Let #T:P_2→P_1# be defined by #T(a+bx+cx^2)=b+2c+(a-b)x#. Check that #T# 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}# and #B_2={1,x}#. Find the kernel of #T#.?

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->W# such that for all vectors u and v in V and all scalars #c#, the following two properties hold:

#1." "T(u+v)=T(u)+T(v)#
#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_1# is given by the formula #T(a+bx+cx^2)=b+2c+(a-b)x#, we can verify that T is a linear transformation as follows:

First let #u=dx^2+ex+f# and #v=gx^2+hx+k# be vectors in #P_2# and let #m# be a scalar.

#------------------------#

  1. Show that T preserves addition:

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

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

#=>=(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)#

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

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

#=>(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 preserves addition.

#------------------------#

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

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

#=>=T(cdx^2+cex+cf)#

#=>=ce+2cd+(cf-ce)x#

and

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

#=>=c(e+2d+(f-e)x)#

#=>=ce+2cd+(cf-ce)x#

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

#:.# T preserves scalar multiplication.

#------------------------#

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

#B_1={x^2,x^2+x,x^2+x+1}# and #B_2={1,x}#

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

We will find a #3xx3# matrix that represents T with respect to the basis #B_1={x^2,x^2+x,x^2+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:

#------------------------#

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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