Question #254764

6. a) Given any real symmetric square matrix A of order n, show that the function RnxRn \to defined by I(x, y) =xTAy is a symmetric bilinear function

Give examples of matrices A for which the corresponding I is not an inner product

State the extra properties on A which would make the corresponding I an inner product


b) Show that R[x] deg ≤ 2 x R[x] deg ≤ 2 defined by I (p(x), q(x)) =p(0)q(0)+p(1/2)q(1/2)+p(1)q(1) is an inner product on R[x] deg ≤ 2

Use Gram-Schmidt procedure to obtain a orthogonal basis of R[x] deg ≤ 2(with respect to the inner product defined above) which contains the polynomial x


1
Expert's answer
2021-11-15T05:09:39-0500

The function defined by I(x,y)=xTAy\Iota(x,y) = x^TAy is a symmetric bilinear function

1) it satisfies the axiom <u,v> = <v,u> in such:


⟨x,y⟩=xT^T Ay=x⋅(Ay)


Since A is symmetric


⟨x,y⟩=x⋅(Ay)=(Ay)⋅x

=(Ay)T^Tx=yT^T AT^T x=yT^T Ax=⟨y,x⟩.


(ii) For any vector x,y,z and any real number r, we have:


rx,y=(rx)TAy=rxTAy=rx,y\langle rx,y\rangle=(rx)^TAy=rx^TAy=r\langle x,y\rangle

and

x+y,z=(x+y)TAz=(xT+yT)Az\langle\>x+y,z\rangle=(x+y)^TAz=(x^T+y^T)Az


=xTAz+yTAz=x,y=y,z=x^TAz+y^TAz=\langle\>x,y\rangle=\langle\>y,z\rangle


Therefore the linearity in the first argument is satisfied


It satisfies the rule of positive- Definiteness in such


If x0x\ne0 then we have

x,x=xTAx>0\langle\>x,x\rangle=x^TAx>0


Examples of matrices A for which the corresponding I\Iota is not an inner product

\to \> A matrix of order m××n

\to A matrix A AT\ne\>A^T

\to A matrix A whose eigenvalues λ0\lambda\> \leq0

\to A matrix A whose determinant det A=0A=0


Properties of A which would make the corresponding I\Iota on inner product

Matrix A should be a real symmetric n×nn×n matrix

Matrix A must be positive definite in that

xTAx>0x^TAx>0

Matrix A has real eigenvalues

The eigenvectors of matrix A corresponding to the eigenvalues are orthogonal

Matrix A is diagonalizable


Part 6(B) sections


I\Iota is inner product if it satisfies conjugate symmetry, linearity and positive Definiteness axioms

(I) <q(x),p(x)>=q(0)p(0)+q(12)q(12)+q(1)p(1)<q(x),p(x)>=q(0)p(0)+q(\frac{1}{2})q(\frac{1}{2})+q(1)p(1)


=p(0)q(0)+p(12)q(12)+p(1)q(1)=p(0)q(0)+p(\frac{1}{2})q(\frac{1}{2})+p(1)q(1)


<q(x),p(x)>=<p(x),q(x)>\therefore<q(x),p(x)>=<p(x),q(x)>


I\Iota is symmetric


(ii) c<p,q>=c[p(0)q(0)+p(12q(12+p(1)q(1)]c<p,q>=c\>[{p(0)q(0)+p(\frac{1}{2}q(\frac{1}{2}+p(1)q(1)}]


=cp(0)q(0)+cp(12)q(12)+cp(1)q(1)=cp(0)q(0)+cp(\frac{1}{2})q(\frac{1}{2})+cp(1)q(1)


=<cpf,q>=<cpf,q>

I\Iota satisfies linearity


(iii) <p,p>=p(0)p(0)+p(12)p(12)+p(1)p(1)>0<p,p>=p(0)p(0)+p(\frac{1}{2})p(\frac{1}{2})+p(1)p(1)>0


If p(x)p(x) is not zero


Part 6(B) section 2


The standard polynomial basis for R[x]deg2R[x]deg\>\le2 is defined [1,x,x2][1,x,x^2]

Let orthogonal basis of I\Iota be (b1,b2,b3)(b_1,b_2,b_3)

b1=1b_1=1

b2=xproj1xb_2=x-proj_1x


b2=x<1,x><1,1>1\therefore b_2=x-\frac{<1,x>}{<1,1>}1


<1,x>=(1)(0)+(1)(12)+(1)(1)=32<1,x>=(1)(0)+(1)(\frac{1}{2})+(1)(1)=\frac{3}{2}


<1,1>=(1)(1)+(1)(1)+(1)(1)=3<1,1>=(1)(1)+(1)(1)+(1)(1)=3


b2=x12b_2=x-\frac{1}{2}


b3=x2<x2,1><1,1>1b_3=x^2-\frac{<x^2,1>}{<1,1>}1- <x2,x12><x12,x12>(x12)\frac{<x^2,x-\frac{1}{2}>}{<x-\frac{1}{2},x-\frac{1}{2}>}(x-\frac{1}{2})



<x2,1>=(1)(0)+(1)(12)2+(1)(1)2=54<x^2,1>=(1)(0)+(1)(\frac{1}{2})^2+(1)(1)^2=\frac{5}{4}


<x12,x12>=(12)(12)+(0)(0)+(12)(12)=12<x-\frac{1}{2},x-\frac{1}{2}>=(\frac{-1}{2})(\frac{-1}{2})+(0)(0)+(\frac{1}{2})(\frac{1}{2})=\frac{1}{2}



<x2,x12>=(0)(12)+(12)2(0)+(1)2(12)<x^2,x-\frac{1}{2}>=(0)(\frac{-1}{2})+(^\frac{1}{2})^2(0)+(1)^2(\frac{1}{2})


=12=\frac{1}{2}


b3=x25121(x12)b_3=x^2-\frac{5}{12}-1(x-\frac{1}{2})


=x2x+112=x^2-x+\frac{1}{12}


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