Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Linear Transformations: Planes, Lines, and Dependence, Assignments of Linear Algebra

Various properties of linear transformations in rn, focusing on the image of planes, lines, and linearly dependent sets. Topics include parametric equations of lines and planes, the effect of linear transformations on linearly dependent sets, and the relationship between linear transformations and invertible matrices.

Typology: Assignments

Pre 2010

Uploaded on 08/18/2009

koofers-user-ji5
koofers-user-ji5 🇺🇸

10 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Newberger Math 247 Spring 03
Homework solutions: Section 1.8 #26, 27, 31, Section 2.3 #34,36,37,38
Section 1.8 #26,27,31.
26. Let uand vbe linearly independent vectors in R3and let Pbe the plane
through u,vand 0. The parametric equation of Pis x=su+tv(with s
and tin R). Show that a linear transformation T:R3R3maps Ponto a
plane through 0, a line through 0or onto just the origin 0in R3. What must
be true about T(u) and T(v) in order for the image of the plane Pto be a
plane?
We want to know the image of Punder T. Any point xon Pwill have the
form x=su+tvwhere sand tare real numbers. So T(x) = sT (u) + tT (v),
and T(P)is the collection of all vectors of the form sT (u)+ tT (v)where sand
tare any real number. Thus the image of Punder Tis Span{T(u), T (v)}.
Since there are three cases for the geometric description of the span of two
vectors in R3, there are three cases for the geometric description of the image
of Punder T. First {T(u), T (v)}could be independent, in which case, the
image of Punder Tis a plane. Second, {T(u), T (v)}could be dependent but
not both zero, in which case the image of Pis a line. Third, both vectors T(u)
and T(v)could be zero, in which case the image of Pis simply 0.
27. a. Show that the line through vectors pand qin Rnmay be written in
parametric form x= (1 t)p+tq.
b. The line segment from pto pis the set of points of the form (1t)p+tq
where 0 t1. Show that a linear transformation Tmaps this line segment
onto a line segment or onto a single point.
a. The line through vectors pand qin Rnis parallel to the vector qpand
through p. Thus it has parametric vector form (qp)t+p= (1 t)p+tq.
b. Apply the transformation to the line segment. We get
T((1 t)p+tq) = (1 t)T(p) + tT(q)
since Tis linear. Thus there are two cases. First if T(p) = T(q)then this
expression simplifies to T(p), which is a single point. Second if T(p)6=T(q),
then the expression describes a line segment parallel to the vector T(q)T(p)
and through T(p).
31. Let T:RnRnbe a linear transformation, and let {v1,v2,v3}be a
linearly dependent set in Rn. Explain why the set {T(v1), T (v2), T (v3)}is
linearly dependent.
Since {v1,v2,v3}is linearly dependent, the homogeneous equation
[v1,v2,v3]x=0
has non-trivial solutions. Let cbe one of these non-trivial solutions. Then
[v1,v2,v3]c=0,
so
c1v1+c2v2+c3v3=0.
1
pf2

Partial preview of the text

Download Linear Transformations: Planes, Lines, and Dependence and more Assignments Linear Algebra in PDF only on Docsity!

Newberger Math 247 Spring 03 Homework solutions: Section 1.8 #26, 27, 31, Section 2.3 #34,36,37,

Section 1.8 #26,27,31.

  1. Let u and v be linearly independent vectors in R^3 and let P be the plane through u, v and 0. The parametric equation of P is x = su + tv (with s and t in R). Show that a linear transformation T : R^3 → R^3 maps P onto a plane through 0 , a line through 0 or onto just the origin 0 in R^3. What must be true about T (u) and T (v) in order for the image of the plane P to be a plane?

We want to know the image of P under T. Any point x on P will have the form x = su + tv where s and t are real numbers. So T (x) = sT (u) + tT (v), and T (P ) is the collection of all vectors of the form sT (u)+tT (v) where s and t are any real number. Thus the image of P under T is Span{T (u), T (v)}. Since there are three cases for the geometric description of the span of two vectors in R^3 , there are three cases for the geometric description of the image of P under T. First {T (u), T (v)} could be independent, in which case, the image of P under T is a plane. Second, {T (u), T (v)} could be dependent but not both zero, in which case the image of P is a line. Third, both vectors T (u) and T (v) could be zero, in which case the image of P is simply 0.

