Answer to Question #254764 in Linear Algebra for Sabelo Xulu

Question

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

Answer 1

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

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

⟨x,y⟩=x$^T$ Ay=x⋅(Ay)

Since A is symmetric

⟨x,y⟩=x⋅(Ay)=(Ay)⋅x

=(Ay)$^T$x=y$^T$ A$^T$ x=y$^T$ Ax=⟨y,x⟩.

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

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

and

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

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

Therefore the linearity in the first argument is satisfied

It satisfies the rule of positive- Definiteness in such

If $x\ne0$ then we have

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

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

$\to \>$ A matrix of order m$×$n

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

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

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

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

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

Matrix A must be positive definite in that

$x^TAx>0$

Matrix A has real eigenvalues

The eigenvectors of matrix A corresponding to the eigenvalues are orthogonal

Matrix A is diagonalizable

Part 6(B) sections

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

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

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

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

$\Iota$ is symmetric

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

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

$=<cpf,q>$

$\Iota$ satisfies linearity

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

If $p(x)$ is not zero

Part 6(B) section 2

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

Let orthogonal basis of $\Iota$ be $(b_1,b_2,b_3)$

$b_1=1$

$b_2=x-proj_1x$

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

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

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

$b_2=x-\frac{1}{2}$

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

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

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

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

$=\frac{1}{2}$

$b_3=x^2-\frac{5}{12}-1(x-\frac{1}{2})$

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