Answer to Question #267036 in Functional Analysis for Gestavo

Question #267036

Proof whether the following operations are inner product operations:

⟨x, y⟩ = 2x1y1 − x1y2 − x2y1 + 2x2y2, x=(x1, x2), y=(y1, y2)



1
Expert's answer
2021-11-17T07:10:01-0500

We must verify linearity,symmetry and positive definiteness properties.

We can define <x,y> through matrix operations as

<x,y>=xTAy, x=(x1x2),xT=(x1 x2),y=(y1y2),A=(2112)<x,y>=x^T\cdot A\cdot y, \space x=\begin{pmatrix} x_1 \\ x_2\end{pmatrix},x^T=(x_1\space x_2),y=\begin{pmatrix} y_1 \\ y_2\end{pmatrix}, A=\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}


and use linearity of transposition operation T^T and matrix multiplication

So we have

<x(1)+x(2),y>=(x(1)+x(2))TAy=((x(1))T+(x(2)))Ay=((x(1))TA+(x(2))TAy=(x(1))TAy+(x(2))TAy=<x(1),y>+<x(2),y><x^{(1)}+x^{(2)},y>=(x^{(1)}+x^{(2)})^T\cdot A\cdot y=\\ ( (x^{(1)})^T+(x^{(2)}))\cdot A\cdot y=((x^{(1)})^T\cdot A+(x^{(2)})^T\cdot A\cdot y=\\ (x^{(1)})^T\cdot A\cdot y+(x^{(2)})^T\cdot A\cdot y=<x^{(1)},y>+<x^{(2)},y>

and

<cx,y>=(cx)TAy=(cxT)Ay=c(xTAy)=c<x,y><c\cdot x,y>=(c\cdot x)^T\cdot A \cdot y=(c\cdot x^T)\cdot A\cdot y=\\ c\cdot (x^T\cdot A\cdot y)=c\cdot <x,y>

Thus linearity by the first argument of <,> is proved.

2) Symmetry property

<x,y>=xTAy=[xTAyR]=(xTAy)T=yTAT(xT)T=[AT=A, Asymmetric,(xT)T=x]=yTAx=<y,x><x,y>=x^T\cdot A \cdot y=[x^T\cdot A \cdot y\in R]=(x^T\cdot A \cdot y)^T=\\ y^T\cdot A^T\cdot (x^T)^T=[A^T=A,\space A-symmetric,(x^T)^T=x]=y^T\cdot A\cdot x=<y,x>

From symmetry of <x,y> linearity by 2-argument is proved also.

3) Positive definiteness property.

This property better to verify by using simplest definition of <x,y>.

Let x0x\ne\overline 0 or x12+x22>0x_1^2+x_2^2> 0 . We have

<x,x>=2x122x1x2+2x22=(x1x2)2+x12+x22>02x_1^2-2x_1x_2+2x_2^2=(x_1-x_2)^2+x_1^2+x_2^2>0

Thus all properties of inner product is proved.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog