# introduction To Linear Transformations (sec. 1.8 Linear Algebra book by Lay)

• June 1st 2007, 09:19 AM
googoogaga
introduction To Linear Transformations (sec. 1.8 Linear Algebra book by Lay)
I have three homework problems that I got stuck on while, trying to solve. I'd really appreciate it if anyone could help me with these. Here they are

For # 14 & 16 use a rectangular coordinate system to plot u= [vertical: 5 2]
, v=[vert: -2 4], and their images under the given transformation T. (Make a large sketch for each exercise.) Describe geometrically what T does to each vector x in R^2.
#14 T(x)= [.5 0 ] [x1] horiz.
[0 -.5] [x2]horiz.
# 16 T(x)= [0 1] [x1] horiz.
[1 0] [x2] horiz.

(T of each exercices is just one matrix of two columns and two rows)

#34 Let T: R^n----> R^m be a linear transformation. Show that if T maps two linearly independent vectors onto a linearly dependent set, then the equation T(x)=0 has a notrivial solution.
• June 1st 2007, 11:11 AM
ThePerfectHacker
Quote:

Originally Posted by googoogaga
#34 Let T: R^n----> R^m be a linear transformation. Show that if T maps two linearly independent vectors onto a linearly dependent set, then the equation T(x)=0 has a notrivial solution.

It is necessary and sufficient to show that $\ker T \not = \{ \bold{0} \}$.

Let $\bold{u},\bold{v} \in \mathbb{R}^n$ be these two linearly independent vectors.

We are given that $T(\bold{u}),T(\bold{v})$ are linearly dependent.

Thus, there exists $k_1,k_2 \in \mathbb{R}$ so that $k_1T(\bold{u})+k_2T(\bold{v}) = \bold{0}$ and $k_1,k_2$ are not both zero.

Hence, $T(k_1\bold{u}+k_2\bold{v}) = \bold{0}$ since $T$ is a linear transformation.

But, $k_1\bold{u}+k_2\bold{v} \not = \bold{0}$ because $k_1,k_2$ are not both zero and $\bold{u},\bold{v}$ are linearly independent.

Hence, there exists a non-zero element vector $\bold{w} = k_1\bold{u}+k_2\bold{v}$ such that $T(\bold{w}) = \bold{0}$.

Thus, $\ker T\not = \{\bold{0}\}$.

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Note: This can be generalized to $n\geq 2$ linearly independent vectors getting mapped into a linearly dependent set.
• June 2nd 2007, 11:02 AM
AfterShock
Quote:

Originally Posted by googoogaga
I have three homework problems that I got stuck on while, trying to solve. I'd really appreciate it if anyone could help me with these. Here they are

For # 14 & 16 use a rectangular coordinate system to plot u= [vertical: 5 2]
, v=[vert: -2 4], and their images under the given transformation T. (Make a large sketch for each exercise.) Describe geometrically what T does to each vector x in R^2.
#14 T(x)= [.5 0 ] [x1] horiz.
[0 -.5] [x2]horiz.
# 16 T(x)= [0 1] [x1] horiz.
[1 0] [x2] horiz.

(T of each exercices is just one matrix of two columns and two rows)

14.)

I'll describe this graphically.

You have the vector v at (-1, 4); then T(v) is at (-1, 2) (both in the direction of the origin.

You have the vector u at (5, 2); then T(u) is at (2.5, 1) (both in the direction of the origin.

Thus, there's a CONTRACTION of a factor of 1/2.

16.) It's easy to see that there's a reflection through the line x_2 = x_1 (or y = x).

Vector v is at (-2, 4) and is in the direction of T(v) at (4, -2).

Vector u is at (5, 2) and is in the direction of T(u) at (2, 5).

Draw a dotted line x_2 = x_1 (or y = x) and you can see the reflection.
• June 2nd 2007, 11:08 AM
AfterShock
Quote:

Originally Posted by googoogaga
#34 Let T: R^n----> R^m be a linear transformation. Show that if T maps two linearly independent vectors onto a linearly dependent set, then the equation T(x)=0 has a notrivial solution.

Suppose that {u,v} are linearly independent in R^n and we have T(u) and T(v) are linearly dependent. There exists weights c_1, c_2 (both of them not zero), such that:

c_1*T(u) + c_2*T(v) = 0

Since T is linear, T(c_1*u + c_2*v) = 0. Thus,

x= c_1*u+ c_2*v will satisfy T(x) = 0. And, we know that it's impossible for x to be the zero vector since that'd imply that a nontrivial linear combination of u and v is zero, and we know that is impossible since u and v are linearly independent.

Therefore, we can conlude that the equation T(x) = 0 has a nontrivial solution.