In R3 Let V1 = (X1,Y1,Z1) and V2 = (X2,Y2,Z2)

The function t : R3 −→ R3 is given by the rule
(x, y, z) 0−→ (y − z, x + z, x + y).
(a) Use Strategy 1.1 in Unit LA4 to show that t is a linear transformation.
(b) Write down the matrix of t with respect to the standard basis for R3.
(c) Determine the matrix of t with respect to the basis
{(1, 0, 0), (1, 1, 0), (0, 1, 1)} for the domain and the standard basis for
the codomain.
(d) Find the kernel of t, describe it geometrically and state its dimension.
(e) Find a basis for the image of t, state the dimension of the image and
describe the image geometrically.
(f) Let s be the linear transformation
s : P3 −→ R3
a + bx + cx2 0−→ (a + c, b, a + b + c).
Find the matrix of s and the matrix of t ◦ s with respect to the
standard basis for the domain P3 and the standard basis for the
codomain R3.

Answers:

a)

t(x,y,z) = (y-z,x+z,x+y)

1)

t(0,0,0) = (0-0,0+0,0+0)=(0,0,0) (checked)

2)

We check LT1

In R3 Let v1 = (x1,y1,z1) and v2 = (x2,y2,z2)

t(v1+v2) = t((x1,y1,z1)+(x2,y2,z2)) = t((x1+x2,y1+y2,z1+z2)) =

( (y1+y2)-(z1+z2),(x1+x2)+(z1+z2),(x1+x2)+(y1+y2))

= ( (y1-z1)+(y2-z2),(x1+z1)+(x2+z2),(x1+y1)+(x2+y2))

= (y1-z1,x1+z1,x1+y1)+(y2-z2,x2+z2,x2+y2)

= t(x1,y1,z1)+t(x2,y2,z2)

= t(v1)+t(v2)

We check LT2

Let v = (x,y,z) and R

t(v) = t(x,y,zt(x,y,z) = (y-z,x+z,x+y) = (y-z,x+z,x+y) = t(v)

t verifies both properties

Then: t is a linear transformation

b)

t(x,y,z) = (y-z,x+z,x+y)

Let A = Thematrix of t with respect to the standard basis for R3

t(1,0,0) = (0-0,1+0,1+0)= (0,1,1)

t(0,1,0) = (1-0,0+0,0+1) = (1,0,1)

t(0,0,1) = (0-1,0+1,0+0) = (-1,1,0)

c)

t(x,y,z) = (y-z,x+z,x+y)

Let B = The matrix of t with respect to the basis
{(1, 0, 0), (1, 1, 0), (0, 1, 1)} for the domain and the standard basis for
the codomain.

t(1,0,0) = (0-0,1+0,1+0) = (0,1,1)

t(1,1,0) = (1-0,1+0,1+1) = (1,1,2)

t(0,1,1) = (1-1,0+1,0+1) = (0,1,1)

d)

We have to find Ker(t) = {vR3 /t(v)=(0,0,0)}

t(v) = t(x,y,z) = (y-z,x+z,x+y) = (0,0,0)

Then

y-z = 0

x+z = 0

x+y = 0

Then:

y = z , x= -y

Assuming the parameter k to the unknown z we obtain:

x = - k , y = k , z = k

Then

Ker(t) = {(-k,k,k): kR}

Geometrically Ker(t) is the line through (0,0,0) and (-1,1,1)

And its dimension is 1

e)

1)

We take the standard basis {(1,0,0),(0,1,0),(0,0,1)}

2)

From part b)

t(1,0,0) = (0,1,1)

t(0,1,0) = (1,0,1)

t(0,0,1) = (-1,1,0)

Then S = {(0,1,1),(1,0,1),(-1,1,0)}

3)

Since (-1,1,0) = (0,1,1) - (1,0,1) then S1 = S – {(-1,1,0)} = {(0,1,1),(1,0,1)}

4)

Since S1 is a linearly independent set then S1 is a basis of Im(t)

Dimension of Im(t) is 2

Geometrically Im(t) it´s a plane through the origin and containing the points (0,1,1) and (1,0,1)

f)

P3 = {p(x):p(x)=a+bx+cx2, a,b,cR}

Standard basis of P3 is: {1,x,x2}

s : P3→ R3
s(a + bx + cx2) =(a + c, b, a + b + c).

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

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

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

Then the matrix of s with respect to the standard basis for the domain P3and the standard basis for the codomain R3 is M where:

From part b)

Let A = Thematrix of t with respect to the standard basis for R3

Then the matrix of t ◦ s with respect to the standard basis for the domain P3 and the standard basis for the codomain R3 is AM

AM =