Chapter 3

Question 1

image of a function

Answer 1

all of the values that a function takes in its target space.

if f is a function from X to Y, then
im(f) = {f(x): x in X}
= {b in Y: b = f(x), for some x in X

The image of a linear transformation T(x) = A(x) is the span of the column vectors of A. It’s denoted by im(A) or im(T).

It’s a subset of the target space of T rather than its domain.

Question 2

parametrization

Answer 2

a parametrization of a curve C in R^2 is a function g from R to R^2 whose image is C.

Question 3

span

Answer 3

for the vectors v1, v2 … vm in R^n, the set of all linear combinations c1v1 + … cmvm of those vectors is called their span.

Question 4

properties of an image

Answer 4

the image of a transformation T (from R^m to R^n) has the following properties:

a. the zero vector in R^n is in the image of T.
b. the image of T is closed under addition: if v1 and v2 are in the image of T, so is v1 + v2.
c. the image of t is closed under scalar multiplication: if v is in the image of T and k is an arbitrary scalar, then kv is in the image of T as well.

Therefore the image of T is closed under linear combinations in general.

Question 5

kernel

Answer 5

the kernel of a linear transformation T(x) = Ax from R^m to R^n consists of all zeros of the transformation; the solutions of the equation T(x) = Ax = 0.

It’s denoted by ker(T). It’s a subset of the domain R^m of T rather than its target space.

Question 6

kernel properties

Answer 6

for a transformation from R^m to R^n:

the zero vector in R^m is in the kernel of T.
The kernel is closed under addition.
The kernel is closed under scalar multiplication.

Question 7

when is ker(A) = {0}?

Answer 7

if and only if rank(A) = m
if ker(A) = {0}, then m t an iff)
for a square matrix A, we have ker(A) = {0} iff A is invertible.

Question 8

various characterizations of invertible matrices

Answer 8

1. A is invertible
2. the linear system Ax = b has a unique solution x for all b in R^n.
3. rref(A) = In
4. rank(A) = n
5. im(A) = R^n
6. ker(A) = {0}

Question 9

subspace of R^n

Answer 9

subset W of the vector space R^n with the following properties:
a. W contains the zero vector in R^n
b. W is closed under addition.
c. W is closed under scalar multiplcation.
(therefore also closed under linear combinations.

kernels and images are subspaces

Question 10

redundant vector; basis

Answer 10

consider vectors v1…vm in R^n

• vector vi is redundant if vi is a linear combination of the preceding vectors.

Question 11

linear independence and dependence;

Answer 11

consider vectors v1…vm in R^n

• vectors are called linearly independent if none of them are redundant; otherwise they’re linearly independent (one of them at least is redundant)

Question 12

basis

Answer 12

consider vectors v1…vm in R^n

• if they span V and are linearly independent, the vectors in a subspace V form a basis

Question 13

basis of the image

Answer 13

to construct a basis of the image of a matrix A, list all the column vectors of A, and omit the redundant vectors from this list

Question 14

linear independence and zero components

Answer 14

consider vectors v1…vm in R^n

if v1 is nonzero and if each of the other vectors vi has a nonzero entry in a component where all preceding vectors have a 0, then all of the vectors are linearly independent

Question 15

linear relation

Answer 15

consider vectors v1…vm in R^n

an equation of the form
c1v1 + … + cmvm = 0

is a linear relation among the vectors v1…vm. There is always the trivial relation where c1 = … = cm = 0. Nontrivial relations may or may not exist.

Question 16

relations and linear dependence

Answer 16

the vectors v1 … vm in R^n are linearly independent if (and only if) there are nontrivial relations among them

Question 17

kernel and relations

Answer 17

the vectors in the kernel of an n x m matrix correspond to the linear relations among the column vectors v1 … vm of A:

Ax = 0 means that x1v1 + … + xmvm = 0

So the column vectors of A are linearly independent iff ker(A) = 0 or if rank(A) = m. This all implies that m <= n.

Thus we can find at most n linearly independent vectors in R^n.

Question 18

various characterizations of linear independence

Answer 18

for a list v1…vm of vectors in R^n, the following statements are equivalent.

i. vectors v1…vm are linearly independent
ii. none of the vectors v1…vm is redundant, meaning that none of them is a linear combination of preceding vectors.
iii. none of the vectors is a linear combination of the other vectors in the list
iv. there is only the trivial relation among the vectors v1…vm, meaning that the equation c1v1…+cmvm = 0 has only the solution c1 = … = cm = 0.

v. ker [v1 .. v1] = {0}
vi. rank[v1 .. vm] = m

Question 19

basis and unique representation

Answer 19

consider the vectors v1…vm in a subspace V of R^n.
The vectors v1…vm form a basis of V iff every vector v in V can be expressed uniquely as a linear combination

v = c1v1 + … + cmvm