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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.
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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.
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.
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).
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.
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.
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.
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.