Test 2 Review Answers
Note this test covers the material from the first test as well as the new material (because you need
to use the techniques and ideas from the previous material throughout).
I will include questions like adding matrices, adding vectors, and finding dot products as well as
inverting matrices and finding rref of a matrix.
Terms to know (some appeared earlier):
Chapter 2 Linear transformations
Inverses of Linear transformations
(and the relationship with inverse matrices)
Chapter 3
Image and Kernel of a linear transformation
Span
Span of column vectors
Subspace
Basis
redundant vector
linear independence
relation
dimension
Key theorems
For any function T, Rm -> Rn, T is a linear transformation iff:
1. For any v,w  Rm ,T(v  w)  T(v )  T(w) and
2. For any scalar k in R and any v  R m ,T(kv )  kT(v )
If A is the matrix of a linear transformation T, then the columns of A are the images of the

standard
unit vectors. That is, the ith column is T(ei) In addition, the columns of A span the
 then the columns of A that row reduce to include pivot elements in B
image of A. If B = rref(A),
are a basis for the image of A. That is they are a minimum spanning set of vectors for the image.
If A is the matrix of a linear transformation from Rm to Rn, then
m = dimension(Im(A)) + dimension(ker(A)).
These statements about a set of vectors V = {v1,v2,...vm } are logically equivalent:
(So you should be able to show any of these implies any other of these.)
1. The vectors of V are linearly independent.
2. There is no non-trivial relation involving the vectors of V.

3. None of the vectors in V are redundant.
4. The vectors of V are a basis of the span of V.
5. If A is a matrix whose column vectors are the vectors of V, ker(A) = {0}

Some practice problems:
1 1 1


1. If A = 1 2 3Find a basis for the Im(A) and a basis for ker(A).

4 5 6

1 0 1


>> The rref(A) = 0 1 2  So the first 2 columns of A are independent and span Im(A) so


0 0 0 

form a basis for it.

If the columns of A are v1,v 2,v 3 respectively, the third column of rref(A) tell us:

v3  1v1  2v2 so 0  1v1  2v 2  v 3 So make a column vector from these coefficients that is in
1
 
the kernel: 2
 Since the kernel is 1 dimensional this is a basis for the kernel.
 
1

0

0
2.Same as above for A = 
0

0
0

0
>>The rref(A) =
0

0

1
0
0
0
2
0
0
0
0 3 0

1 4 0
0 0 1

0 0 1
0 3 0

1 4 0
so if v1,v2,...v6 are the column vectors of A, then
0 0 1

0 0 0
x1 

 
x 2 
x 3 

v2,v4 ,v6 are a basis of the Im(A). For ker(A), if v    ker(A) then A  v  0 so letting
x 4 
x 5 
 
x 6 

1
0
0
0
2
0
0
0

x1  1r  0s  0t
x 2  0r  2s  3t
x  0r  1s  0t
so a basis for
x1  r, x3  s,x5  t (the free variables) multiplying gives us 3
x 4  0r  0s  4t
x 5  0r  0s  1t
x 6  0r  0s  0t
10 0 
   
023
01 0 

ker(A) is  , , 
00 4
00 1 
   
00 0 

3. For any matrix A, what is the relationship between the rank(A) and the Im(A) and ker(A)?
>>rank(A) = dim(Im(A)) and if A is nxm, then m =rank(A) + dim(ker(A)).

4. Show that if vectors v1,v2,v3 are linearly independent,
then so are the vectors v1,v1  v2,v1  v2  v3
>>Proof by contradiction:
Assume vectors v1,v 2,v 3 are linearly independent, and the vectors v1,v1  v2,v1  v2  v3 are not

independent. Then by theorem, there is a non-trivial relation among v1,v1  v2,v1  v2  v3. This

mean there exist scalars c1,c 2,c 3 that are not all zero such that
c1v1  c 2 (v1  v 2 )  c 3 (v1  v 2  v 3 )  0 .


Multiplying this out and combining similar terms gives us:

(c1  c 2  c 3 )v1  (c 2  c 3)v 2  c 3v 3  0 This is a relation involving v1,v 2,v 3 which are

independent, so it must be trivial (the coefficients must all be 0. So starting with the last, c3  0.
But then the second coefficient being 0 tells us c2  0  0 , so c 2  0 .
But substituting both these into the formula for the first coefficient, we get: c1  0  0  0 so

c1  0 Oh my, this contradicts the statement above: c1,c2,c3 are not all zero. Done.



1

 
5. Find a matrix A such that im(A) = the plane with normal (perpendicular) vector 3


2

>>Lets try to find 2 independent vectors that are both perpendicular to the above. Then they will
be a basis for the plane under discussion. Guess wisely:



3 
 
1

0 
,


0 
 
2 

3


are two vectors that are independent. They are, (using dot products) perpendicular to the
above. So they are a basis for the plane withb this normal vector. Then just set up A as the
3 0 


A  1 2 

0 3

matrix with these as its column vectors:
6. So prove any of the 5 statements at the bottom of page 1 implies any other of these statements.

In class we proved 1 iff 2. 3 is the definiton of 1 so 1 iff 3 is true.
To prove 1 implies 4, note that if the vectors of V are independent, then they are a minimal set of
vectors spanning the span of V. So they are a basis.
Show 4 implies 5 by contradiction. Assume 4 (meaning the vectors of V are a basis of their
span). Also assume ker(A) is not just the zero vector (so it has dimension >0). Then since m =
dim(Im(A)) + dim(ker(A)), we conclude dim(Im(A)) < m. So the matrix A row reduces to a
matrix with at least one free variable. That column vector is redundant in the original A. So the
vectors do not form a basis (they are not minimal). Contradiction.
Finally, prove 5 implies 3 (which is the definition of 1), by contradiction. Assume ker(A) = just
the zero vector. Also assume one of the vectors of V is redundant. But then the column of the
redundant vector will not have a pivot element when A is row reduced. So dim(Im(A)) < m. So
m-dim(Im(A)) =dim(Ker(A) > 0 Contradiction.
7. If A is a square matrix and ker(A) is the 0 vector, what can you say about ker(A2) and about
im(A2)? For a vector v to be in the ker(A2), A(Av) = 0, so Av is in the ker(A) which is only the 0
vector. So Av = 0. But then, again since ker(A) = 0, we conclude v = 0. So ker(A2) = 0. If A
maps Rn to Rn then since Im(A) has dimension n, A must be onto. Since ker(A2) = 0, A2 must
also have dim(Im(A2)) = n so it is also onto and Im(A2)=Rn
8. In R4, find a basis for the solution set of x+y+z+w = 0. What is the dimension of this space?
We know dim(solution set) < 4 because not every vector of R4 is in it. (Try x = y=z=w=1). But
with just a little educated guessing, you can find 3 independent vectors in the solution space (so
the dim is at least 3). Try: (writing the 3 as column vectors of a matrix):
1 0 0 


1 1 0  So we conclude we have a basis and the dimension is 3.
0 1 1 


0 0 1

9. Find a matrix A where Im(A) is the solution set of #8 and a matrix B whose kernel is the
solution set of #8.
So for the first part, use the matrix above. Its image is the solution set. For B use [1,1,1,1]