Let T:P_2→P_1T:P2→P1 be defined by T(a+bx+cx^2)=b+2c+(a-b)xT(a+bx+cx2)=b+2c+(a−b)x. Check that TT is a linear transformation. Find the matrix of the transformation with respect to the ordered bases B_1={x^2,x^2+x,x^2+x+1}B1={x2,x2+x,x2+x+1} and B_2={1,x}B2={1,x}. Find the kernel of TT.?

1 Answer
Morgan
Feb 11, 2018

See below.

Explanation:

A linear transformation from a vector space V to a vector space W is a function T:V->WT:V→W such that for all vectors u and v in V and all scalars cc, the following two properties hold:

1." "T(u+v)=T(u)+T(v)1. T(u+v)=T(u)+T(v)
2." "T(cu)=cT(u)2. T(cu)=cT(u)

That is to say that T preserves addition (1) and T preserves scalar multiplication (2).

If T:P_2->P_1T:P2→P1 is given by the formula T(a+bx+cx^2)=b+2c+(a-b)xT(a+bx+cx2)=b+2c+(a−b)x, we can verify that T is a linear transformation as follows:

First let u=dx^2+ex+fu=dx2+ex+f and v=gx^2+hx+kv=gx2+hx+k be vectors in P_2P2 and let mm be a scalar.

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  1. Show that T preserves addition:

T(u+v)=T((dx^2+ex+f)+(gx^2+hx+k))T(u+v)=T((dx2+ex+f)+(gx2+hx+k))

=>=T((d+g)x^2+(e+h)x+(f+k))⇒=T((d+g)x2+(e+h)x+(f+k))

=>=(e+h)+2(d+g)+((f+k)-(e+h))x⇒=(e+h)+2(d+g)+((f+k)−(e+h))x

and

T(u)+T(v)=T(dx^2+ex+f)+T(gx^2+hx+k)T(u)+T(v)=T(dx2+ex+f)+T(gx2+hx+k)

=>=(e+2d+(f-e)x)+(h+2g+(k-h)x)⇒=(e+2d+(f−e)x)+(h+2g+(k−h)x)

=>e+h+2d+2g+fx-ex+kx-hx⇒e+h+2d+2g+fx−ex+kx−hx

=>(e+h)+2(d+g)+((f+k)-(e+h))x⇒(e+h)+2(d+g)+((f+k)−(e+h))x

as found above. Therefore, we conclude that T(u+v)=T(u)+T(v)T(u+v)=T(u)+T(v).

:.∴ T preserves addition.

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2".". Show that T preserves scalar multiplication:

T(cu)=T(c(dx^2+ex+f))T(cu)=T(c(dx2+ex+f))

=>=T(cdx^2+cex+cf)⇒=T(cdx2+cex+cf)

=>=ce+2cd+(cf-ce)x⇒=ce+2cd+(cf−ce)x

and

cT(u)=cT(dx^2+ex+f)cT(u)=cT(dx2+ex+f)

=>=c(e+2d+(f-e)x)⇒=c(e+2d+(f−e)x)

=>=ce+2cd+(cf-ce)x⇒=ce+2cd+(cf−ce)x

as found above. Therefore, we conclude that T(cu)=cT(u)T(cu)=cT(u).

:.∴ T preserves scalar multiplication.

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Find the matrix representation of the transformation with respect to the ordered bases:

B_1={x^2,x^2+x,x^2+x+1}B1={x2,x2+x,x2+x+1} and B_2={1,x}B2={1,x}

I am not terribly familiar with this concept, but here is my attempt:

We will find a 3xx33×3 matrix that represents T with respect to the basis B_1={x^2,x^2+x,x^2+x+1}B1={x2,x2+x,x2+x+1}; that is, find the matrix A so that

T([(a), (b), (c)])=A[(a), (b), (c)]

Since we know that T(a+bx+cx^2)=b+2c+(a-b)x, we will use this equation to find a 3xx3 matrix.

T(x^2)=T(0+0x+1x^2)=2(1)=2
T(x^2+x)=T(0+1x+1x^2)=1+2(1)+(0-1)x=3-x
T(x^2+x+1)=T(1+1x+1x^2)=1+2(1)+(1-1)x=3

We will use the coefficients of the vector space polynomials to obtain our 3xx3 matrix.

The coefficients of T(x^2) give the vector [(2),(0),(0)]

The coefficients of T(x^2+x) give the vector [(3),(-1),(0)]

The coefficients of T(x^2+x+1) give the vector [(3),(0),(0)]

These four vectors give the 3xx3 matrix:

[(2,3,3),(0,-1,0),(0,0,0)]

If this does not agree with your expectations for what the answer should be, this may be helpful:

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Find the kernel of T.

The kernal of a linear transformation T is the set of all vectors v such that T(v)=0 (i.e. the kernel of a transformation between vector spaces is its null space).

To find the null space we must first reduce the 3xx3 matrix found above to row echelon form. We obtain:

[(1,0,3/2),(0,1,0),(0,0,0)]

Which provides the equations:

x_1=3/2

x_2=0

0x_3=0

We see that x_3 is a free variable, so we can parameterize by setting x_3=t.

Then "ker"(T)=[(3/2),(0),(t)].

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