Answer to Question #254802 in Linear Algebra for Sabelo Xulu

Question #254802

4.a) Let R[x] deg≤2 be the vector space of all polynomial functions in a single variable x with real coefficient and degree at most 2

Let R[x] deg≤2\toR[x] deg≤2 be defined by T(p(x)) =p(x-1)

Let U = <1,x,x2> be the usual basis for R[x] deg≤2 and B = <1+x+x2,2x+x2,x+x2> be another basis for R[x] deg≤2

Let, for any polynomial p(x) \isin R[x] deg≤2, coordB(p(x)) R3 is the coordinates of p(x) with respect to the basis B for instance

coordV(x+x2) =(0, 1,1)

and

coordB(x+x2)=(0, 0,1)

Also recall the agreement every vector X=(x1, x2, x3) \isin R3 is considered as a 3x1 matrix (X1X2X3)\begin{pmatrix} X1 \\ X2 \\ X3 \end{pmatrix} i.e, as a column vector

a) Find the matrix MatU\to U (T) representing the linear transformation T with respect to the basis of U

b) Verify the equation MatU\to U (T)coordU(p(x)) =coordU(T(p(x)), for any p(x) \isin R[x] deg≤2

c) Obtain the change of basis matrix PB\toU from B to U

d) Hance, or otherwise, obtain coordB(p(x))for each p(x) \isin R[x] deg≤2


1
Expert's answer
2021-10-25T18:07:58-0400

a) T((P(x))=P(x1)T((P(x))=P(x-1)


T(1)=0,T(x)=x1,T(1)=0 , T(x)=x-1,

T(x)T(x) 2 =x=x 21-1

With respect to U

0=a(1)+b(x)+c(x0= a(1)+b(x)+c(x 2))

a=0,b=0,c=0,a=0\:,b=0,c=0 ,

[0]u=[0,0,0]T[0]_u =[0,0,0]^T


x1=a(1)+b(x)+c(x2)x-1=a(1)+b(x)+c(x^2)

a=1,b=1,c=0a=-1,b=1,c=0

[x1]u=[1,1,0]T[x-1]_u=[-1,1,0]^T


x21=a(1)+b(x)+c(x2)x^2-1=a(1)+b(x)+c(x^2)

a=1,b=0,c=1a=-1,b=0,c=1

[x21]u=[1,0,1]T[x^2-1]_u=[-1,0,1]^T


Matrix U_U \to u(T)=(011010001)=\begin{pmatrix} 0&-1 & -1\\ 0&1 & 0\\ 0&0&1 \end{pmatrix}


Part B

Coordu(P(x))=[010](P(x))= \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}


Coord u(T(P(x)))=(T(P(x)))=

Coordu (P(x1))=[110](P(x-1))=\begin{bmatrix} -1 \\ 1\\ 0 \end{bmatrix}



(011010001)\begin{pmatrix} 0&-1 & -1\\ 0&1 & 0\\ 0&0&1 \end{pmatrix} (010)\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix} =(110)\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}


Verified


Part C

Expressing elements of B in terms of the elements of U

1+x+x2=a(1)+b(x)+c(x2)1+x+x^2=a(1)+b(x)+c(x^2)

a=1,b=1,c=1a=1,b=1,c=1

[1+x+x2]u=[1,1,1]T[1+x+x^2]_u=[1,1,1]^T


2x+x2=a(1)+b(x)+c(x2)2x+x^2=a(1)+b(x)+c(x^2)

a=0,b=2,c=1a=0,b=2,c=1

[2x+x2]u=[0,2,1]T[2x+x^2]_u=[0,2,1]^T


x+x2=a(1)+b(x)+c(x2)x+x^2=a(1)+b(x)+c(x^2)

a=0,b=1,c=1a=0,b=1,c=1

[x+x2]u=[0,1,1]T[x+x^2]_u=[0,1,1]^T


Change of base matrix PBU=P_B\to_U=


[100121111]\begin{bmatrix} 1&0 & 0\\ 1&2 & 1\\ 1&1&1 \end{bmatrix}


Part D

Expressing elements of Coordu (P(x))(P(x)) in terms of basis B

x=a(1+x+x2)+b(2x+x2)x=a(1+x+x^2)+b(2x+x^2)

+c(x+x2)+c(x+x^2)

x=a+(a+2b+c)x+(a+b+c)x2x=a+(a+2b+c)x+(a+b+c)x^2

    a=0,a+2b+c=1\implies a=0,a+2b+c=1

    2b+c=1.......(i)\implies 2b+c=1.......(i)

a+b+c=0a+b+c=0

    b+c=0......(ii)\implies b+c=0......(ii)

Solving (i)(i) and(ii)(ii)

b=1,c=1b=1,c=-1

CoordB(P(x))=(011)_B(P(x))=\begin{pmatrix} 0 \\ 1\\ -1 \end{pmatrix}


Alternatively (method 2)


Coordu (P(x))=(010)(P(x))=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}

Change of basis matrix PuB=P_u\to_B=


(MatrixPBu)1(Matrix P_B\to_u)^{-1}


(100121111)\therefore \begin{pmatrix} 1&0 & 0 \\ 1&2 & 1\\ 1&1&1 \end{pmatrix} 1^-1 ( coordB(P(x))_B(P(x)))


(100011112)\begin{pmatrix} 1&0& 0\\ 0&1&-1\\ -1&-1&2 \end{pmatrix} (010)\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} =(011)\begin{pmatrix} 0\\ 1\\ -1 \end{pmatrix}


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