  1. a. Show that the line through vectors p and q in Rn^ may be written in parametric form x = (1 − t)p + tq. b. The line segment from p to p is the set of points of the form (1 − t)p + tq where 0 ≤ t ≥ 1. Show that a linear transformation T maps this line segment onto a line segment or onto a single point.

a. The line through vectors p and q in Rn^ is parallel to the vector q − p and through p. Thus it has parametric vector form (q − p)t + p = (1 − t)p + tq. b. Apply the transformation to the line segment. We get T ((1 − t)p + tq) = (1 − t)T (p) + tT (q)

since T is linear. Thus there are two cases. First if T (p) = T (q) then this expression simplifies to T (p), which is a single point. Second if T (p) 6 = T (q), then the expression describes a line segment parallel to the vector T (q) − T (p) and through T (p).

  1. Let T : Rn^ → Rn^ be a linear transformation, and let {v 1 , v 2 , v 3 } be a linearly dependent set in Rn. Explain why the set {T (v 1 ), T (v 2 ), T (v 3 )} is linearly dependent.

Since {v 1 , v 2 , v 3 } is linearly dependent, the homogeneous equation [v 1 , v 2 , v 3 ]x = 0

has non-trivial solutions. Let c be one of these non-trivial solutions. Then

[v 1 , v 2 , v 3 ]c = 0 ,

so c 1 v 1 + c 2 v 2 + c 3 v 3 = 0. 1

2

Now we are interested in {T (v 1 ), T (v 2 ), T (v 3 )}. From the above we get that

T (c 1 v 1 + c 2 v 2 + c 3 v 3 ) = T ( 0 ).

But this means c 1 T (v 1 ) + c 2 T (v 2 ) + c 3 T (v 3 ) = 0 ,

so the homogeneous equation

[T (v 1 ), T (v 2 ), T (v 3 )]x = 0

has c as a solution as well, i.e. it has non trivial solutions. Thus the vectors {T (v 1 ), T (v 2 ), T (v 3 )} are linearly dependent.

Section 2.3 #36, 37, 38.

  1. Let T be a linear transformation that maps Rn^ onto Rn. Show that T −^1 exists and maps Rn^ onto Rn. Is T −^1 also one-to-one?

Since T is linear there is a matrix A such that T (x) = Ax. Since T is onto, A is invertible by Theorem 8. By Theorem 9, T −^1 exists and equals T −^1 (x) = A−^1 x. Since A−^1 is the standard matrix for T −^1 , and A−^1 is invertible, T −^1 is one-to-one by the invertible matrix theorem.

  1. Suppose T and U are linear transformations from Rn^ to Rn^ such that T (U (x)) = x for all x in Rn. Is it true that U (T (x)) = x for all x in Rn?

It is true. Since T and U are linear, there are matrices A and B such that T (x) = Ax and U (x) = Bx. Thus we have that T (U (x)) = ABx = x for all x in Rn. This means AB = I. Now the theorem at the top of p (blue box) states that A and B are inverses. This means BA = I as well. So U (T (x)) = BAx = Ix = x, so U (T (x)) = x.

  1. Suppose a linear transformation T : Rn^ → Rn^ has the property that T (u) = T (v) for some pair of distinct vectors u and v in Rn. Can T map Rn onto Rn?

T cannot be onto. Since T is linear there is a matrix A such that T (x) = Ax. Since T (u) = T (v), we have Au = Av, which implies Au − Av = 0 , or equivalently A(u−v) = 0. Since u and v are distinct, u 6 = v, and so u−v 6 = 0. Thus Ax = 0 has a non-trivial solution, namely x = u − v. This means that A is not invertible, and hence T is not onto, by the invertible matrix theorem.