangle = \langle \mathbf{u},\mathbf{v} \rangle + \langle \mathbf{u},\mathbf{w} \rangle$
3) $c \langle \mathbf{u},\mathbf{v} \rangle = \langle c.\mathbf{u},\mathbf{v} \rangle$
4) $\langle \mathbf{v},\mathbf{v} \rangle \geq 0, \text{ and } \langle \mathbf{v},\mathbf{v} \rangle = 0 \text{ if and only if } \mathbf{v} = 0$

My question: How do you prove:
$\langle (\mathbf{u} - b\mathbf{v} ),(b - c)\mathbf{v}\rangle = 0$
using inner product axioms? I can't make a connection with the inner product axioms and this.

As you seen above the author is saying that this can be proved using inner product axioms. Is it possible to kindly show that $\langle (\mathbf{u} - b\mathbf{v} ),(b - c)\mathbf{v}\rangle = 0$?

2. ## Re: Question:Inner product axioms and orthogonal projection/distance

Originally Posted by x3bnm
There's a theorem in my Linear Algebra book which states:

Let $\mathbf{u}$ and $\mathbf{v}$ be two vectors in an inner product space $V$, such that $\mathbf{v} \neq 0$. Then
$d(\mathbf{u}, \text{proj}_v{\mathbf{u}}) < d(\mathbf{u}, c\mathbf{v}), \quad c \neq \frac{\langle \mathbf{u},\mathbf{v}\rangle}{\langle \mathbf{v},\mathbf{v} \rangle}.$

Here $d$ means the distance between two vectors.

The proof is given like this:

Let $b = \langle \mathbf{u},\mathbf{v} \rangle / \langle \mathbf{v},\mathbf{v} \rangle$. Then you write:

$\parallel \mathbf{u} - c\mathbf{v} \parallel^2 = \parallel (\mathbf{u} - b\mathbf{v} ) + (b - c)\mathbf{v} \parallel^2,$

where $(\mathbf{u} - b\mathbf{v} )$ and $(b - c)\mathbf{v}$ are orthogonal. You can verify this by using the inner product
axioms to show that

$\langle (\mathbf{u} - b\mathbf{v} ),(b - c)\mathbf{v}\rangle = 0$

....rest of the proof.

And Inner product axioms are:
Let $\mathbf{u}, \mathbf{v},$ and $\mathbf{w}$ be vectors in a vector space $V$, let $c$ be any scalar. An inner product on $V$
is a function that associates a real number $\langle \mathbf{u},\mathbf{v} \rangle$ with each pair of vectors $\mathbf{u}$ and $\mathbf{v}$
and satisfies the following axioms.

1) $\langle \mathbf{u},\mathbf{v} \rangle = \langle \mathbf{v},\mathbf{u} \rangle$
2) $\langle \mathbf{u},\mathbf{v} + \mathbf{w} \rangle = \langle \mathbf{u},\mathbf{v} \rangle + \langle \mathbf{u},\mathbf{w} \rangle$
3) $c \langle \mathbf{u},\mathbf{v} \rangle = \langle c.\mathbf{u},\mathbf{v} \rangle$
4) $\langle \mathbf{v},\mathbf{v} \rangle \geq 0, \text{ and } \langle \mathbf{v},\mathbf{v} \rangle = 0 \text{ if and only if } \mathbf{v} = 0$

My question: How do you prove:
$\langle (\mathbf{u} - b\mathbf{v} ),(b - c)\mathbf{v}\rangle = 0$
using inner product axioms? I can't make a connection with the inner product axioms and this.

As you seen above the author is saying that this can be proved using inner product axioms. Is it possible to kindly show that $\langle (\mathbf{u} - b\mathbf{v} ),(b - c)\mathbf{v}\rangle = 0$?
note that the first and the third axiom imply that $c \langle \mathbf{u},\mathbf{v} \rangle = \langle \mathbf{u},c\mathbf{v} \rangle.$ so you want to prove that $(b-c) \langle \mathbf{u} - b\mathbf{v}, \mathbf{v}\rangle = 0,$ which is equivalent to $\langle \mathbf{u} - b\mathbf{v}, \mathbf{v}\rangle = 0$ because $b \neq c.$ finally, the identity $\langle \mathbf{u} - b\mathbf{v}, \mathbf{v}\rangle = 0$ is nothing but the definition of $b$.

3. ## Re: Question:Inner product axioms and orthogonal projection/distance

Originally Posted by NonCommAlg
note that the first and the third axiom imply that $c \langle \mathbf{u},\mathbf{v} \rangle = \langle \mathbf{u},c\mathbf{v} \rangle.$ so you want to prove that $(b-c) \langle \mathbf{u} - b\mathbf{v}, \mathbf{v}\rangle = 0,$ which is equivalent to $\langle \mathbf{u} - b\mathbf{v}, \mathbf{v}\rangle = 0$ because $b \neq c.$ finally, the identity $\langle \mathbf{u} - b\mathbf{v}, \mathbf{v}\rangle = 0$ is nothing but the definition of $b$.
Thanks a lot NonCommAlg.