# Thread: inner product on R^2

1. ## inner product on R^2

We'll define the following operation for vectors u and v in $\mathbb{R}^2$:

$=u^TAv$

Where $A= \left[\begin{array}{cc}1 & 2 \\ 2 & 4\end{array}\right]$

Which axioms of inner products are satisfied for <,>? Is it an inner product in $\mathbb{R}^2$?

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I'm not sure how to answer this question but I know that it must satisfy all the axioms in order to be an inner product. These are the properties of inner products:

<u,v>=<v,u> (symmetry property)
<ku,v>=k<u,v> (homogeneity property)
<v,v> ≥ 0 and <v,v>=0 if and only if v=0 (positivity property)

I know that matrix multiplication not commutative, so I'm guessing the symmetry property is violated. I think the only condition it satisfies is the homogeneity. But I'm not sure how to go about proving the other ones. Any ideas?

2. Originally Posted by demode
We'll define the following operation for vectors u and v in $\mathbb{R}^2$:

$=u^TAv$

Where $A= \left[\begin{array}{cc}1 & 2 \\ 2 & 4\end{array}\right]$

Which axioms of inner products are satisfied for <,>? Is it an inner product in $\mathbb{R}^2$?

-----------------
I'm not sure how to answer this question but I know that it must satisfy all the axioms in order to be an inner product. These are the properties of inner products:

<u,v>=<v,u> (symmetry property)
<ku,v>=k<u,v> (homogeneity property)
<v,v> ≥ 0 and <v,v>=0 if and only if v=0 (positivity property)

I know that matrix multiplication not commutative, so I'm guessing the symmetry property is violated. I think the only condition it satisfies is the homogeneity. But I'm not sure how to go about proving the other ones. Any ideas?

Begin by checking positiveness, which you listed last but i's usually listed the first property an inner product must fulfill:

$v=\binom {x_1}{x_2}\,,\,\,v^tAv=(x_1\,\,x_2)\begin{pmatrix} 1&2\\2&4\end{pmatrix}\binom {x_1}{x_2}=$ $(x_1\,\,x_2)\binom {x_1+2x_2}{2x_1+4x_2}$ $=x_1^2+2x_1x_2+2x_1x_2+4x_2^2$ $=x_1^2+4x_1x_2+4x_2^2 = (x_1+2x_2)^2$ ,

and then you can see that it does NOT fulfill the condition (why?), so this is not an inner product (this was plain clear from the beginning if you knew about positive definite and semidefinite matrices but I assumed you don't know this yet).

Tonio

3. Originally Posted by tonio
Begin by checking positiveness, which you listed last but i's usually listed the first property an inner product must fulfill:

$v=\binom {x_1}{x_2}\,,\,\,v^tAv=(x_1\,\,x_2)\begin{pmatrix} 1&2\\2&4\end{pmatrix}\binom {x_1}{x_2}=$ $(x_1\,\,x_2)\binom {x_1+2x_2}{2x_1+4x_2}$ $=x_1^2+2x_1x_2+2x_1x_2+4x_2^2$ $=x_1^2+4x_1x_2+4x_2^2 = (x_1+2x_2)^2$ ,

and then you can see that it does NOW fulfill the condition
Typo- Tonio clearly meant "it does NOT fulfill the condition"

(why?), so this is not an inner product (this was plain clear from the beginning if you knew about positive definite and semidefinite matrices but I assumed you don't know this yet).

Tonio

4. Thanks Tonio,

if I may ask, what process did you use to get from $(x_1\,\,x_2)\binom {x_1+2x_2}{2x_1+4x_2}$ to $x_1^2+2x_1x_2+2x_1x_2+4x_2^2$?

5. Originally Posted by demode
Thanks Tonio,

if I may ask, what process did you use to get from $(x_1\,\,x_2)\binom {x_1+2x_2}{2x_1+4x_2}$ to $x_1^2+2x_1x_2+2x_1x_2+4x_2^2$?
It is common matrix multiplication of conformant matrices (which just means that they have appropriate dimensions to multiply by matrix multiplication).

CB

6. Originally Posted by tonio
and then you can see that it does NOT fulfill the condition (why?), so this is not an inner product (this was plain clear from the beginning if you knew about positive definite and semidefinite matrices but I assumed you don't know this yet).

Tonio
Why?? As I see it $(x_1+2x_2)^2$ fulfill the positivity condition because $(x_1+2x_2)^2 \geq 0 \forall x_1, x_2$ and $(x_1+2x_2)^2 = 0$ if the vectors are zero so $x_1, x_2$ are zero.

Now, can you check my work for the homogeneity property:

Let $k$ be any scalar.

$v=\binom {x_1}{x_2}\,,\,\,u=(y_1\,\,y_2)$

$\left\langle ku,v \right\rangle =ku^TAv=(ky_1 \,\, ky_2) \binom {x_1+2x_2}{2x_1+4x_2} \binom {x_1}{x_2}$

$= (ky_1\,\,ky_2)\binom {x_1+2x_2}{2x_1+4x_2}$

$=ky_1 (x_1+2x_2) + ky_2 (2x_1+4x_2)$

$=ky_1x_1+ky_12x_2+ky_22x_1+ky_24x_2$

$= k(y_1x_1+y_12x_2+y_22x_1+y_24x_2) = k \left\langle u,v \right\rangle$

So it satisfies the homogeneity condition

7. Originally Posted by demode
Why?? As I see it $(x_1+2x_2)^2$ fulfill the positivity condition because $(x_1+2x_2)^2 \geq 0 \forall x_1, x_2$ and $(x_1+2x_2)^2 = 0$ if the vectors are zero so $x_1, x_2$ are zero.

Not really: try the vector $(-2,1)$ ...

Tonio

Now, can you check my work for the homogeneity property:

Let $k$ be any scalar.

$v=\binom {x_1}{x_2}\,,\,\,u=(y_1\,\,y_2)$

$\left\langle ku,v \right\rangle =ku^TAv=(ky_1 \,\, ky_2) \binom {x_1+2x_2}{2x_1+4x_2} \binom {x_1}{x_2}$

$= (ky_1\,\,ky_2)\binom {x_1+2x_2}{2x_1+4x_2}$

$=ky_1 (x_1+2x_2) + ky_2 (2x_1+4x_2)$

$=ky_1x_1+ky_12x_2+ky_22x_1+ky_24x_2$

$= k(y_1x_1+y_12x_2+y_22x_1+y_24x_2) = k \left\langle u,v \right\rangle$

So it satisfies the homogeneity condition
.

8. I see! Can you show me how the symmetry property can be proven? This is my last question.

9. Originally Posted by demode
I see! Can you show me how the symmetry property can be proven? This is my last question.

Take two vectors $u=(x_1,y_1)\,,\,v=(x_2,y_2)\in\mathbb{R}^2$ and check that $=u^tAv=v^tAu=$ . Not hard though slightly annoying work.

Tonio

10. You may also note the fact that $(v^tAu)^t = u^tAv$ however $v^tAu \in \mathbb{R}$ and therefore its transpose is simply itself, that is:

$ = v^tAu = (v^tAu)^t = u^tAv = $.